Quantitative Methods for Business and Economics (Jakub Kielbasa)
187
1 Quantitative Methods for Business and Economics Jakub Kielbasa This book is written for students for free, so feel free to use, cite, steal example etc. It is a learning tool, but please cite it when you do use it. If you come across any mistakes, please contact [email protected] . The wording of all theory, as well as all the content of all examples and exercises were the sole work of the author. ISBN: 5555000008952 A huge thanks to my editor Alex Tharby, for helping me edit and publish this textbook.
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Quantitative Methods for Business and Economics (Jakub Kielbasa)
Transcript of Quantitative Methods for Business and Economics (Jakub Kielbasa)
1
Quantitative Methods
for
Business and Economics
Jakub Kielbasa
This book is written for students for free, so feel free to use, cite, steal example etc. It is a learning tool, but please cite it when you do
use it. If you come across any mistakes, please contact [email protected].
The wording of all theory, as well as all the content of all examples and exercises were the sole work of the author.
ISBN: 5555000008952
𝑑𝑙𝑎 𝑤𝑎𝑠 A huge thanks to my editor Alex Tharby, for helping me edit and publish this textbook.
2
Contents Chapter 1: Vital Mathematical Knowledge p. 4
1.1 Revision of Negative Numbers p. 6
1.2 Multiplying and Dividing Fractions p. 7
1.3 Adding and Subtracting Fractions p. 8
1.4 Notes on Fractions p. 12
1.5 Defining a Variable p. 13
1.6 Indices p. 14
1.7 BIMDAS p. 18
1.8 Equations p. 21
1.9 Factorisation p. 25
1.10 Inequalities and Absolute Values p. 26
1.11 The Number Zero p. 28
Chapter One Summary p. 28
Chapter One Questions p. 29
Chapter 2: Linear Algebra p. 30 2.1 Linear Equations p. 31
2.2 Main Features of Linear Equations p. 32
2.3 Negative Gradients p. 33
2.4 Graphing Lines from Equations p. 34
2.5 Obtaining the Equation of a Line p. 36
2.6 Intersecting Lines p. 38
2.7 Microeconomic Applications p. 40
2.8 Elasticity p. 42
2.9 Interpreting Elasticity p. 44
Chapter Two Summary p. 46
Chapter Two Questions p. 47
Chapter 3: Simultaneous Equations and Matrices p.49 3.1 Simultaneous Equations p. 50
3.2 Two Simultaneous Equations p. 50
3.3 Three Simultaneous Equations p. 52
3.4 The Matrix p. 54
3.5 Solving Two Equation Matrices p. 55
3.6 Solving Three Equation Matrices p. 59
3.7 Notes on Solutions to Matrices p. 61
3.8 Applications p. 62
3.9 The Determinant of a 2 × 2 Matrix p. 65
3.10 The Determinant of a 3 × 3 Matrix p. 65
3.11 Using the Jacobian Determinant p. 68
Chapter Three Summary p. 70
Chapter Three Questions p. 71
Chapter 4: Non-Linear Functions p. 73 4.1 Defining Non-Linear Functions p. 74
4.2 Defining a Quadratic Function p. 75
4.3 Quadratic Graphs p. 76
4.4 Sketching a Quadratic Function p. 77
4.5 The Cubic Function p. 80
4.6 The Exponential Function p. 82
4.7 The Logarithmic Function p. 84
4.8 Logarithmic Graphs p. 88
3
4.9 The Natural Number 𝑒 p. 89
4.10 The Hyperbolic Function p. 90
4.11 Economic Applications p. 91
Chapter Four Summary p. 93
Chapter Four Questions p. 94
Chapter 5: Single Variable Differentiation p. 96 5.1 What is Differentiation? p. 97
5.2 Differentiation by First Principles p. 99
5.3 Differentiation Rules: Power Rule p. 103
5.4 Differentiation Rules: Chain Rule p. 104
5.5 Differentiation Rules: Product Rule p. 105
5.6 Differentiation Rules: Quotient Rule p. 108
5.7 Differentiation Rules: 𝑒 Rule p. 109
5.8 Differentiation Rules: ln Rule p. 110
5.9 The Second Derivative p. 111
5.10 The Gradient Function p. 112
5.11 Simple Applications p. 113
Chapter Five Summary p. 114
Chapter Five Questions p. 115
Chapter 6: Applications of Differentiation p. 117 6.1 Graphical Optimisation p. 118
6.2 Mathematical Optimisation p. 118
6.3 The Nature of an Optimal Point p. 120
6.4 Inflection Points p. 123
6.5 Combining all Theory p. 124
6.6 Applications – Profit p. 126
6.7 Applications – Break-Even p. 129
6.8 Applications: Marginal and Average Values p. 130
6.9 Differentiation and Elasticity p. 133
6.10 Elasticity and Total Revenue p. 135
Chapter Six Summary p. 137
Chapter Six Questions p. 138
Chapter 7: Multiple Variable Differentiation p. 140 7.1 Additional Variables p. 141
7.2 Simple Partial Differentiation p. 142
7.3 Complex Partial Differentiation p. 144
7.4 Second Order Partial Derivatives p. 146
7.5 Application of Partial Differentiation p. 147
7.6 Total Differentiation p. 148
7.7 Optimisation with Many Variables p. 151
7.8 Economic Applications p. 155
Chapter Seven Summary p. 158
Chapter Seven Questions p. 158
4
Chapter 8: Financial Mathematics p. 160 8.1 Index Numbers and Averages p. 161
8.2 Series and Sums p. 163
8.3 Simple Interest p. 164
8.4 Compound Interest p. 168
8.5 Annual Interest Rates p. 170
8.6 Net Present Value p. 172
8.7 Internal Rate of Return p. 174
Chapter Eight Summary p. 177
Chapter Eight Questions p. 177
Solutions p. 180
5
Chapter 1
Vital Mathematical Knowledge
Things you must know 1.1 Revision of Negative Numbers 6
1.2 Multiplying and Dividing Fractions 7
1.3 Adding and Subtracting Fractions 8
1.4 Notes on Fractions 12
1.5 Defining a Variable 13
1.6 Indices 14
1.7 BIMDAS 18
1.8 Equations 21
1.9 Factorisation 25
1.10 Inequalities and Absolute Values 26
1.11 The Number Zero 27
Chapter One Summary 28
Chapter One Questions 29
6
1.1 revision of negative numbers
Adding and subtracting negative numbers is very
similar to adding and subtracting positive
numbers.
Working on the number line (a line with all
negative and positive numbers) will help you revise
the methods of adding and subtracting negative
numbers.
For the problem 4 − 3, find the number 4 on the
number-line, then go back three places to get 1.
Similarly with negative numbers, for the problem
−1 − 2, find the number −1 on the number-line,
then go back two places to get −3.
Example 1: solve
−4 + 5 − 2 − 4 + 6
Solution: find the number −4 on the number-line,
then add 5, take 2, take 4, then add 6. The answer
is 1.
After you get the hang of it, you will never need to
use the number-line again.
Example 2: solve:
−5 − 6 − 7 − 2 + 15
Solution: do it without the number-line
= −5
Theory:
Adding a negative number is the same as
subtracting the positive value of that same
number. E.g. 5 + −3 = 5 − 3
Subtracting a negative number is the same as
adding the positive value of that number. E.g.
4 − −6 = 4 + 6
The two rules above are very important.
Example 3: solve
3 − −4 + 2
Solution: the two negatives (one after the other)
can be changed into a positive sign:
= 3 + 4 + 2
Then solve:
= 3 + 4 + 2 = 9
Example 4: solve
−4 + −3 − (−5)
Solution: this is a bit more complex. Do one
section at a time; the −4 remains as it is, because
it cannot be changed using the rules, but the
+(−3) can be changed to −3. Then the −(−5)
can be changed to +5 to get:
= −4 − 3 + 5 = −2
Example 5: solve
− −5 − 6 + −8 − −9
Solution: simplify each part separately. −(−5)
becomes +5, −6 remains as it is, +(−8) becomes
−8, and − −9 becomes +9.
= +5 − 6 − 8 + 9 = 0
A similar theory applies to multiplying negative
numbers.
Theory:
The multiplication of two negative numbers
equals a positive number.
The multiplication of a positive and a negative
number equals a negative number.
This is best explained with examples.
Example 6: solve
−6 × (−5)
Solution: multiplying two negative numbers equals
a positive number. Since
6 × 5 = 30
then −6 × −5 = 30
−4 −3 −2 −1 0 1 2 3 4
7
Example 7: solve
(−6) × 4
Solution: a negative number multiplied by a
positive number equals a negative number. Since
6 × 4 = 24, then −6 × 4 = −24
Theory: mathematicians use the dot “∙” to mean
multiplication, e.g. 6 ∙ 4 is the same as 6 × 4
Similarly, if two terms are in brackets without
anything between the brackets, this is taken as a
multiplication. E.g. 6 4 = 6 × 4
Example 8: solve
−5 (−7)
Solution: two negative numbers are multiplied
together which must equal a positive number.
5 ∙ 7 = 35 so −5 −7 = 35.
Dividing with negative numbers also follows the
rule of two negatives equals a positive.
Theory:
The division of two negative numbers equals a
positive number.
The division of one negative and one positive
number equals a negative number.
Example 9: solve
−15
−3
Solution: dividing two negative numbers equals a
positive number.
15
3= 5 then
−15
−3= 5
Example 10: solve
−24
3
Solution: division with both a negative number
and a positive number gives a negative number.
24
3= 8 so
−24
3= −8
Note: there is a reason brackets were used so
often in this section. Later in this chapter, you will
see the consequence of not using brackets
appropriately.
Exercises: 1. Simplify the following:
𝑎) −2 + 3 − 4
𝑏) −2 + −4 − −1
𝑐) −4 − −4 − 4
𝑑) 5 + 1 − 4 + (−3)
𝑒) −6 − 7 − 3 − −21 + 1
2. Simplify the following: 𝑎) 3 ∙ −2
𝑏) −6 −6
𝑐) −36 ÷ −4
𝑑) 81/ −9
𝑒) −20 −3
1.2 multiplying and dividing fractions
Most students hate fractions, but they are part of
the real world. For example, 50𝑐 is half a dollar,
$100,000 is one tenth of a million dollars,
$4million is 4
1000 of a billion dollars. Most things
can have fractions, so you must be able to
manipulate them.
Theory: to multiply two fractions, multiply the two
tops (numerators) to give a new top, then multiply
the two bottoms (denominators) to give a new
bottom.
Example 1: solve
2
3∙
4
5
Solution: multiply the two tops: 2 ∙ 4 = 8, then
multiply the two bottoms: 3 ∙ 5 = 15, and these
form the new numerator and the new
denominator:
=2 ∙ 4
3 ∙ 5=
8
15
8
Example 2: solve
6
5∙
7
3
Solution:
=6 ∙ 7
5 ∙ 3=
42
15
Dividing fractions is very similar except it has one
difference.
Theory: when dividing fractions, invert the second
fraction then proceed just like multiplication.
Inverting a fraction is flipping the fraction; making
the bottom the new top, and making the top the
new bottom.
Inverting 𝑎
𝑏 makes it
𝑏
𝑎.
Example 3: solve
2
5÷
3
4
Solution: invert the second fraction, and change
the division sign to a multiplication sign:
=2
5∙
4
3
Proceed with the multiplication:
=2 ∙ 4
5 ∙ 3=
8
15
Example 4: solve
2
3/
1
16
Solution: invert the second fraction, then proceed
as for multiplication.
2
3∙
16
1=
32
3= 10
2
3
Exercises: 1. Simplify the following fractions:
𝑎) 3
4
2
3
𝑏) 5
7
12
5
𝑐) 3
−8
3
1
𝑑) 5
4÷
2
3
𝑒) 7
2/ −
3
10
2. Simplify the fractions into a single fraction
𝑎) 14
−3
−5
12
𝑏) 14
−3 /
5
−12
𝑐) −3
3 /
23
17
𝑑) −5
9/
−3
−7
𝑒) 17
23 −1
12
1.3 adding and subtracting fractions
The hardest aspect of fractions is adding and
subtracting fractions. Multiplying and dividing
fractions is easier.
The concept used for adding/subtracting fractions
is the Common Denominator approach. The
denominator is the bottom of a fraction.
Before the theory is given, an intuition of the
concept is needed. Apologies for the patronising
subject, but it works well.
You purchased a pie, and you take it home. You
cut the pie into quarters, and since you are hungry,
you eat half the pie. After two hours, you are
hungry again and take a quarter of the original pie,
so that only one quarter is left. The question is
how much of the pie have you eaten?
Write out the fractions:
1
2+
1
4
That is, first you ate half the pie, then you ate a
quarter. You might be tempted to simply add the
top numbers and the bottom numbers. If we do
that, we get:
1 + 1
2 + 4=
2
6=
1
3
Have you really eaten only one third of the pie?
No, you ate three quarters. The theory of common
denominator helps find the true answer.
9
Intro example 1: from the pie example, solve
1
2+
1
4
The common denominator process is finding the
same denominator for all fractions. In this
example, the 1/2 must be changed to a fraction
with a 4 on the bottom, however if the 2 is erased
and replaced with a 4, the value will no longer be
one half. What is wanted is to change the form of
the fraction 1
2 so the bottom number is 4, while
keeping the value of the fraction unchanged.
Another way of writing 1
2 is
2
4 or
3
6 or
4
8 but since 4
needs to be on the bottom, 2
4 is used. The problem
becomes:
2
4+
1
4
The first part still has the value of a half, and the
second part is still one quarter. The form of the
first fraction has changed but the value is the
same.
Once all fractions have the same denominator, all
the top numbers (numerators) can be written over
the Common Denominator:
2
4+
1
4=
2 + 1
4
Then simplify the top:
2 + 1
4=
3
4
Exactly as is intuitive.
Theory: the Common Denominator approach
changes the form of all fractions so that the
denominators are all the same, but leaves the
values of all fractions unchanged.
Sometimes it is not so obvious how the fractions
need to be changed. For example, solving:
1
3+
1
7
What is the common denominator here? How can
1
3 be changed into something with 7 as a
denominator? Or alternatively, how can 1
7 be
changed to have 3 as a denominator? There is a
simple solution.
Theory: a Common Denominator is found by
multiplying the denominators (bottoms) of all the
fractions to be added or subtracted.
Intro example 2: solve
1
3+
1
7
A Common Denominator is the multiplication of
the two denominators 3 and 7:
3 ∙ 7 = 21
The form of 1
3 needs to be changed to have 21 on
the bottom but the value must be left unchanged.
Multiply the 1
3 by
7
7:
1
3∙
7
7
The number 7
7 has value 1, so the value of
1
3 is
unchanged however the form is changed:
1
3∙
7
7 =
7
21
Similarly, 1
7 needs to be multiplied by some number
to get 21 on the bottom, while leaving the value of
the fraction unchanged. That number is 3
3:
1
7∙
3
3 =
3
21
Rewrite the original problem with the equivalent
fractions:
7
21+
3
21=
7 + 3
21=
10
21
Theory: to add/subtract fractions, change the form
of each fraction without changing its value, so that
all the denominators are the same. Then
add/subtract the numerators (top numbers),
having this over the common denominator.
10
Example 1: solve
1
9+
1
5
Solution: a Common Denominator is 9 ∙ 5 = 45.
The first fraction 1
9 needs to be multiplied by
5
5 to
have 45 on the bottom, but leaving the value of
the fraction unchanged. The 5 in 5
5 comes from the
denominator of the other fraction.
1
9∙
5
5 =
5
45
Similarly for 1
5, to change the form of this fraction
to get 45 on the bottom, it needs to be multiplied
by 9
9 where the 9 comes from the denominator of
the other fraction:
1
5∙
9
9 =
9
45
Rewrite the original problem and solving:
5
45+
9
45=
5 + 9
45=
14
45
Notice that the 1
5 was multiplied by
9
9 where the 9
came from the denominator of the other fraction.
Similarly for 1
9 which was multiplied by
5
5 where the
5 came from the denominator of the other
fraction.
Example 2: solve
3
11+
4
5
Solution: a common denominator is 11 ∙ 5 = 55.
Then, change the 3
11 to have 55 as a denominator
without changing its value, by multiplying it by 5
5
(where the 5 comes from the denominator of the
other fraction):
3
11∙
5
5 =
15
55
Similarly for the other fraction, to change 4
5 to have
a 55 on the bottom, it needs to be multiplied by 11
11
(where the 11 comes from the denominator of the
other fraction):
4
5∙
11
11 =
44
55
Rewrite the original problem:
15
55+
44
55=
15 + 44
55=
59
55
Subtracting fractions uses the same process,
except it has a negative sign in between.
Example 3: solve
7
11−
1
4
Solution: a Common Denominator is 11 ∙ 4 = 44.
To change the 11 to a 44, multiply by 4
4 (as this
does not change the value of the fraction):
7
11∙
4
4 =
28
44
Then change the form (but not the value) of the
second fraction:
1
4∙
11
11 =
11
44
Rewrite the original problem:
28
44−
11
44=
28 − 11
44=
17
44
So far, we have only been using two fractions at a
time, but a very similar theory applies for adding
and subtracting more than two fractions.
Theory: when adding or subtracting many
fractions, to change the form of each fraction
without changing its value, it is multiplied by 𝑥/𝑥
where 𝑥 is the multiplication of all the other
denominators.
Example 4: solve
3
11+
3
4−
1
5
Solution: a Common Denominator is the
multiplication of all the denominators, which is
11 ∙ 4 ∙ 5 = 220. The fraction 3
11 needs to be
11
multiplied by some number to get 220 as a
denominator (bottom). This number is 20
20 which is
obtained by the multiplication of all the other
denominators (i.e. 4 ∙ 5 = 20):
3
11∙
20
20 =
60
220
Similar theory applies for the other two fractions.
The 3
4 needs to be multiplied by
55
55. The number 55
is obtained by multiplying the denominators of all
the other fractions (i.e. 5 ∙ 11 = 55):
3
4∙
55
55 =
165
220
Lastly:
1
5∙
44
44 =
44
220
Rewrite the original fraction:
60
220+
165
220−
44
220=
60 + 165 − 44
220
=181
220
Go over this last example to make sure you know
how the form of each fraction was changed
without changing its value.
Example 5: Solve
1
3−
7
13−
1
10
Solution: a Common Denominator is 3 ∙ 13 ∙ 10 =
390. To change the form of the first fraction
without changing its value, it must be multiplied by
130
130, where the 130 comes from the multiplication
of the other denominators (i.e. 13 ∙ 10 = 130):
1
3∙
130
130 =
130
390
The second fraction must be multiplied by 30
30
where the 30 is from the multiplication of all the
other denominators (i.e. 3 ∙ 10 = 30):
7
13∙
30
30 =
210
390
The third fraction must be multiplied by 39
39 which
comes from 3 ∙ 13 = 39:
1
10∙
39
39 =
39
390
Do not use the negative signs yet, as so far, we are
just changing the form of the fraction. The
problem is rewritten with all the relevant signs
added in after the form of each fraction has been
changed to have a Common Denominator.
130
390−
210
390−
39
390=
130 − 210 − 39
390
Simplify the numerator (top):
= −119
390
Exercises: 1. Simplify the following fractions:
𝑎) 1
4+
1
3
𝑏) 2
5−
1
8
𝑐) 7
3+
12
2+
1
6
𝑑) 21
13−
1
11
𝑒) 9
13+
15
39−
2
1
2. Simplify the fractions into a single fraction:
𝑎) 1
2+
1
3+
1
4+
1
5
𝑏) 1
2−
1
3+
1
4−
1
5
𝑐) 21
11+
3
17−
2
23
𝑑) 99
12+
3
2−
12
9+
7
8
𝑒) 10
3−
15
4+
17
9+
12
7
3. If a shareholder owns 3/11 of a company, and his brother owns 9/83 of that same company, what proportion of the company do the brothers own combined?
4. Jane, the CEO of a company, owns 9/11 of the company. She is ready to retire, so to have cash, she sells 5/12 of the shares she owns. What proportion of the company does she still own after the sale. What if Jane had sold 5/12 of the total company stock from her portfolio; how much would she have left then?
12
1.4 notes on fractions
Fractions can be written in many different ways, so
you must be familiar with all of them.
Theory: for a negative fraction, the negative sign
can be in three different places without changing
the value of the fraction:
−𝑎
𝑏=
−𝑎
𝑏=
𝑎
−𝑏
For example:
−2
5=
−2
5=
2
−5
All three are the same. Applying this to a larger
context:
2
3−
5
3=
2
3+
−5
3=
2
3+
5
−3
All three coloured parts are equal. It may be
convenient to write it one of these three ways
when you are doing a question with fractions.
Now, a more common kind of fraction is one with
a number out front, such as:
21
3
This means “two wholes plus one third”. This can
also be rewritten as:
2 +1
3
Write the 2 as 2
1:
2
1+
1
3
Then find a Common Denominator:
6
3+
1
3=
6 + 1
3=
7
3
This is the long way which shows the theory. The
short-cut is shown below.
Theory: to simplify a fraction involving whole
numbers, multiply the number out front with the
denominator (bottom), then add that result to the
top.
Example 1: change the following to a single
fraction
41
3
Solution: multiply the “whole” number with the
denominator (4 × 3 = 12), then adding this to the
numerator 12 + 1 = 13 giving:
41
3=
12 + 1
3=
13
3
Example 2: find the single fraction equivalent of
46
11
Solution: multiply the “whole” number out front
with the denominator (4 × 11 = 44), then add the
result to the numerator (top): (44 + 6 = 50)
46
11=
50
11
The reverse theory is the same, but the process is
slightly different.
Theory: to find how many “whole” numbers there
are in a fraction, work out how many
denominators will fit into the numerator without
exceeding the numerator. Then remove that
number from the numerator, and any remaining
numbers are written over the original
denominator.
Example 3: find how many “wholes” and a
remaining fraction results from
43
6
Solution: 6 goes into 43 seven times (to give 42)
as any more would exceed 43, so the number out
front in 7, and the remainder is 1 (as 43 − 42 =
1), so this is written as a fraction:
43
6= 7
1
6
13
Example 4: change the following fraction into
“whole” numbers and a remaining fraction:
97
10
Solution: 10 can go into 97 nine times as if we had
ten times, then it would exceed the numerator
(97). This gives 7 as a remainder, written over the
original denominator:
97
10= 9
7
10
Exercises: 1. Give the following fractions as a single fraction:
𝑎) 21
2 𝑑) 3
1
3
𝑏) 122
3 𝑒) 9
3
9
𝑐) 97
8 𝑓) 11
13
27
2. Change the following into whole numbers and a remaining fraction:
𝑎) 12
3 𝑑)
27
13
𝑏) 15
4 𝑒)
91
12
𝑐) 21
4 𝑓)
103
13
3. Simplify the following into a single fraction:
𝑎) −12
5
3
7 −
1
3
𝑏) 13
2
1
12 2
2
3 1
3
7
𝑐) 12
3 +
−2
7−
−3
−10
1.5 defining a variable
Theory: A variable is something that can change.
Temperature is a variable, because it can change.
The Woolworths share price can change, therefore
it is a variable. Variables are found everywhere in
life.
Different letters are used to distinguish different
variables. In business, a variable is often something
like quantity sold (usually 𝑄). Companies do not
know how many of a certain product they are
going to sell (say computers), so variables are used
that allow those companies to forecast how much
profit will be made if 𝑄 units are sold – where 𝑄 is
a variable number of computers.
However, it is not always that simple. Profit is also
a variable but the amount of profit depends on
how many computers are sold. So profit is called a
‘dependent variable’ because it is a variable (i.e. it
can change), but the extent to which it changes
‘depends’ on another variable (i.e. the number of
computers sold). Also, the number of computers
the company sells is called an ‘independent
variable’ because it is not determined by other
variables.
For now, we need to learn the basics of
manipulating variables and constants.
Theory: “collecting like terms” is the process of
bring together things that are the same. Things
that are different cannot be grouped together.
Intro example 1: There are two boxes of fruit:
Box 1: 7 bananas, 5 apples and 8 peaches.
Box 2: 3 bananas, 6 apples and 12 peaches.
To find out the total number of each fruit, the
bananas can be added to each other, but they
cannot be added to the apples or the peaches (this
is logical). To simplify the variables, the two boxes
can be rewritten as:
Box 1: 7𝑏 + 5𝑎 + 8𝑝
Box 2: 3𝑏 + 6𝑎 + 12𝑝
Where 𝑏 =bananas, 𝑎 = apples and 𝑝 = peaches.
Add Box 1 to Box 2:
𝑇𝑜𝑡𝑎𝑙 = 7𝑏 + 5𝑎 + 8𝑝 + 3𝑏 + 6𝑎 + 12𝑝
Since only apples can be added to apples, then
5𝑎 + 6𝑎 = 11𝑎. This is the same for bananas and
peaches. That is, only like terms can be added.
𝑇𝑜𝑡𝑎𝑙 = 10𝑏 + 11𝑎 + 20𝑝
14
Example 1: collect all like terms in the following
3𝑥 + 5𝑥 + 𝑦 − 𝑥 + 5𝑦 + 𝑧
Solution: collect all the 𝑥’s together, all the 𝑦′s
together and all the 𝑧’s together:
7𝑥 + 6𝑦 + 𝑧
The 𝑥’s cannot be added to the 𝑦’s or 𝑧’s as they
are not like terms.
Example 2: collect the like terms in
5𝑥 − 𝑦 + 4𝑥 + 3 − 7𝑦 − 9
Solution: collect all the 𝑥’s together, all the 𝑦’s
together, and all the lone numbers together.
9𝑥 − 8𝑦 − 6
Note: lone numbers are only like other lone
numbers, and are not like anything with a variable
in it.
Example 3: collect all like terms
5𝑥 + 3𝑦 − −4𝑥 + 7𝑦 − 13
Solution: get rid of the brackets:
5𝑥 + 3𝑦 + 4𝑥 + 7𝑦 − 13
Collect like terms:
9𝑥 + 10𝑦 − 13
Applying variables to fractions; finding a Common
Denominator of fractions with variables is easiest
when you think of them as numbers. That is, work
with them as if they were numbers.
Example 4: find the Common Denominator and
put the following two fractions under a single
denominator:
1
𝑥+
2
𝑦
Solution: a Common Denominator is the
multiplication of all the denominators, so in this
case will be 𝑥𝑦.
Then, the first fraction needs to be multiplied by 𝑦
𝑦
to get 𝑥𝑦 on the bottom (the 𝑦 in 𝑦
𝑦 comes from
the denominator of the other fraction):
1
𝑥∙
𝑦
𝑦 =
𝑦
𝑥𝑦
It is easiest to ignore the fact that they are letters,
and just do what you would do if they were
numbers.
The other fractions also needs 𝑥𝑦 on the bottom,
so must be multiplied by 𝑥
𝑥:
2
𝑦∙
𝑥
𝑥 =
2𝑥
𝑥𝑦
Rewrite the original problem and solve:
𝑦
𝑥𝑦+
2𝑥
𝑥𝑦=
𝑦 + 2𝑥
𝑥𝑦
The terms on top cannot be added as they are not
like terms.
Exercises: 1. Collect the like terms in the following
𝑎) 5𝑥 + 4𝑦 − 3𝑥 − 9𝑥 + 4𝑦 − 3
𝑏) 13𝑥 − 12𝑥 − 11𝑥 − 10 − 9𝑦
𝑐) − 8𝑥 − 4𝑦 − −3𝑥 − 4𝑦 − −1
𝑑) 8𝑥 + 8𝑥 + 8𝑦 + 8𝑧 − 8 + 12𝑥 − 13𝑦
2. Collect like terms using all the theory learnt so far:
𝑎) 1
2𝑥 +
1
3𝑥 −
2
5𝑥 𝑑)
𝑥
2+
𝑥
3−
2𝑥
5
𝑏) 12𝑥
5+
13𝑥
6 𝑒)
12𝑥
3−
−2𝑥
4+
2𝑦
3−
13
12
𝑐) 3
𝑥+
4
𝑦 𝑓) 2 +
5
𝑥+
5
𝑦
3. Simplify the following fractions into a single fraction:
𝑎) 12𝑥
−5−
−2𝑦
3 +
2𝑥
30+
4𝑦
10
𝑏) 5
3−
5
𝑦−
3
𝑧+ 2
𝑐) 3
2𝑥+
3
5𝑦
1.6 indices
When mathematicians talk about an index, they
are talking about numbers or variables to the
power of other numbers or variables.
Theory: an index (plural indices) is when a number
(called a base) is put to the power of another
number (called an index).
𝐵𝐴𝑆𝐸𝑖𝑛𝑑𝑒𝑥
15
Examples of indices include 22 , 53 , 𝑥2 , 1 + 𝑦 9,
𝑥2+𝑦 .
An index means that the base number is multiplied
by itself the number of times of the index.
For the index 36
The 3 is multiplied by itself 6 times:
3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 = 36
For 𝑥3, it is:
𝑥 ∙ 𝑥 ∙ 𝑥 = 𝑥3
There are seven rules you need to learn before you
can successfully manipulate indices:
Theory: RULE 1: when two numbers with the same
base are multiplied together, the indices can be
added to one another.
𝑥𝑎 ∙ 𝑥𝑏 = 𝑥𝑎+𝑏
Example 1: simplify the indices in
𝑥3 ∙ 𝑥5 = 𝑥3+5 = 𝑥8
Solution: write this out in extended form:
𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 = 𝑥8
The brackets do not matter as everything is
multiplied.
Theory: RULE 2: when two numbers are divided
and have the same base, the index on the bottom
of the fraction is subtracted from the index on the
top of the fraction.
𝑥𝑎
𝑥𝑏= 𝑥𝑎−𝑏
Example 2: simplify the indices in:
𝑥7
𝑥3
Solution: use the above theory
𝑥7
𝑥3= 𝑥7−3 = 𝑥4
Rewrite in extended form:
Three of the 𝑥’s on top cancel with three 𝑥’s on
the bottom.
Theory: RULE 3: any number with an index of zero
equals one.
𝑥0 = 1
This is something you simply have to know.
Theory: RULE 4: the 𝑛𝑡 root of a number is the
same as having the number to the power of 1/𝑛.
This is a bit complex in words, but simple in
practice.
𝑥𝑛
= 𝑥1/𝑛
Example 3: rewrite the following as an index
𝑥2
Solution: the square root of 𝑥 is simply 𝑥 to the
power of half.
𝑥2 = 𝑥1/2
Note: the square root sign 2
is often written
with the number 2 omitted: .
Example 4: write the following as an index
𝑥4
Solution: it is 𝑥 to the power of one on four:
𝑥4 = 𝑥1/4
Theory: RULE 5: when a number to the power of
one index is put to the power of another index, the
two indices are multiplied together.
𝑥𝑎 𝑏 = 𝑥𝑎∙𝑏
Example 5: simplify the following to a single index
𝑥3 2
Solution: multiply the indices together:
𝑥3 2 = 𝑥3∙2 = 𝑥6
This can also be written out in extended form so
you see where it comes from:
𝑥3 𝑥3 = 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 = 𝑥6
𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙 ∙ 𝒙
𝒙 ∙ 𝒙 ∙ 𝒙= 𝒙𝟒
16
As the two brackets represent 𝑥3 squared.
Theory: RULE 6: a number with a negative index
can be made to have a positive index by inverting
the fraction.
𝑥−𝑎 =1
𝑥𝑎
Example 6: rearrange the following to have a
positive index:
𝑥−3
Solution: invert the fraction, and change the sign
on the index:
𝑥−3
1=
1
𝑥3
The 𝑥−3 is really a fraction with 𝑥−3 “on one”, so
to have a positive power, the fraction is inverted
and the sign of the index is changed.
Example 7: change the following to have a
negative index
𝑥4
Solution: invert the fraction, and change the sign
of the index:
𝑥4
1=
1
𝑥−4
Example 8: change the following to have a positive
index
1
𝑥−5
Solution: invert the fraction, then change the sign
of the index:
1
𝑥−5= 𝑥5
Theory: RULE 7: when the product of two or more
bracketed terms are put to an index, that index
applies to each term within the brackets.
𝑥𝑦 𝑎 = 𝑥𝑎𝑦𝑎
Example 9: expand
2𝑥𝑦 3
Solution: since all terms in the brackets are
multiplied together, every term is raised to the
index:
23𝑥3𝑦3
Simplify the 23:
8𝑥3𝑦3
Example 10: expand
−𝑥𝑦 4
Solution: despite the negative sign inside the
brackets, all the terms are still multiplied together.
The brackets can be rewritten as:
−1 ∙ 𝑥𝑦 4
Then it can be seen that every term in the brackets
is put to the index:
−1 4𝑥4𝑦4
Simplifying −1 4 = 1 gives:
𝑥4𝑦4
All these rules can work forwards and backwards.
The following examples involve the use of multiple
rules.
Example 11: simplify the following to only have
positive indices
𝑥0.5𝑦3𝑥2𝑦−4
Solution: only the like terms can be simplified, so
using RULE 1 on the 𝑥’s and 𝑦’s respectively:
= 𝑥0.5+2𝑦3−4 = 𝑥2.5𝑦−1
To have all indices positive, the 𝑦−1 must be
brought to the bottom using RULE 6, then the sign
of the index is changed:
=𝑥2.5
𝑦1=
𝑥2.5
𝑦
Example 12: simplify the following leaving all
indices positive
17
8𝑦53
𝑥3𝑦2 𝑥3
𝑦𝑥2
Solution: get rid of the root signs first (remember
that if there is no number outside the root sign, it
is assumed to be a square root):
= 8𝑦5 1/3𝑥3𝑦2𝑥1/3
𝑦1/2𝑥2
Get rid of the brackets using RULE 5 with RULE 7:
=81/3𝑦5/3𝑥3𝑦2𝑥1/3
𝑦1/2𝑥2
Then use RULE 1 on the top for the 𝑥’s (as well as
simplifying the addition of fractions):
=81/3𝑦5/3𝑥10/3𝑦2
𝑦1/2𝑥2
And again use RULE 1 for the 𝑦’s on the top (a
Common Denominator needs to be used):
=81/3𝑥10/3𝑦11/3
𝑦1/2𝑥2
The 11/3 comes from 5/3 + 2. Then using RULE 2
for both 𝑥 and 𝑦:
= 81/3𝑥4/3𝑦19/6
Where the 19/6 comes from 11/3 − 1/2; and the
4/3 comes from 10/3 − 2.
Finally, change the 81/3 into 2, as it is of the same
value:
= 2𝑥4/3𝑦19/6
Go over this last example, as the fractions were
difficult.
Example 13: collect all like terms and make all
indices positive for
2
−1𝑥3𝑦0.5
20.5𝑥0𝑦 3
Solution: use RULE 4 to change all the root signs:
= 20.5 −1𝑥3𝑦0.5
20.5𝑥0𝑦 30.5
Get rid of the brackets using RULE 5:
=2−0.5𝑥3𝑦0.5
20.5𝑥0𝑦 30.5
Use RULE 3 to change anything to the power of
zero to 1, which can then be removed from the
bottom:
=2−0.5𝑥3𝑦0.5
20.5𝑦 30.5
Doing this the long way to begin with, use RULE 2
to bring everything on the bottom up top while
changing the sign of each of the indices:
= 2−0.5𝑥3𝑦0.52−0.5𝑦−1 3−0.5
Then use RULE 1 to bring the like terms together.
Remember, the bases must be the same!
= 2−1𝑥3𝑦−0.53−0.5
Now, to make all the indices positive, bring any
bases with negative indices to the bottom while
changing the sign of the index:
=𝑥3
2𝑦0.530.5
This is as far as simplification can go, as all the
bases are different.
In many questions, numbers may need to be
simplified to contain indices. For example, the
number 81 may need to be simplified to either 92
which you should be familiar with, or maybe even
34.
Theory: be aware that each of the following
numbers can be written as the indicated base, to
the power of a whole number:
BASE 2: 2, 4, 8, 16, 32, 64, 128 ,256
BASE 3: 3, 9, 27, 81, 243
BASE 4: 4, 16, 64, 256
BASE 5: 5, 25, 125, 625
Example 14: simplify the following as much as
possible
3𝑥272𝑥 𝑦−63
9𝑥𝑦2
18
Solution: get rid of the root sign using RULE 4:
3𝑥272𝑥 𝑦−6 1/3
9𝑥𝑦2
Then use RULE 5 to get rid of the brackets:
3𝑥272𝑥𝑦−2
9𝑥𝑦2
Change the 27 to 33 and 9 to 32:
3𝑥 33 2𝑥𝑦−2
32 𝑥𝑦2
Use RULE 5 again to get rid of the brackets:
3𝑥36𝑥𝑦−2
32𝑥𝑦2
Then use RULES 1 and 2 to collect all the indices
with base 3:
3𝑥+6−2𝑥𝑥𝑦−2
𝑦2
Simplify the index with base 3 and using RULE 6 to
get the 𝑦−2 to the bottom and with a positive
index:
3−𝑥+6𝑥
𝑦4
Theory: the square root of a positive number will
always have two solutions: one positive and one
negative.
In the first section in this chapter, one of the
exercises was −6 −6 which you should have
worked out to be 36. Similarly, 6 6 = 36.
Example 15: simplify 36
Solution: 36 = ±6
The ± sign means that there are two solutions,
one positive and one negative. It is pronounced
“plus or minus”.
Example 16: simplify 144
Solution: 144 = ±12
Exercises:
1. Simplify the following to have all positive indices
and fraction answers.
𝑎) 𝑥2𝑥−3𝑥−6𝑥15 𝑓) 𝑥2𝑦3𝑥4𝑦5𝑥6𝑧7
𝑏) 𝑥3𝑥5 𝑥
𝑥−2 𝑔)
𝑥−4 𝑥25𝑦3
𝑥−1𝑦𝑥2𝑦1.5𝑥
𝑐) 𝑥2𝑦383
𝑥−2𝑦−42 )
𝑥3.5 𝑥−63𝑥0.25
𝑥−2𝑦−3 𝑥𝑦
𝑑) 5 𝑥4 −1 10𝑥23
𝑦4
3𝑥2𝑦2𝑥−2 𝑖)
27𝑥33𝑥−5𝑦3
𝑥𝑦𝑥−0.2
𝑒) 25 𝑥44
625𝑥23𝑦−5
625𝑥2𝑥−1 𝑗)
81𝑥34𝑥−1/2𝑦−1/6
𝑥0.1𝑦0.4𝑥0.2
1.7 BIMDAS
Theory: the order in which you conduct
mathematical operation (such as addition,
multiplication etc.) is very important. BIMDAS is a
method that gives the order in which operations
must be done. BIMDAS stands for:
Brackets
Indices
Multiplication
Division
Addition
Subtraction
The multiplication/division have no precedence
over one another, and neither does the pair of
addition/subtraction. Because of this, it is
sometimes called BIDMAS.
Intro example: for a number to an index, which is
then added to another number, the index takes
precedence (as Index takes precedence over
Addition in BIMDAS). When solving:
24 + 5
to be able to add the 5, the index must be
simplified first:
16 + 5 = 2
If you were thinking that the 2 and 5 can be added
to one another, and then put that to the power of
4 you would be incorrect, as that does not follow
the order of BIMDAS.
19
Example 1: using BIMDAS, simplify the following
5 ∙ 34 − 7 ∙ (−2)
Solution: following BIMDAS, get rid of the Brackets
first by multiplying the 7 into the −2 :
= 5 ∙ 34 − −14
Since two negative signs equals a positive sign:
= 5 ∙ 34 + 14
Expand the Indices:
= 5 ∙ 81 + 14
Multiplication/Division:
= 405 + 14
Addition/Subtraction:
= 419
BIMDAS must be followed to simplify correctly.
Example 2: using BIMDAS, simplify the following
−8 ÷ 4 + 32 × 4 ÷ 2 + 7/2
Solution: since there are no brackets, simplify the
Index:
= −8 ÷ 4 + 9 × 4 ÷ 2 + 7/2
Get rid of the Multiplication/Division:
= −2 + 18 + 3.5
Finally, Add/Subtract the numbers:
= 19.5
Example 3: using BIMDAS, simplify the following to
a single fraction
−3 ÷ 5 + 3 × 22/32 − 1
Solution: before using BIMDAS, it is easiest to
rewrite the problem with the “÷” changed:
= −3
5+
3 × 22
32− 1
Get rid of the brackets:
= −3
5+
3 × 22
32− 1
Get rid of the indices:
= −3
5+
3 × 4
9− 1
Then the multiplication:
= −3
5+
12
9− 1
This is now a Common Denominator problem:
= −3
5
9
9 +
12
9
5
5 −
1
1
45
45
= −27
45+
60
45−
45
45
=−27 + 60 − 45
45= −
12
45
Moving on to a harder concept that is common
throughout this book; informally, it is called the
crab-claw method, and it is used to simplify the
multiplication of two (or more) brackets. It is also
called the FOIL rule (First, Outer, Inner, Last). It will
make sense soon.
For the multiplication of the following brackets:
𝑥 + 2 𝑦 + 4
Draw in lines to match the letters of FOIL or to look
like a “crab-claw” (the reason is explained below).
Theory: to expand the multiplication of two
brackets, every term in one bracket must be
multiplied with every term in the other bracket
(i.e. First, Outer, Inner, Last – FOIL)
Looking back at the crab-claw that was drawn,
each term in the first bracket is multiplying every
term in the second bracket. That is, 𝑥 multiplies
the 𝑦 (to give 𝑥𝑦), then that same 𝑥 also multiplies
the 4 (to give 4𝑥). The 2 multiplies the 𝑦 to give 2𝑦
and the 2 also multiplies the 4 to give 8. Adding
these together:
𝑥 + 2 𝑦 + 4 = 𝑥𝑦 + 4𝑥 + 2𝑦 + 8
Example 4: expand the brackets for
𝑦 = 2𝑥 + 1 −𝑥 − 1
Solution: draw in the crab claw:
𝑦 = 2𝑥 + 1 (−𝑥 − 1)
𝑦 = 𝑥 + 2 𝑦 + 4
20
Follow the lines:
𝑦 = 2𝑥 −1 + 2𝑥 −𝑥 + 1 −𝑥
+ 1 −1
Simplify:
𝑦 = −2𝑥 − 2𝑥2 − 𝑥 − 1
Collect like terms:
𝑦 = −3𝑥 − 2𝑥2 − 1
The following is where many students make
mistakes, so don’t be one of them!
Theory: when a bracket containing two terms
separated by an addition/subtraction sign is put to
an index, write out the bracket that many times. If
given:
𝑥 + 3 2
Write the bracket out twice:
𝑥 + 3 2 = 𝑥 + 3 𝑥 + 3
This is index theory. However, many students think
they can simply put the index into each of the two
terms inside the bracket:
𝑥 + 3 2 = 𝑥2 + 32
This is incorrect as there is an addition (or
subtraction) sign separating two terms inside the
brackets. This is a VERY common mistake. The
correct answer to this problem is 𝑥2 + 6𝑥 + 9.
Find it for yourself using the crab-claw method.
Example 5: expand
𝑥 − 7 2
Solution: write out the brackets twice:
= 𝑥 − 7 𝑥 − 7
Draw in the crab-claw:
Follow the lines:
= 𝑥2 − 7𝑥 − 7𝑥 + 49
Collect like terms:
= 𝑥2 − 14𝑥 + 49
Example 6: expand the bracket
−2𝑥 + 4 2
Solution: write out the two brackets:
= −2𝑥 + 4 −2𝑥 + 4
Draw in the crab-claw:
Follow the lines:
= 4𝑥2 − 8𝑥 − 8𝑥 + 16
Collect like terms:
= 4𝑥2 − 16𝑥 + 16
Example 7: using BIMDAS, expand and collect like
terms for
−3𝑥 + 1 2 + 2 − 𝑥 + 1 2 + 52
Solution: This seems complex, but do it one small
step at a time. To get rid of the Brackets, take the
−3𝑥 + 1 2 and expand this first (ignoring
everything else):
−3𝑥 + 1 2 = −3𝑥 + 1 −3𝑥 + 1
Using the crab-claw:
Follow the lines gives:
= 9𝑥2 − 3𝑥 − 3𝑥 + 1
= 9𝑥2 − 6𝑥 + 1
Replace this in the original statement (in brackets):
9𝑥2 − 6𝑥 + 1 + 2 − 𝑥 + 1 2 + 52
Move on to the other bracket:
𝑥 + 1 2 = 𝑥 + 1 𝑥 + 1
Use the crab-claw method again:
Follow the lines:
= 𝑥2 + 𝑥 + 𝑥 + 1
= 𝑥2 + 2𝑥 + 1
𝑥 + 1 𝑥 + 1
−3𝑥 + 1 −3𝑥 + 1
= −2𝑥 + 4 (−2𝑥 + 4)
= 𝑥 − 7 𝑥 − 7
21
Substitute back into the original statement:
9𝑥2 − 6𝑥 + 1 + 2 − 𝑥2 + 2𝑥 + 1 + 52
The brackets must still be there as that negative
sign must go into every term in the brackets:
9𝑥2 − 6𝑥 + 1 + 2 − 𝑥2 − 2𝑥 − 1 + 52
All the brackets have been removed, so move on
to Indices:
9𝑥2 − 6𝑥 + 1 + 2 − 𝑥2 − 2𝑥 − 1 + 25
The index of 𝑥2 cannot be simplified as 𝑥 is a
variable so it is left alone. Move on to
Multiplication/Division (of which there is none),
and then onto Addition/Subtraction by collecting
like terms:
= 8𝑥2 − 8𝑥 + 27
Example 8: show that the expanded forms of the
two statements below are not the same:
Statement 1: 𝑥 + 3 𝑥 + 5 − 15𝑥 + 4
Statement 2: 𝑥 + 3 𝑥 + 5 − 15𝑥 + 4
Solution:
For statement 1: use BIMDAS to get rid of the
Brackets:
𝑥 + 3 𝑥 + 5 − 15𝑥 + 4
The only thing multiplying 𝑥 + 5 is 3, so only half
a crab-claw is required. The reason is that 𝑥 + 3 is
not in brackets so only the 3 needs to be crab-
clawed:
= 𝑥 + 3𝑥 + 15 − 15𝑥 + 4
Collecting like terms:
= −11𝑥 + 19
For statement 2: use BIMDAS to get rid of the
Brackets first, by using the crab-claw:
Which gives:
= 𝑥2 + 3𝑥 + 5𝑥 + 15 − (15𝑥 + 4)
The other brackets need to have the negative sign
introduced into each term:
= 𝑥2 + 3𝑥 + 5𝑥 + 15 − 15𝑥 − 4
Since there are no Indices (that can be simplified),
nor multiplication/division, the only thing left is to
collect like terms:
= 𝑥2 − 7𝑥 + 11
Despite the two statements having the same
numbers, the placement of brackets is very
important as obviously, the two statements are
not the same.
The more practice you have with BIMDAS, the
easier it will become.
Exercises: 1. Simplify the following using BIMDAS
𝑎) 2 5 + 7 − 12 − 23
𝑏) 4 − 1 5 + 32 − 16 + 16 ÷ 8
𝑐) 12 ÷ 4 + 3 ÷ 3 × 2 + 1
𝑑) 7 3 − 22 ÷ −1 + 5𝑥 − 4 + 22
2. Simplify the following using the crab-claw 𝑎) 𝑥 + 1 𝑥 + 1
𝑏) 𝑥 + 3 𝑥 − 4
𝑐) 𝑥 + 2 2 + 1
𝑑) 2𝑥 + 2 𝑥 − 1
𝑒) −3𝑥 − 1 −𝑥 + 2
3. Simplify the following using the crab-claw and BIMDAS
𝑎) 3 − 42 − 𝑥2 + 𝑥 + 13 2
𝑏) 𝑥 + 1 2 − 5 + 22 2 ÷ 3 + 1
𝑐) 5 − 1 − 13 2 + 𝑥 + −1 2 2 + 1 − 𝑥
𝑑) 𝑥 − 1 2 + 2𝑥 + 1 2 + 12
𝑒) 2 𝑥 − 1 2 + 3𝑥 − 22 − 4 3 − 𝑥
4. Simplify the following using all the theory from this chapter.
𝑎) 3𝑏2 3 + 4𝑏 𝑏2
𝑏3
𝑏) 5𝑎 2𝑎 + 3
9 + 𝑎2
𝑐) 7𝑏 1
𝑏3+ 2𝑏2 − 𝑏2
𝑑) 3𝑏 𝑏2
𝑏0.5 𝑏+ 1
2
1.8 equations
You have already been introduced to equations
without it being said explicitly.
𝑥 + 3 𝑥 + 5 − 15𝑥 + 4
𝑥 + 3 𝑥 + 5 − 15𝑥 + 4
22
Theory: an equation is anything with an equals
sign.
Examples of equations include:
5 + 1 = 6
3𝑥 − 4 = 7
𝑦 = 𝑥2 + 6𝑥 − 3
All these have equals signs so all are equations.
Equations are used widely to find solutions to
problems.
Intro example 1: a customer wants to buy 45
laptop computers from a retail company. The
retailer has only 13 in stock. The problem is to find
how many computers need to be ordered. Write
the problem as an equation:
13 + 𝑥 = 45
The left side is what is in stock (13) plus what
needs to be ordered (𝑥), and the right side (45) is
what the customer wants. This is a very simple
example, and when 13 is taken from 45, the result
will be 𝑥, the number of computers that need to
be ordered (𝑥 = 45 − 13 = 32).
Manipulating equations is difficult to begin with,
but the more you practice, the more natural it
becomes. There are a few rules and a few hints
which will make rearranging equations very
simple.
Theory: make sure there is a reason for
rearranging an equation.
This sounds obvious, but it is often forgotten. Most
of the time, rearranging an equation is to isolate a
variable. In the above example, 𝑥 was isolated to
give the number of computers to be ordered.
Theory: RULE 1: whatever is done to one side of
the equation must also be done to the other side.
If one side of an equation is changed without
changing the other, it will no longer be equal.
Example 1: isolate 𝑥 in
𝑥 + 5 = 9
Solution: to isolate 𝑥, the 5 needs to be removed.
A “+5” is “reversed” by subtracting 5 from the left
side; but whatever is done to one side has to be
done to the other side as well. So 5 is taken from
both sides:
𝑥 + 5 − 5 = 9 − 5
This can then be simplified to:
𝑥 = 4
As on the left side, 5 − 5 = 0 and on the right side,
9 − 5 = 4. To make sure, go back to the original
equation and substitute 𝑥 = 4. Does 4 + 5 = 9?
Yes.
Example 2: isolate 𝑥 in
𝑥
4= 10
Solution: to isolate 𝑥, the 4 needs to be removed.
The “reverse” of a “÷ 4” is to multiply the left side
by 4. But once again, whatever is done to one side
must be done to the other, so both sides must be
multiplied by 4:
4 ∙𝑥
4= 4 ∙ 10
On the left side, the 4’s cancel to give 1𝑥 or simply
𝑥 on the left side:
𝑥 = 40
Substitute this into the original equation:
40
4= 10 ? Yes
Theory: RULE 2: when isolating a variable, use the
reverse order of BIMDAS to get rid of the elements
around the variable to be isolated.
This simply means work backwards to BIMDAS
when deciding what to get rid of on the side you
are trying to isolate 𝑥.
23
Example 3: isolate 𝑥 in the following:
𝑥
3+ 5 = 7
Solution: to isolate 𝑥, the 3 and the 5 need to be
“reversed”, but they must be done one at a time.
Using the reverse of BIMDAS, get rid of the 5 as it
is an addition. To “reverse” the +5, subtract 5
from both sides:
𝑥
3+ 5 − 5 = 7 − 5
𝑥
3= 2
To “reverse” the ÷ 3 multiply both sides by 3
(remember that 3 =3
1):
3
1∙𝑥
3= 3 ∙ 2
𝑥 = 6
If 𝑥 = 6 is substituted into the original equation:
6
3+ 5 = 2 + 5 = 7
Which is correct.
Example 4: isolate 𝑥 in the following:
−3𝑥
5+ 7 = −4
Solution: to isolate 𝑥, work backwards through
BIMDAS. Get rid of the +7 by subtracting 7 from
both sides:
−3𝑥
5+ 7 − 7 = −4 − 7
−3𝑥
5= −11
The negative sign on the left can be put either up
top, or down bottom (they are the same, and both
will give the same answer):
3𝑥
−5= −11
To get rid of the ÷ −5 , multiply both sides by
−5 (remember −5 =−5
1):
−5
1∙
3𝑥
−5= −11 ∙ −5
3𝑥 = 55
Divide both sides by 3 to get rid of the × 3 on the
left side:
3𝑥
3=
55
3
𝑥 =55
3= 18
1
3
Note: the negative sign could have been moved to
give −3𝑥
5 without changing the answer. Also, the 3
could have been “reversed” first, and then the −5,
as BIMDAS does not give precedence to
Division/Multiplication.
Example 5: isolate 𝑥 in the following
𝑥
7+ 𝑦 − 3 = 9
Solution: using the reverse of BIMDAS, the +𝑦 or
−3 must be removed. Below, the −3 is removed
first by adding 3 to both sides, but the 𝑦 could
have been removed first without changing the
solution:
𝑥
7+ 𝑦 − 3 + 3 = 9 + 3
𝑥
7+ 𝑦 = 12
Get rid of the +𝑦 (treat this as you would any
other number). To reverse this +𝑦, subtract 𝑦
from both sides:
𝑥
7+ 𝑦 − 𝑦 = 12 − 𝑦
𝑥
7= 12 − 𝑦
Finally, “reverse” the ÷ 7 by multiplying both sides
by 7:
𝑥 = 7 ∙ 12 − 𝑦
Brackets are needed around the 12 − 𝑦 as RULE 1
above is really:
Theory: RULE 1: whatever is done to the whole of
one side of an equation must be done to the
whole of the other side.
24
You might have thought to write:
𝑥 = 7 ∙ 12 − 𝑦
This is incorrect as the whole of the right side must
be multiplied by 7, however here, only the 12 is
being multiplied by 7.
Given the correct form:
𝑥 = 7 ∙ 12 − 𝑦
It can be expanded using half a crab-claw:
𝑥 = 84 − 7𝑦
Example 6: isolate 𝑥 in
−2
5𝑥 + 3𝑥 −
2
3𝑦 + 5 = 7
Solution: firstly, put everything not with an 𝑥 onto
the right side, by adding 2
3𝑦 to both sides and
subtracting 5 from both sides:
−2
5𝑥 + 3𝑥 −
2
3𝑦 +
2
3𝑦 + 5 − 5 = 7 +
2
3𝑦 − 5
−2
5𝑥 + 3𝑥 = 2 +
2
3𝑦
Change each side to have a Common Denominator
(the left side Common Denominator does not have
to be the same as the Common Denominator on
the right):
−2𝑥
5+
3𝑥
1=
2
1+
2𝑦
3
On the left side, the Common Denominator is 5, so
3𝑥/1 needs to be changed to have 5 on the
bottom, by multiplying by 5/5:
−2𝑥
5+
3𝑥
1
5
5 =
2
1+
2𝑦
3
−2𝑥
5+
15𝑥
5=
2
1+
2𝑦
3
Change the left side under a single division:
−2𝑥 + 15𝑥
5=
2
1+
2𝑦
3
Simplify the top of the left side:
13𝑥
5=
2
1+
2𝑦
3
Find a Common Denominator for the right side
(1 × 3 = 3):
13𝑥
5=
6
3+
2𝑦
3
13𝑥
5=
6 + 2𝑦
3
The reason why a Common Denominator is found
for both sides is that it makes it easier to “reverse”
the ÷ 5 making it easier to simplify the right side:
13𝑥
5∙ 5 =
6 + 2𝑦
3 ∙ 5
Remember that ∙ 5 is really like multiplication by 5
1:
13𝑥
5∙
5
1=
6 + 2𝑦
3 ∙
5
1
13𝑥 =5 6 + 2𝑦
3
Finally, to get rid of the × 13, divide the whole of
both sides by 13:
13𝑥
13=
5 6 + 2𝑦
3÷
13
1
To get rid of the ÷ sign on the right, invert the
second fraction and multiply:
𝑥 =5 6 + 2𝑦
3∙
1
13
𝑥 =5 6 + 2𝑦
39
That is the answer. Half a crab claw could be used
to put that 5 into the brackets, which would give:
𝑥 =30 + 10𝑦
39
A common application of equations is finding
exchange rates.
Example 7: if the currency conversion rate from
American dollars (𝑈𝑆$) to Euros (€) is:
𝑈𝑆$1.50 = €1
How many Euros would be obtained from
converting 𝑈𝑆$173?
Solution: to solve this sort of question, find how
many Euros you will get per American dollar. This
25
means both sides must be divided by 1.50, so as to
get a single 𝑈𝑆$ on the left side:
𝑈𝑆$1 = €1
1.5
Since 𝑈𝑆$173 is being converted, multiply both
sides by 173:
𝑈𝑆$173 = €1
1.5 173 ≈ €115.33
Theory: for currency conversions, obtain the
exchange rate for a single unit of the currency you
have, then multiply both sides of that conversion
equation by the number of units of currency
initially had.
Exercises: 1. Solve for 𝑥 in the following equations:
𝑎) 2𝑥 − 3 = 4𝑥 + 7
𝑏) 12𝑥 − 15 +10
2= 2𝑥
𝑐) 17𝑥 − 3𝑥 = 15𝑥 − 2𝑥 + 13.5𝑥
𝑑) 19
3𝑥 − 15 = 17 − 5
𝑒) −2𝑥
3+ 5𝑥 −
1
3𝑥 − 3 = 15 − 𝑥
𝑓) 𝑥 +𝑥
2−
𝑥
3+
1
4= 5
2. Isolate 𝑥 in the following equations, leaving answers as a single fraction (where applicable):
𝑎) 3𝑦 − 2𝑥 = 𝑥 + 15
𝑏) 𝑦 = 3𝑥 − 15 − 𝑦
𝑐) 𝑦 + 1 + 𝑥 = 𝑦 − 1
𝑑) 3𝑥
4+
5
3𝑦 − 4 = 0
𝑒) 17𝑥 − 13𝑦 − 11 = 7
𝑓) 4𝑦 − 1
𝑥 + 2= 5
3. Given the initial amount of currency, and given the exchange rate, determine the equivalent value in the other currency.
𝑎) 𝐴$181 𝐴𝑈𝑆$0.80 = 𝑁𝑍$1.00
𝑏) 𝑈𝑆$47 𝑈𝑆$0.94 = 𝐶𝐴𝑁$1.04
𝑐) ¥1801 𝑈𝑆$1.03 = ¥96.1
𝑑) 𝐴𝑈𝑆$94 𝐴𝑈𝑆$1.23 = 𝐶𝐴𝑁$0.92
1.9 factorisation
Factorisation is the opposite of expansion of
brackets. The application of factorising will be seen
in subsequent chapters. Factorising does not
change the equation, but rather is a method of
manipulating an equation.
Intro example 1: for the equation
𝑦 = 2𝑥 + 2
Factorise out the number 2. Factorising out 2 from
every part of the right side is analogous to
“dividing” each term by the number 2, but then
bringing 2 out front of brackets:
𝑦 = 2 𝑥 + 1
This may be confusing but if half a crab-claw was
used to get rid of the brackets, the original
equation will be obtained.
Example 1: factorise 𝑥 out of the following
𝑦 = 4𝑥 + 3𝑥2
Solution: “divide” each of the parts on the right
side by 𝑥, then bring 𝑥 out front of brackets:
𝑦 = 𝑥 4 + 3𝑥
Example 2: factorise anything common in the
equation
𝑦 = 4𝑥2 + 8𝑥3
Solution: each part on the right has an 𝑥2 (the 𝑥3
is the same as 𝑥2𝑥), and also a 4 (as 8 = 4 ∙ 2), so
this means that 4𝑥2 can be factorised out from
each part on the right side:
𝑦 = 4𝑥2 1 + 2𝑥
Theory: for a statement which is separated by
additions/subtractions, and each part of the
statement has a certain common aspect, then each
separate part of the statement can be “divided” by
the common aspect, then having that common
aspect being put outside of the brackets.
Example 3: factorise
𝑦 = 3𝑥3 − 12𝑥 + 6𝑥2
26
Solution: the common aspect in each part is 3𝑥, so
put 3𝑥 out front of brackets and then “divide”
each part by 3𝑥:
𝑦 = 3𝑥 𝑥2 − 4 + 2𝑥
A common use of factorisation is finding the
solution(s) to complex statements being equal to
zero. However, after the statements are
factorised, there are at least two big parts being
multiplied together.
Theory: when two statements are multiplied
together and they equal zero, one of the
statements must be equal to zero.
Example 4: solve
0 = 3𝑥2 + 6𝑥
Solution: factorise a common 3𝑥 out of each part:
0 = 3𝑥 𝑥 + 2
Now there are two parts being multiplied to give
zero. The two “parts” are colour coded below:
0 = 3𝑥 𝑥 + 2
For the whole right side to equal zero, either
3𝑥 = 0 making 𝑥 = 0 or 𝑥 + 2 = 0 making
𝑥 = −2. The two solutions are 𝑥 = 0 and 𝑥 = −2.
Exercises: 1. Factorise
𝑎) 2𝑥 + 4𝑥𝑦 + 8𝑥2 𝑏) 3𝑥2 + 9𝑥 − 12𝑦 𝑐) 3𝑥3 + 2𝑥2 𝑑) 13𝑥2 + 𝑥
𝑒) −3𝑥
4+
𝑥2
4
𝑓) 18𝑥2𝑦 − 91𝑥𝑦2 𝑔) 12𝑥2𝑦2 − 48𝑥𝑦
2. Solve for 𝑥 using factorisation: 𝑎) 18𝑥2 − 9𝑥 𝑏) 8𝑥 − 4𝑥2 𝑐) 19𝑥2 − 38𝑥 𝑑) 17𝑥 − 18𝑥2 − 𝑥 𝑒) 12𝑥4 − 37𝑥3
1.10 inequalities and absolute values
An equality is when two things are equal.
Inequalities are when two things are not equal.
Theory: inequalities are noted with the signs “<”
and “>”. The sign > means the left side is greater
than the right side, and the sign < means the left
side is less than the right side. Similarly, the sign ≥
means “greater than or equal to”, and the
converse sign ≤ means “less than or equal to”.
The easiest way of remembering which side is
greater than and which side is less than, is that the
side of the open part of the signs is greater than
the pointed end. E.g. 5 > 3 as 5 is on the side of
the open end meaning that 5 is greater than 3.
Example 1: put in the inequality sign for
6∎8
Solution: since 6 is less than 8, the inequality sign
has to have the open part to the right:
6 < 8
Manipulating inequalities is a little different from
equations.
Theory: addition and subtraction of inequalities is
identical to that of equalities.
Example 2: take 5 from both sides of
11 > 7
Solution:
11 − 5 > 7 − 5
6 > 2
Theory: multiplication and division of inequalities
is identical to equalities only when multiplied or
divided by a positive number.
Example 3: divide both sides of the inequality by 5
50 > −40
Solution:
50
5>
−40
5
10 > −8
27
Theory: when multiplying or dividing an inequality
by a negative number, the inequality sign must be
reversed.
Example 4: multiply both sides by −3
8 > 5
Solution:
8 −3 > 5 −3
−24 > −15
This is not true, so the inequality sign must be
reversed to give the correct answer:
−24 < −15
Moving on to absolute values.
Theory: an absolute value makes any statement
which is negative into a positive statement.
Absolute values are denoted with straight lines
around the statement:
𝑥 means “the absolute value of 𝑥”.
Example 5: find the absolute value of
−5
Solution: −5 = 5
Example 6: Find the absolute value of
7
Solution: 7 = 7
This is not the extent of absolute values, but it is
what you will need for understanding this book.
Exercises: 1. Divide both sides of the inequality by the number in
brackets 𝑎) 15 > 4 2
𝑏) 13 < 15 −2
𝑐) − 12 < −6 −1
𝑑) 1
5>
1
8 −5
2. Isolate 𝑥 in the following inequalities: 𝑎) 2𝑥 > 15 − 𝑥
𝑏) 13𝑥 +1
3𝑥 < 0 − 15
𝑐) 11𝑥 − 12𝑥 < 12 − 11
𝑑) 2𝑥 − 3𝑥 +15
2𝑥 −
𝑥
3> 21 − 3𝑥
3. Find the absolute values of the following (Hint: simplify before finding the absolute value).
𝑎) | − 4|
𝑏) |79|
𝑐) |5 − 4|
𝑑) |32 − 42|
𝑒) |4 − 6 − 1|
1.11 the number zero
Zero is a special number, with special properties.
Theory:
Any number multiplied by zero has a result
equal to zero. E.g. 5 ∙ 0 = 0
Zero divided by any number (other than
another zero) is zero. 0
15= 0
Zero divided by zero does not exist, but is not
necessarily equal to zero. 0
0≠ 0
Example 1: solve the following problem when
𝑥 = 0:
𝑥
𝑥
Solution: this is impossible, and further
investigation into the original function is required.
This above example does not mean that 𝑥
𝑥 cannot
be simplified. What it means is that when 𝑥 = 0, 𝑥
𝑥
is undefined. However, 𝑥
𝑥 is equal to 1, whenever 𝑥
is not equal to zero.
Example 2: solve
15𝑥2 0
Solution: because there is a multiplication by zero,
the value of 𝑥 is irrelevant:
15𝑥2 0 = 0
Theory: as mentioned in the previous section,
when two statements are multiplied together and
they equal zero, one of the statements must be
zero.
28
Example 3: solve for:
0 = 𝑥 + 3 𝑥 − 1
Solution: the two parts (bracketed statements) are
colour coded below:
0 = 𝑥 + 3 𝑥 − 1
Either 𝑥 + 3 = 0 which gives 𝑥 = −3 or
𝑥 − 1 = 0 giving 𝑥 = 1. So the solutions are:
𝑥 = −3 𝑎𝑛𝑑 𝑥 = 1
Exercises: 1. Solve for 𝑥 in the following: 𝑎) 3 − 5𝑥 + 12 0 = 3 − 16𝑥 0
𝑏) 15𝑥 + 0 − 15 = 0
𝑐) 30 𝑥 + 0 2 + 10 𝑥 = 30𝑥2 − 20
𝑑) 0
0+ 1 = 𝑥
chapter one summary
Arithmetic with negative numbers: 𝑎 + −𝑏 = 𝑎 − 𝑏 𝑎 − −𝑏 = 𝑎 + 𝑏
−𝑎 −𝑏 = 𝑎 ∙ 𝑏 𝑎 −𝑏 = −𝑎 ∙ 𝑏
−𝑎
−𝑏=
𝑎
𝑏
−𝑎
𝑏=
𝑎
−𝑏= −
𝑎
𝑏
Different ways of denoting a multiplication: 𝑎 ∙ 𝑏 = 𝑎 × 𝑏 = 𝑎 𝑏
Multiplication of fractions: 𝑎
𝑏∙𝑐
𝑑=
𝑎 ∙ 𝑐
𝑏 ∙ 𝑑
Division of fractions: 𝑎
𝑏÷
𝑐
𝑑=
𝑎
𝑏×
𝑑
𝑐=
𝑎 ∙ 𝑑
𝑏 ∙ 𝑐
Addition and subtraction of fractions uses the Common Denominator approach. This approach changes the form of all fractions so that the denominators are all the same, but leaves the values of all fractions unchanged. A Common Denominator is found by multiplying the denominators (bottoms) of all the fractions to be added or subtracted. The form of each individual fraction is changed by multiplying by 𝑥/𝑥 where 𝑥 is the multiplication of all the other denominators. The numerators (tops) are then put over the common denominator.
Simplifying a fraction involving whole numbers: multiply the whole number out front by the denominator (bottom), then add that result to the top. To find how many “whole” numbers there are in a fraction, work out how many denominators will fit into the numerator without exceeding the numerator. Then remove that many denominators from the numerator, and any remaining numbers are left over the original denominator.
“Collecting like terms” is the process of bring together things that are the same.
INDEX RULE 1: 𝑥𝑎 ∙ 𝑥𝑏 = 𝑥𝑎+𝑏
INDEX RULE 2: 𝑥𝑎
𝑥𝑏 = 𝑥𝑎−𝑏
INDEX RULE 3: 𝑥0 = 1
INDEX RULE 4: 𝑥𝑛
= 𝑥1/𝑛 INDEX RULE 5: 𝑥𝑎 𝑏 = 𝑥𝑎∙𝑏
INDEX RULE 6: 𝑥−𝑎 =1
𝑥𝑎
INDEX RULE 7: 𝑥𝑦 𝑎 = 𝑥𝑎𝑦𝑎 Common numbers and their base: BASE 2: 2, 4, 8, 16, 32, 64, 128, 256 BASE 3: 3,9,27,81,243
BASE 4: 4,16,64,256 BASE 5: 5,25,125,625
The square root of a positive number will always have two solutions: one positive and one negative.
BIMDAS is a method that gives the order in which operations must be done. BIMDAS is Brackets, Indices, Multiplication, Division, Addition, Subtraction. The multiplication/division have no precedence over one another, and neither does the pair of addition/subtraction.
To expand the multiplication of two brackets, every term in one bracket must be multiplied with every term in the other bracket (i.e. First, Outer, Inner, Last – FOIL) EQUATION RULE 1: whatever is done to the whole of one side of the equation must also be done to the whole of the other side. EQUATION RULE 2: when isolating a variable, use the reverse order of BIMDAS to get rid of the elements around the variable to be isolated.
For currency conversions, obtain the exchange rate for a single unit of the currency you have, then multiply both sides of that conversion equation by the number of units of currency initially had.
To factorise a statement which is separated by additions/subtractions, and each part of the statement has a certain common aspect, then each separate part of the statement can be “divided” by the common aspect, then having that common aspect being put outside of the brackets. When two statements are multiplied together and they equal zero, one of the statements must be equal to zero.
Inequalities are noted with the signs “<” and “>”. The sign > means the left side is greater than the right side, and the sign < means the left side is less than the right side.
Addition and subtraction of inequalities is identical to that of equalities. Multiplication and division of inequalities is identical to equalities only when multiplied or divided by a positive number. When multiplying or dividing an inequality by a negative number, the inequality sign must be reversed.
An absolute value makes any statement which is negative into a positive statement. Absolute values are denoted with straight lines around the statement: 𝑥 means “the absolute value of 𝑥”.
29
Any number multiplied by zero has a result equal to zero. Zero divided by any number (other than another zero) is
zero. 0
15= 0
Zero divided by zero does not exist, but is not necessarily
equal to zero. 0
0≠ 0
chapter one questions
1. Simplify the following negative numbers: 𝑎) −3 −5 𝑏) −2 2
𝑐) 5 ∙ −5 ∙ −2 𝑑) 3 ∙ −2
2. Simplify the following fractions:
𝑎) 1
2∙
1
3 𝑏)
2
3∙
1
8
𝑐) 1
2/
1
3 𝑑) −
4
3
2
5
𝑒) 9
−7∙−3
7 𝑓)
18
2÷
1
3
𝑔) 27
13×
1
2÷
3
8 )
5
3×
4
3÷
3
3×
2
3÷
1
3
3. Simplify the following fractions:
𝑎) 1
2+
1
3 𝑏)
2
3+
1
8
𝑐) 3
4+
4
3 𝑑)
2
11−
1
2
𝑒) 2
13+
1
5 𝑓)
1
2−
1
3+
1
4
𝑔) 6
7+
5
6−
4
5 ) 3 +
2
3−
3
7
4. Simplify the following fractions:
𝑎) 1
3∙
2
−7 −
1
2 𝑏)
2
2∙
1
6 +
1
5
𝑐) 1
𝑥+
1
3 𝑑)
1
−2−
−2
3 −
1
2
𝑒) 4
5/
3
4−
1
6 𝑓) 1 +
1
2+
1
3+
1
4+
1
𝑥
𝑔) 2 𝑥
2 /
3
4 ∙
5
1 +
𝑥
3 )
1
𝑥+
1
𝑦
2
3
5. Simplify using index rules, leaving positive indices: 𝑎) 𝑥𝑦2 3
𝑏) 2𝑥 9𝑦𝑥4 3
𝑐) 𝑥84
𝑥−3𝑦−4 −2
𝑑) 4𝑥0.5𝑥−1.25𝑦
3𝑥𝑦
𝑒) 81𝑥290.5 3𝑥𝑦2 3
27𝑥𝑦−2
𝑓) 2𝑥𝑦 4𝑦0.125 8𝑦−2𝑥−1 2
𝑔) 216𝑥−3𝑦83
36𝑥𝑦𝑥𝑦𝑥1.5𝑦−0.5
) 25𝑥4𝑦−0.25 −0.5 −𝑥𝑦 3
𝑥𝑦 125𝑦33
6. Simplify the following using BIMDAS: 𝑎) 3 + 5 ∙ 7 − (−3 + 5) ∙ 2
𝑏) 4 −2 + 4 ∙ 32 − 2 ∙ 2
𝑐) 3 − 2 ∙ 2 − 3 2 + 23
𝑑) 14𝑥 − 𝑥 3 + 4 2 + 3 − 1 𝑥
𝑒) 2 − 3 ∙ 32 − 1 2 + 4
𝑓) 2 − 6 3 + 33 + −2 − 1 + 2
𝑔) −2 −3 − 43 + 2 ∙ −3
) 2𝑥 − 3 𝑥 + 2𝑦 − 1 + 2 3𝑦 − 𝑥 + 3
7. Expand and simplify the following: 𝑎) 𝑥 + 2 2 𝑏) 𝑥 − 3 2
𝑐) 2𝑥 + 1 2 𝑑) 1 − 𝑥 2
𝑒) 𝑥 + 3 2 + 2𝑥 − 4
𝑓) 2 − 2𝑥 − 1 2 + 3𝑥 − 40.5
𝑔) 3 𝑥 − 1 2 + 1
) − 2 − 𝑥 + 1 2 − 14
8. Isolate 𝑥 in the following equations:
𝑎) 𝑥 − 4 = 1 𝑏) 3𝑥 + 1 = 7
𝑐) 2𝑥 − 1 = 4𝑥 − 3 𝑑) 3𝑥 − 7 = 4
𝑒) 2𝑥 + 2
3= 5 𝑓) 7𝑥 +
2
3= 1
𝑔) 12𝑥 + 3 = −6 + 2𝑥 ) 3𝑥 + 𝑦 − 7 = 2
𝑖) 4𝑥 − 2
𝑦= 2 𝑗) 𝑥 − 7 = 5𝑥 − 3𝑦 + 2
9. Factorise out anything possible in the following: 𝑎) 2𝑥 + 2𝑥𝑦 + 2 𝑏) 2𝑥2 + 6𝑥𝑦 − 8𝑦2𝑥
𝑐) 13𝑦𝑥 − 12𝑦2 + 𝑦 𝑑) 3𝑥2 + 9𝑥3 + 12𝑥5
𝑒) 4𝑄𝑃 + 𝑟𝑃𝑄 − 𝑥𝑃𝑦𝑄 𝑓) 1
3𝑥2 +
1
3𝑥4 +
1
6
𝑔) 15𝑥3 + 25𝑥 − 50𝑥4 ) 1
𝑥+
1
𝑥2−
1
𝑥3
10. Simplify and solve for 𝑥 in the following: 𝑎) 𝑥2 − 𝑥 = 0 𝑏) 𝑥2 + 3𝑥 = 0
𝑐) 2𝑥 − 6𝑥2 = 0 𝑑) 4𝑥2 + 4𝑥 = 0
𝑒) 𝑥3 − 𝑥2 = 0 𝑓) 3𝑥4 − 5𝑥3 = 0
11. Simplify, then factorise, the following: 𝑎) 2𝑥 + 3 2 + 3
𝑏) 9 − 𝑥 2
− 𝑥2 + 3
𝑐) 216𝑥6 13 − 3 + 𝑥2 + 3 + 𝑥−2
𝑑) −3 − 2 𝑥 − 32 + 1 −3𝑥2 + 6𝑥3
12𝑥
𝑒) 27𝑥63− 𝑥2
2
+𝑥(𝑥−3 −2 𝑥4.5
4𝑥
𝑓) −125𝑥93
+ 3𝑥3𝑥−2
1𝑥2
− 5𝑥2 + 3𝑥
12. Solve for 𝑥 in the following equations: 𝑎) 𝑥 − 3 𝑥 − 2 = 0
𝑏) 𝑥 + 7 𝑥 − 4 = 0
𝑐) 2𝑥 − 4 𝑥 − 3 = 0
𝑑) −3𝑥 − 1 6𝑥 + 15 = 0
𝑒) 3𝑥5 − 6𝑥4
𝑥3= 0
𝑓) 4 − 𝑥 2
− 𝑥2 = 0
13. Find the absolute value of the following: 𝑎) −3 𝑏) 4 𝑐) | − 1|
𝑑) −15 𝑒) −0.1 𝑓) |2 − 3|
30
Chapter 2
Linear Algebra How to describe and draw lines in maths
2.1 Linear Equations 31
2.2 Main Feature of Linear Equations 32
2.3 Negative Gradients 33
2.4 Graphing Lines from Equations 34
2.5 Obtaining the Equation of a Line 36
2.6 Intersecting Lines 38
2.7 Microeconomic Applications 40
2.8 Elasticity 42
2.9 Interpreting Elasticity 44
Chapter Two Summary 46
Chapter Two Questions 47
31
2.1 linear equations
Theory: linear equations have the general form:
𝑦 = 𝑚𝑥 + 𝑐
Where 𝑚 is the gradient, 𝑐 is a constant, and 𝑥 and
𝑦 are variables.
For example 𝑦 = 2𝑥 + 3. Both 𝑚 and 𝑐 are
definite numbers (𝑚 = 2, 𝑐 = 3).
In simple terms, 𝑚 is the gradient of the line –
which is another way of saying the slope of the
line. Steeper lines have larger values of 𝑚. The 𝑐 is
a constant which is added to everything else (it is
just a number). The 𝑥 and the 𝑦 are variables; 𝑦 is
the dependent variable, because its value depends
on 𝑥, which is the independent variable (see
chapter 1).
Graphs of linear equations are straight lines.
Below is a graph of two lines which might help you
understand this theory.
Line 𝐴 is steeper than line 𝐵, so it has a larger
gradient (slope=gradient). Thus, line 𝐴 has a larger
value of 𝑚 than line 𝐵.
The graph below shows how the value of 𝑐 affects
the position of a line.
These lines have the same slope (𝑚), but the
values of 𝑐 differ. The value of 𝑐 is a constant that
determines how much the graph of the line is
shifted up or down (line 𝐷 has a larger value of 𝑐
than line 𝐸).
Theory: a mathematical definition of gradient is:
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑚 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛
The graph below shows how this definition is used:
Going from point 𝐹 to point 𝐺, ask, how much
does the line rise for every unit it goes across
(runs)?
Example 1: determine the gradient of the
following line:
Plan: use the equation 𝑚 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛
Solution:
𝑚 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛=
3
10= 0.3
Theory: the gradient of a line is the same at all
points along it.
The values of 𝑚 and 𝑐 are actual numbers, such as:
𝑦 = 4𝑥 + 5 𝑚 = 4 𝑐 = 5
𝑦 = 81𝑥 − 7 𝑚 = 81 𝑐 = −7
3
10
𝑥
𝑦
𝑥
𝑦
𝑟𝑢𝑛
𝑟𝑖𝑠𝑒 𝐹
𝐺
𝑥
𝑦
𝐸
𝐷
𝑥
𝐴
𝐵
𝑦
32
𝑦 = −1
4𝑥 +
3
4 𝑚 = −
1
4 𝑐 =
3
4
The constants 𝑚 and 𝑐 can be positive or negative,
and can be whole numbers, fractions or decimals.
The 𝑦 = 𝑚𝑥 + 𝑐 equation is the general form of
a line, but there are many other forms. Despite all
these other forms, it is best practice to manipulate
equations of lines to get them into this easier,
general form. From Chapter 1, you should be able
to rearrange equations easily:
Example 2: rearrange to get into the general form
−7𝑥 + 4𝑦 = 𝑦 − 3
Plan: use the reverse of BIMDAS to isolate 𝑦.
Solution: add 7𝑥 to both sides.
−7𝑥 + 7𝑥 + 4𝑦 = 𝑦 − 3 + 7𝑥
Take 𝑦 from both sides (to have all the 𝑦’s on the
left side):
+4𝑦 − 𝑦 = 𝑦 − 𝑦 − 3 + 7𝑥
3𝑦 = 7𝑥 − 3
Then divide the whole of both sides by 3.
3𝑦
3=
7𝑥 − 3
3
𝑦 =7
3𝑥 −
3
3
𝑦 =7
3𝑥 − 1
Example 3: rearrange to get into the general form
3𝑦 − 4
𝑥 + 1= 6
Plan: apply the reverse of BIMDAS to isolate 𝑦.
Solution: multiply both sides by (𝑥 + 1):
3𝑦 − 4
𝑥 + 1(𝑥 + 1) = 6(𝑥 + 1)
3𝑦 − 4 = 6𝑥 + 6
Add 4 to both sides:
3𝑦 − 4 + 4 = 6𝑥 + 6 + 4
3𝑦 = 6𝑥 + 10
Divide both sides by 3:
3𝑦
3=
6𝑥 + 10
3
𝑦 = 2𝑥 +10
3
The gradient (𝑚) and the constant (𝑐) of each line
is unknown until it is rearranged into the general
form.
Exercises: 1. Get the following equations in general form:
𝑎) − 𝑥 − 𝑦 = 7 𝑏) 2𝑥 − 2𝑦 − 4 = 0 𝑐) 3𝑦 + 4𝑥 = 17 𝑑) − 3 𝑥 − 2𝑦 = 𝑦 − 2 𝑒) 𝑦 = 2𝑥 − 0.5𝑦 + 8
𝑓) 𝑦 + 3
𝑥 + 3= 3
2. Determine the exact gradient and 𝑐 − 𝑣𝑎𝑙𝑢𝑒 of the following equations:
𝑎) 8𝑦 + 3𝑥 − 7 = 0 𝑏) − 3𝑥 − 5𝑦 = 7 − 𝑥
𝑐) 𝑥
−3+
4
−1𝑦 = 7
𝑑) −−5
4𝑥 + 𝑦
6
7 = 𝑦 − 1
𝑒) 𝑦 = 9
𝑓) 2𝑦 − 1
𝑥 + 1= 5
2.2 main features of linear equations
The graph below shows a couple of lines.
Theory: the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is where the line cuts
the 𝑥 − 𝑎𝑥𝑖𝑠. Similarly, the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is
where the line cuts the 𝑦 − 𝑎𝑥𝑖𝑠.
There are two aspects to note when graphing
lines:
1. steeper lines have larger gradients (𝑚 values).
2. lines cutting the 𝑦 − 𝑎𝑥𝑖𝑠 higher up have larger
𝑐 value; the 𝑐 value shifts the line up or down.
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
𝑥
𝑦
33
Example 1: from the following graph, order the
lines from the largest 𝑚 value to the lowest 𝑚
value. Then order the lines from the largest 𝑐 value
to the smallest 𝑐 value.
Plan: use the fact that 1) steeper lines have larger
gradients, and 2) lines cutting the 𝑦 − 𝑎𝑥𝑖𝑠 higher
have larger 𝑐 values.
Solution: Looking only at the slopes of the lines, 𝐵
is steeper than 𝐷 which is steeper than 𝐴. So the
order of gradients is 𝐵, 𝐷, 𝐴.
Now, look at where the lines cut the 𝑦 − 𝑎𝑥𝑖𝑠; 𝐴
cuts it higher than 𝐵 which cuts it higher than 𝐷,
so the order of 𝑐 values is 𝐴, 𝐵, 𝐷.
Exercises: 1. Order the lines from the one with the largest
gradient to the one with the lowest gradient:
2. Order the above lines from the one with the largest
𝑐 − 𝑣𝑎𝑙𝑢𝑒 to the one with the lowest.
2.3 negative gradients
Up to this point, all linear equations have had a
positive gradient.
Theory: lines with a positive gradient go from the
bottom left to the top right, and lines with a
negative gradient go from the top left to the
bottom right.
The graph below shows lines with negative
gradients:
Both lines have a negative gradient, however line
𝐴 is steeper than line 𝐵. So the gradient 𝑚 is
bigger for 𝐴 than for 𝐵 in absolute terms. That is,
line 𝐴 has gradient −3 and line 𝐵 has gradient −1.
Remember from Chapter 1 that −1 > −3, but in
absolute terms (ignoring the negative signs) 3 > 1.
Theory: the steeper a line with a negative gradient,
the smaller is the gradient 𝑚 (as it is more
negative) and at the same time, the larger the
absolute value of 𝑚.
Example 1: for the three lines graphed below,
determine 1) which has a larger 𝑚 value, 2) which
has a larger absolute value of 𝑚, and 3) which has
a larger 𝑐 value.
Plan: the line which is steeper has a higher
absolute 𝑚 value, and at the same time has a
more negative 𝑚 value than lines less steep. The
line with the higher 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 has a larger 𝑐
value.
𝐵
𝐴
𝐷
𝑥
𝑦
𝐴: 𝑦 = −3𝑥 − 1
𝐵: 𝑦 = −𝑥 + 3
𝑥
𝑦
𝐵
𝐴
𝐷
𝑥
𝑦
𝐵
𝐴 𝐷
𝑥
𝑦
34
Solution: From the graph, 𝐴 is steeper than 𝐵
which is steeper than 𝐷. So the absolute values of
the gradient 𝑚 is 𝐴, 𝐵, 𝐷. Because 𝐴 is steeper
than the other two, it must have a more negative
gradient; similarly 𝐵 is steeper than 𝐷 so it must
have a more negative 𝑚 value than 𝐷. Thus, the
normal 𝑚 value order is the opposite of the one
above: 𝐷, 𝐵, 𝐴.
Finally, 𝐵 cuts the 𝑦 − 𝑎𝑥𝑖𝑠 higher than 𝐷, which
cuts it higher than 𝐴, so the order for the size of 𝑐
is 𝐵, 𝐷, 𝐴.
Exercises: For the following lines:
a) Order the lines from the one with the lowest
gradient to the one with the largest gradient: b) Order the lines from the shallowest to the steepest
(in absolute terms). Compare a) and b). c) Order the lines from the one with the largest
𝑐 − 𝑣𝑎𝑙𝑢𝑒 to the one with the lowest.
2.4 graphing lines from equations
Lines can be easily plotted on a set of axes.
Theory: to plot a line, points are needed which are
called coordinates. Coordinates are written as
(𝑥, 𝑦), where 𝑥 is always written first, and 𝑦
second.
For the coordinate (3,7), 𝑥 = 3 and 𝑦 = 7.
To be able to draw a line, you need at least two
points (sets of coordinates).
It is good practice to have equations of lines in the
general form, but the following method allows
plotting of lines in any initial form.
Theory: to find:
1. the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 (where the line crosses the
𝑥 − 𝑎𝑥𝑖𝑠), replace all values of 𝑦 with zero,
then solve for 𝑥.
2. the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, replace all values of 𝑥 with
zero, then solve for 𝑦.
This may be a little bit confusing at first.
Example 1: plot the line
𝑦 = 2𝑥 + 6
Plan: set all 𝑦′𝑠 equal to zero then solve for 𝑥 to
find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. Then set all the 𝑥's equal
to zero, and solve for 𝑦 to find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡.
Solution: for 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0:
0 = 2𝑥 + 6
2𝑥 = −6
𝑥 = −3
This gives the coordinate −3,0 . Remember that
at the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, the value of 𝑦 = 0, and
coordinates are written with 𝑥 first then 𝑦: (𝑥, 𝑦).
To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0:
𝑦 = 2 0 + 6
𝑦 = 6
This gives the coordinate (0,6), as at the
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, the value of 𝑥 = 0.
These two points can be plotted on a set of axes,
then joined with a ruler.
Theory: the 𝑥 − and 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 are usually
the easiest points to find to be able to plot a line.
(−3,0)
(0,6)
𝑥
𝑦
𝐷
𝐴
𝐵
𝑥
𝑦
35
Example 2: Plot 𝑦 = 3𝑥 − 7
Plan: find:
1) the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑦 = 0 and
solving for 𝑥
2) the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑥 = 0 and
solving for 𝑦.
Plot these two points then join them with a ruler.
Solution: for the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0:
0 = 3𝑥 − 7
3𝑥 = 7
𝑥 =7
3
Thus the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is:
7
3, 0
For the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0:
𝑦 = 3 0 − 7
𝑦 = −7
Which gives the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 0, −7 .
Plot these two points and join with a ruler:
Example 3: plot the equation
3𝑦 − 2𝑥 = 5
Plan: find:
1) the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑦 = 0 and
solving for 𝑥
2) the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑥 = 0 and
solving for 𝑦.
Plot these two points then join them with a ruler.
Solution: to find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0:
3 0 − 2𝑥 = 5
−2𝑥 = 5
𝑥 = −2.5
The coordinate of the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is (−2.5,0).
To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0:
3𝑦 − 2 0 = 5
3𝑦 = 5
𝑦 =5
3
The coordinates of the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is 0,5
3 .
Plot these two points, then join with a ruler:
Note, that if asked for the gradient of this last
equation (3𝑦 − 2𝑥 = 5), you would have to
rearrange the equation into the general form. Try
rearranging it, and plotting the rearranged
equation.
Example 4: find the gradient and plot the line
3𝑦 + 4
2𝑥 − 1= −1
Plan: rearrange into the general form to find the
gradient. Then find two points:
1) the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑦 = 0 and
solving for 𝑥
2) the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting 𝑥 = 0 and
solving for 𝑦.
Plot these two points then join them with a ruler.
Solution: rearrange into the general form:
3𝑦 + 4
2𝑥 − 1(2𝑥 − 1) = −1(2𝑥 − 1)
3𝑦 + 4 − 4 = −2𝑥 + 1 − 4
3𝑦
3=
−2𝑥 − 3
3
𝑦 = −2
3𝑥 −
3
3
𝑦 = −2
3𝑥 − 1
(−2.5,0)
0,5
3
𝑥
𝑦
(0, −7)
7
3, 0
𝑥 𝑦
36
Now that it is in the general form, the gradient is
𝑚 = −2
3
To plot the line, use either the original equation, or
the equation in the general form. The following
uses the original equation:
1. To find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑦 = 0:
3 0 + 4
2𝑥 − 1= −1
4
2𝑥 − 1= −1
4 = −2𝑥 + 1
3 = −2𝑥
𝑥 = −1.5
Giving 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 coordinates (−1.5,0).
2. To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑥 = 0:
3𝑦 + 4
2 0 − 1= −1
3𝑦 + 4
−1= −1
3𝑦 + 4 = 1
3𝑦 = −3
𝑦 = −1
This gives coordinates of (0, −1). Plot these two
points:
Exercises: 1. Find the gradient of each of the equations, then
plot the function: 𝑎) 𝑦 = 2𝑥 − 3
𝑏) 3𝑦 = 6𝑥 − 9
𝑐) 2𝑦 + 3𝑥 = 7
𝑑) 12𝑥 − 𝑦 = 3
𝑒) 7𝑦 − 3 = −2 + 𝑥
𝑓) 𝑦 − 𝑥 = 4 − 𝑥
𝑔) 3𝑦 − 1
2𝑥 + 1= 3
) 8𝑦 + 𝑥
𝑥 − 𝑦= 4
2.5 obtaining the equation of a line
In many cases you will need to find the equation of
a line without being given a graph.
Theory: when given two points, approximately plot
the points on a set of axes (if not given the graph),
then find how much the line “runs” along the
𝑥 − 𝑎𝑥𝑖𝑠, and then how much it “rises” along the
𝑦 − 𝑎𝑥𝑖𝑠 between the two points.
Applying the 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula, the gradient can
be found.
Using the general form of a line, substitute in the
gradient and any one of the points given, and solve
for 𝑐. Rewrite the equation of the line.
You will always be given either two coordinates, or
a coordinate and a gradient. The second case is a
subset of the first case, so if you know how to do
the first case, you will be able to do the second.
Intro example 1: find the equation of the line
passing through the points (1,3) and (3,7).
Solution: approximately plot the points:
Then draw a horizontal line from the point that is
most left, and a vertical line at the point most
right.
Ask yourself: how much does the horizontal line
“run” before it reaches the intersection? The
answer is two units along the 𝑥 − 𝑎𝑥𝑖𝑠. Then ask:
(1,3)
(3,7)
2
4
intersection
𝑥
𝑦
(0, −1)
−1.5,0
𝑥
𝑦
37
from this intersection, how much does the vertical
line “rise”? The answer is four units on the
𝑦 − 𝑎𝑥𝑖𝑠.
Use the 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula to find the gradient:
𝑚 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛=
4
2= 2
Rewrite the general form of a line:
𝑦 = 𝑚𝑥 + 𝑐
Then replace the gradient with the value found:
𝑦 = 2𝑥 + 𝑐
To find 𝑐, substitute either one of the two point in
for 𝑥 and 𝑦. Using the point (1,3), substitute 1 for
𝑥 and 3 for 𝑦 in the general form of the line:
3 = 2(1) + 𝑐
3 = 2 + 𝑐
1 = 𝑐
So the equation of the line is:
𝑦 = 2𝑥 + 1
Using the other point (3,7) will give the same
answer. Try it!
Theory: a gradient can be found using the
𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula, which can be written in
mathematical terms as:
𝑚 =𝑦2 − 𝑦1
𝑥2 − 𝑥1=
𝑟𝑖𝑠𝑒
𝑟𝑢𝑛
where the subscripts refer to the two points
(𝑥1 , 𝑦1) and 𝑥2 , 𝑦2 . This formula allows you to
find the gradient between two points without
plotting the points.
Example 1: find the equation of the line going
through the points: (−16,4) and (−9, 2)
Plan: use the 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula to find the
gradient, then substitute that gradient and one of
the points into 𝑦 = 𝑚𝑥 + 𝑐 to solve for 𝑐.
Solution: approximately plot the points:
The line “runs” between −16 and −9 , which is 7,
and “rises” between 4 and 2 which is 2. Now, we
ask, for every 7 that the line runs, how much does
it rise?
If the line falls, as in this case, the rise is negative!
So:
𝑚 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛=
−2
7 (leave it as a fraction)
This could also have been done using the
mathematical 𝑟𝑖𝑠𝑒/𝑟𝑢𝑛 formula:
𝑚 =𝑦2 − 𝑦1
𝑥2 − 𝑥1=
4 − 2
−16 − −9 =
2
−7
Substitute this gradient into the general form:
𝑦 = −2
7𝑥 + 𝑐
Next, substitute in either of the original points. The
point (−9,2) is used below, but the other point
(−16,4) will give the same answer.
Substitute 𝑥 = −9, 𝑦 = 2:
2 = −2
7 −9 + 𝑐
2 =18
7+ 𝑐
14
7−
18
7= 𝑐
𝑐 = −4
7 (leave this as a fraction)
So the equation of this line is:
𝑦 = −2
7𝑥 −
4
7
You must know how to manipulate fractions to be
able to find equations of lines (see Chapter 1).
(−16,4)
(−9,2)
7
2
𝑥
𝑦
38
Example 2: find the equation of a line which has a
gradient of −3 and passes through the point
(−1,6).
Plan: use the general form of a line:
𝑦 = 𝑚𝑥 + 𝑐
Substitute the given facts (𝑚 = −3, and the point
(−1,6)) and solve for 𝑐.
Solution: substitute everything into the general
form of a line:
6 = −3 −1 + 𝑐
6 = 3 + 𝑐
𝑐 = 3
So the equation of the line is 𝑦 = −3𝑥 + 3.
This example did not require you to sketch the
line, as it involved only substitution.
Exercises: 1. Determine the equations of the lines passing
through each of the two points (respectively): 𝑎) (1,10)(2,13)
𝑏) (1,6)(3,16)
𝑐) (−1,3)(1,7)
𝑑) (1,1)(3, −3)
𝑒) (−2,1)(−5,0)
𝑓) −1,21
15 −7,
63
15
2. Determine the equation of a line passing through one point, having the gradient shown:
𝑎) 2,7 𝑚 = 4
𝑏) 6,2
3 𝑚 = −3
𝑐) −1, −1 𝑚 = −1
𝑑) 15,8 𝑚 =3
7
2.6 intersecting lines
All this mathematical theory is leading to real
world applications. If you draw two lines on a set
of axes, chances are they will cross.
Theory: lines with different gradients will intersect
only once.
The two lines above cross only once. Another way
of thinking about this is that if you look along the
blue line, the only point that it touches the red line
is at the intersection; similarly, if you look along
the red line, the only time it touches the blue line
is at the intersection.
On this graph, the two lines are labelled with
subscripts on the 𝑦 and the 𝑥 to distinguish
between the two lines. The only point on the
graph where the lines do not need to be
distinguished is at the intersection, as it is common
to both lines. So at the intersection:
𝑦1 = 𝑦2 𝑎𝑛𝑑 𝑥1 = 𝑥2
Theory: when linear equations with different
gradients intersect, there is only one set of
coordinates that is similar on both lines. At the
intersection, the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of both lines are the
same and at the same time the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of
both lines are the same.
This only applies to the intersection, as no other
point will satisfy this condition.
Example 1: find the intersection of the following
lines
𝑦 = 2𝑥 + 5
𝑦 = −𝑥 − 3
Plan: the only point where the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of both
lines are equal, and at the same time, the
𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of both lines are equal is at the
intersection. Set the two 𝑦's equal to each other
and solve for 𝑥; this gives the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 at the
𝑥
𝑦
𝑦1 = 5𝑥1 + 1
𝑦2 = −4𝑥2 + 11
39
intersection. Substitute this into either of the
original equations to give the 𝑦 − 𝑣𝑎𝑙𝑢𝑒.
Solution: rewrite the above equations with
subscripts to distinguish between the two
equations:
𝑦1 = 2𝑥1 + 5
𝑦2 = −𝑥2 − 3
Set the two 𝑦’s equal to each other:
𝑦1 = 𝑦2
Then substitute the 𝑦's for their respective
equations:
2𝑥1 + 5 = −𝑥2 − 3
At the intersection, the 𝑥’s are the same so the
subscripts can be removed and 𝑥 solved:
2𝑥 + 5 = −𝑥 − 3
3𝑥 = −8
𝑥 = −8
3
This gives the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the intersection, but
the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is still required to get the
coordinates. The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 comes from one of the
original equations (it doesn’t matter which one, as
at the intersection, they both have the same 𝑥 −
and 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠). Using 𝑦1:
𝑦1 = 2 −8
3 + 5
𝑦 = −1
3
So the intersection is at −8
3, −
1
3 .
Try using the other equation to see if you get the
same solution.
Subscripts were used to distinguish between the
two functions, however, this is not necessary.
Example 2: find the intersection of:
4𝑦 − 3𝑥 − 11 = 1
𝑦 − 4 = −2𝑥
Plan: rearrange into the general form, then set the
𝑦's equal to each other. Substitute with the
equations, and solve for 𝑥. Substitute 𝑥 into one of
the equations to find 𝑦.
Solution: rearrange the first equation:
4𝑦 − 3𝑥 = 12
4𝑦 = 3𝑥 + 12
𝑦 =3
4𝑥 + 3
And the other equation:
𝑦 = −2𝑥 + 4
Set 𝑦1 = 𝑦2 and substitute the equations:
𝑦1 = 𝑦2
3
4𝑥 + 3 = −2𝑥 + 4
11
4𝑥 = 1
𝑥 =4
11
This is the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the intersection, and now
for the 𝑦 − 𝑣𝑎𝑙𝑢𝑒:
𝑦 =3
4𝑥 + 3
𝑦 =3
4
4
11 + 3
𝑦 =36
11
Thus, the two lines intersect at 4
11,
36
11 .
If you are struggling with fractions, go back to
Chapter 1.
Exercises: 1. Determine where the following pairs of equations
intersect: 𝑎) 𝑦 = −4𝑥 + 3
𝑦 = −6𝑥 + 7
𝑏) 𝑦 = 3𝑥 + 8
𝑦 = 7𝑥 + 4
𝑐) 𝑦 = −7𝑥 + 12
𝑦 = 2𝑥 + 21
𝑑) 𝑦 = −4𝑥 − 12
𝑦 = −1.5𝑥 − 7
𝑒) 𝑦 = 0.5𝑥 + 15
3𝑦 = −𝑥 + 15
𝑓) 5𝑦 − 3𝑥 = 31
8𝑦 − 2𝑦 = 16
𝑔) 2𝑦 = −3𝑥 − 19
9𝑦 = −3𝑥 − 12
40
) 7𝑦 + 12𝑥 = 72
4𝑦 = −3𝑥 + 36
2.7 microeconomic applications
Economic theory:
DEMAND: this is what you do as a consumer. You
demand goods and services. For example, when
you go to a surf shop to purchase clothes, you are
demanding those clothes. If board shorts cost
$1000 each, would you buy many? Probably not,
as it’s too expensive. What about if they cost $2
each. Would you buy many? Yes, most probably.
So at low prices, you would buy more, and at
higher prices, you would buy less. This explains the
downward slope of the demand function:
SUPPLY: when you go to purchase board-shorts,
the shop supplies them, making them the supplier.
If the price of board-shorts was $1000 (totally
ignoring demand) the shop would supply a lot, to
make a big profit. If the price went down to $2,
then the shop wouldn’t make much profit, so they
wouldn’t supply very many. This is why the supply
curve slopes upwards. At higher prices, shops
would want to sell more so supply more.
Intro example: you are thinking of starting a
business selling guitars so you do a little research
into the market for guitars.
Looking at the demand side of the market, you
estimate that if you started your business, the
demand curve would be:
𝑄𝑑 = −2𝑃 + 700
Now looking at the supply side of the market,
meaning how many guitars you would make at
certain prices, the supply function is given by:
𝑄𝑠 = 4𝑃 − 500
You would like to know the equilibrium price and
quantity (i.e. when the supply and demand curves
intersect).
Solution: at equilibrium, there is no excess supply
or excess demand (don’t worry too much if you
don’t understand excess supply/demand).
Despite there not being any 𝑥's or 𝑦's, simply apply
the same theory of lines; set the 𝑄𝑠 = 𝑄𝑑 (similar
to 𝑦1 = 𝑦2 in the previous section).
𝑄𝑠 = 𝑄𝑑
Then substitute the functions of 𝑃:
4𝑃 − 500 = −2𝑃 + 700
Rearrange to find 𝑃:
6𝑃 = 1200
𝑃 = 200
This is the price the guitars should be sold for, but
the quantity of guitars to be made is still unknown.
This quantity comes from either 𝑄𝑠 or 𝑄𝑑 (either
one as it is at equilibrium):
𝑄𝑠 𝑃 = 200 = 4 200 − 500
𝑄𝑠 𝑃 = 200 = 300
The interpretation is that 300 guitars should be
produced and sold for $200 each.
You may also be asked to draw the two lines on a
set of axes. If you are asked to do this, in
economics, the price of a good (𝑃) is on the
𝑃
𝑄
𝑆 1000
2
𝑄1 𝑄2
𝑃
𝑄
𝐷 2
1000
𝑄1 𝑄2
41
𝑦 − 𝑎𝑥𝑖𝑠, and the quantity of that good (𝑄) is on
the 𝑥 − 𝑎𝑥𝑖𝑠. This is a standard which is used by
economists, so to be consistent, you will have to
rearrange both the supply and demand equation
to have the price (𝑃) by itself.
For the demand equation:
𝑄𝑑 = −2𝑃 + 700
𝑄𝑑 − 700 = −2𝑃
𝑄𝑑 − 700
−2= 𝑃
𝑃 = −0.5𝑄𝑑 + 350
For the supply equation:
𝑄𝑠 = 4𝑃 − 500
𝑄𝑠 + 500 = 4𝑃
𝑃 = 0.25𝑄𝑠 + 125
Plot both equations individually using the method
earlier in the chapter. For the demand equation:
1. To find the 𝑃 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑄 = 0:
𝑃 = −0.5 0 + 350
𝑃 = 350
Giving a coordinate of 0,350 , remembering
that 𝑃 is on the 𝑦 − 𝑎𝑥𝑖𝑠 and 𝑄 on the
𝑥 − 𝑎𝑥𝑖𝑠.
2. To find the 𝑄 intercept, set 𝑃 = 0:
0 = −0.5𝑄𝑑 + 350
0.5𝑄𝑑 = 350
𝑄𝑑 = 700
Giving a coordinate of (700,0).
For the supply equation, you should get
𝑄𝑠 = −500 and 𝑃 = 125. Make sure you get this.
Then plot both demand and supply on the same
axes:
Theory: demand and supply can only exist when
quantity and price are positive or zero. You cannot
have a negative quantity, or a negative price
(hence the dotted extensions).
Also, in economics, 𝑃 is on the 𝑦 − 𝑎𝑥𝑖𝑠 and 𝑄 is
on the 𝑥 − 𝑎𝑥𝑖𝑠.
Example 1: find the equilibrium quantity and price
for the demand and supply equations
𝑄𝑑 = −0.5𝑃 + 25
𝑄𝑠 =1
4𝑃 − 2
Then sketch the two lines, and draw in the
equilibrium.
Plan: set 𝑄𝑑 = 𝑄𝑠 then solve for 𝑄 and 𝑃. Sketch
each of the two equations by finding the four axis
intercepts.
Solution: set 𝑄𝑑 = 𝑄𝑠
−0.5𝑃 + 25 =1
4𝑃 − 2
Rearrange to isolate 𝑃:
27 =1
4𝑃 + 0.5𝑃
Use fraction theory from Chapter 1 to find a
Common Denominator:
27 =𝑃
4+
2𝑃
4
27 =3𝑃
4
108 = 3𝑃
𝑃 =108
3= 36
(0,350)
(0,125)
(−500,0)
(700,0)
𝑄
𝑃 𝑆
𝐷
42
To find the quantity, use either demand or supply
(it does not matter). Demand is used below:
𝑄𝑑 = −0.5 36 + 25
𝑄𝑑 = −18 + 25 = 7
The equilibrium is 𝑄 = 7, 𝑃 = 36.
To sketch the lines, find the two intercepts
(remember that 𝑃 is on the vertical axis, and 𝑄 is
on the horizontal axis).
Demand: to find the 𝑃 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑄 = 0:
𝑄𝑑 = −0.5𝑃 + 25
0 = −0.5𝑃 + 25
0.5𝑃 = 25
𝑃 = 50
This gives coordinates of (50,0).
To find the 𝑄 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑃 = 0:
𝑄𝑑 = −0.5𝑃 + 25
𝑄𝑑 = 25
Supply: to find the 𝑃 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑄 = 0:
𝑄𝑠 =1
4𝑃 − 2
0 =1
4𝑃 − 2
0.25𝑃 = 2
𝑃 = 8
To find the 𝑄 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, set 𝑃 = 0:
𝑄𝑠 =1
4 0 − 2
𝑄𝑠 = −2
Plot demand and supply separately, then join the
respective dots:
Exercises: 1. Find the equilibrium price and quantity given the
following demand and supply equations: 𝑎) 𝑃𝑑 = −7𝑄𝑑 + 50
𝑃𝑠 = 0.5𝑄𝑠 = 5
𝑏) 𝑄𝑑 =𝑃𝑑
−9+
77
9
𝑄𝑠 = 4𝑃𝑠 − 12
𝑐) 𝑄𝑑 − 0.25𝑃𝑑 + 23
𝑃𝑠 = 0.2𝑄𝑠 + 8
𝑑) 𝑃𝑠 = 0.8𝑄𝑠 + 15
𝑃𝑑 = −1.2𝑄𝑑 + 115
𝑒) −0.7𝑄𝑠 + 𝑃𝑠 = 6
7𝑄𝑑 + 𝑃𝑑 = 61
2. Find the equilibrium price and quantity given the following demand and supply equations, then plot the functions.
𝑎) 𝑃𝑠 = 0.9𝑄𝑠 + 12
𝑃𝑑 + 0.9𝑄𝑑 = 117
𝑏) 𝑃𝑠 = 2.5𝑄𝑠 + 26
𝑃𝑑 = −2𝑄𝑑 + 158
𝑐) 𝑃𝑠 = 1.7𝑄𝑠 + 21
𝑃𝑑 + 2.3𝑄𝑑 = 182
𝑑) 𝑃𝑠 = 1.3𝑄𝑠 + 24
𝑃𝑑 = −1.5𝑄𝑑 + 148
𝑒) −0.9𝑄𝑠 + 𝑃𝑠 = 4
7𝑄𝑑 = −𝑃𝑑 + 61
2.8 elasticity
The concept of elasticity is best explained with
intuition. Let’s say that you own a business selling
cars. You have a car that is not selling well, so you
decide to reduce the price by $3,000. That’s a
large reduction, right? Well, we don’t really know.
If the original price of the car was $6,000, then
yes, the reduction is huge. But if the original price
was $50,000 then the reduction is not very big at
all. What we really want to know it the percentage
change in price. In the first instance, the
percentage change is price is −50% (it’s negative
because the price has decreased), and in the
second instance, the percentage change in price is
−6%.
Theory: percentage changes are found like any
other percentage is found:
0,50
25,0 (−2,0)
(0,8)
𝑄
𝑃
7,36
𝑆
𝐷
43
%∆𝑃 =∆𝑃
𝑃× 100 =
𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑖𝑐𝑒
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑖𝑐𝑒× 100
If there is a $3,000 decrease, the ∆𝑃 = −3000.
And if the original price 𝑃 = 50,000, then:
%∆𝑃 =−3000
50,000× 100 = −6%
Back to the elasticity concept: will you sell more
cars if you reduce the price by 6% or 50%? You
probably said 50%. Yes that would be sort–of
correct: you would sell more cars, but by what
percentage will your car sales increase? If you sell
3 cars when you decrease price by 6%, and sell 4
cars when you decrease price by 50%, what is a
better option? (You don’t need to answer this, just
think about the concept).
Theory: elasticity of demand is the responsiveness
of quantity demanded to a change in price.
Another way of thinking about this is: how
sensitive is the quantity demanded to changes in
price.
An intuitive example is; if the shop reduced the
price of beer by $1, would the number of cartons
of beer sold increase a little or increase a lot? This
is what elasticity determines. It uses percentages
to get rid of the error we encountered in the car
example before.
Theory: the elasticity formula is:
𝜀𝑑 =𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦
𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑖𝑐𝑒=
%𝛥𝑄
%𝛥𝑃
The %𝛥𝑄 and %𝛥𝑃 can be replaced with their
respective definitions (%∆𝑃 =∆𝑃
𝑃× 100 and
%∆𝑄 =∆𝑄
𝑄× 100):
𝜀𝑑 = 𝛥𝑄𝑄 × 100
𝛥𝑃𝑃 × 100
= 𝛥𝑄
𝛥𝑃 ×
𝑃
𝑄
NOTE:
1. the 100
100 cancels off
2. the 𝛥𝑄 and 𝛥𝑃 are symbols which are
different from 𝑄 and 𝑃. The 𝛥𝑄 and 𝛥𝑃 mean
change in 𝑄 and change in 𝑃, respectively.
The form of elasticity used most often is:
𝜀𝑑 = 𝛥𝑄
𝛥𝑃 ×
𝑃
𝑄
This last equation is simpler than you think. The
first part 𝛥𝑄
𝛥𝑃 is simply the slope of the line (𝑚)
when you have the equation in the general form
(with 𝑄 by itself). The second part is a point along
the line.
Intro example 1: a business selling motorboats to
James Bond decides to hire an economist to
estimate the demand curve. The economist comes
back and gives the following function before he is
shot:
𝑄𝑑 = −0.0004𝑃 + 10
Initially, the business sells 6 boats per year at
$10,000 each, but then decides to increase the
price by 25%. The question is, how responsive are
boat sales to changes in price?
Solution: a 25% price increase means the new
price will be $12,500 = 10,000 × 1.25 .
Substituting this into the demand equation:
𝑄𝑑(12,500) = −0.0004(12,500) + 10
𝑄𝑑(7,500) = 5
So 6 boats are sold initially, and this is reduced to
5 after the 25% price increase. Now, what is the
change in 𝑄𝑑?
∆𝑄𝑑 = 5 − 6 = −1
What is the percentage change in 𝑄𝑑?
%𝛥𝑄 =𝛥𝑄
𝑄× 100 =
−1
6× 100
%𝛥𝑄 ≈ −16.67%
Putting all this information into the original
elasticity equation:
𝜀𝑑 =%𝛥𝑄
%𝛥𝑃≈
−16.67
25≈ −0.67
44
Now the much easier way of doing this question.
Using the easier form of elasticity:
𝜀𝑑 = 𝛥𝑄
𝛥𝑃 ×
𝑃
𝑄
Look at the original demand equation:
𝑄𝑑 = −0.0004𝑃 + 10
Is the 𝑄 by itself? Yes, so remembering that 𝛥𝑄
𝛥𝑃 is
the slope of the demand function, which is
−0.0004. Substitute this into the elasticity
equation:
𝜀𝑑 = −0.0004 × 𝑃
𝑄
The 𝑃
𝑄 needs to be found and it is simply the
original point. The original price is $10,000 and
original quantity sold is 6, so substitute these in:
𝜀𝑑 = −0.0004 × 10,000
6 = −
2
3≈ −0.67
The same answer as before. Plus this is a more
precise answer. In the next section, the meaning of
this number will be discussed. Also, the elasticity
of demand will always be negative.
Example 1: a producer of laptops has a supply
curve approximated by
𝑄𝑠 = 4𝑃 − 400
The price the laptops are sold to a retailer is
$1000 each. How sensitive is the quantity supplied
to changes in price?
Plan: use the elasticity formula, but now supply
instead of demand:
𝜀𝑠 = 𝛥𝑄
𝛥𝑃 ×
𝑃
𝑄
Solution: the gradient 𝑚 of the supply curve, when
𝑄 is by itself is 4. Substitute this into the elasticity
formula:
𝜀𝑠 = 4 × 𝑃
𝑄
The price of the laptops is $1000, however 𝑄 is
still required; it is found using the supply curve:
𝑄𝑠(1000) = 4(1000) − 400
= 3600
Substitute 𝑄 = 3600, 𝑃 = 1000 into the elasticity
formula:
𝜀𝑠 = 4 × 1000
3600
=10
9≈ 1.11
The elasticity of supply is approximately 1.11.
The elasticity of supply will always be positive.
Exercises: 1. Determine the elasticity of demand for the given
information: 𝑎) 𝑃𝑑 = −0.5𝑄𝑑 + 50 𝑎𝑡 𝑃 = 20
𝑏) 𝑄𝑑 = −3.5𝑃𝑑 + 75 𝑎𝑡 𝑄 = 31
𝑐) 𝑄𝑑 = −4.15𝑃𝑑 + 103 𝑎𝑡 𝑄 = 27
2. Determine the elasticity of supply for the given information: 𝑎) 𝑃𝑠 = 3 + 𝑄𝑠 𝑎𝑡 𝑃 = 15
𝑏) 𝑄𝑠 = 0.4𝑃𝑠 − 44 𝑎𝑡 𝑄 = 17
𝑐) 𝑄𝑠 = 1.2𝑃𝑠 − 37.5 𝑎𝑡 𝑄 = 21
2.9 interpreting elasticity
Theory: the negative sign of the elasticity of
demand means that as prices changes in one
direction (either a positive or negative change),
then the quantity demanded will move in the
opposite direction.
The positive value of the elasticity of supply means
that as prices change, the quantity will change in
the same direction as the price changes.
Example 1: given the demand equation
𝑃𝑑 = 100 − 0.5𝑄𝑑
determine the elasticity at 𝑄𝑑 = 20 and also
determine by what percentage and in what
direction quantity will change when price is
increased by 10%.
Plan: find the elasticity using
𝜀𝑑 = 𝛥𝑄
𝛥𝑃 ×
𝑃
𝑄
45
Then use the other definition of elasticity to
determine the %∆𝑄
𝜀𝑑 =%𝛥𝑄
%𝛥𝑃
Solution: isolate 𝑄 in the demand equation:
𝑄 = 200 − 2𝑃
The gradient is −2 so substitute this into the first
elasticity equation:
𝜀𝑑 = −2 × 𝑃
𝑄
Since 𝑄 = 20, 𝑃𝑑 is found using the demand
equation:
𝑃𝑑 𝑄 = 20 = 100 − 0.5 20 = 90
𝜀𝑑 = −2 90
20 = −9
Now substitute the 10% increase in price and the
elasticity into the other elasticity equation:
−9 =%∆𝑄
10
%∆𝑄 = −90
This means that as price is increased by 10%, the
quantity demanded will decrease (i.e. move in the
opposite direction due to the negative sign) by
90%.
The interpretation of the actual number of an
elasticity tells you the degree to which quantity
changes when price is changed.
Theory: to interpret elasticity, take the absolute
value of the elasticity (if it is negative, make it
positive, and if it is positive, leave it as a positive
number), then apply the following rules:
ε < 1 : INELASTIC – in this case, if the price of a
good is changed by a certain percentage, the
quantity will change by a smaller percentage.
ε = 1 : UNITARY ELASTIC – if the price of a good
is changed by a certain percentage, the
quantity will change by that same percentage.
ε > 1 : ELASTIC – if the price of a good is
changed by a certain percentage, the quantity
will change by a larger percentage.
The James Bond question from the previous
section had an elasticity of demand 𝜀𝑑 ≈ −0.67,
so to interpret this number, take the absolute
value of this number (i.e. giving 0.67). Since
0.67 < 1, it means that this good is inelastic to
price changes. That is, if price of the motorboats
changed by a certain percentage, the percentage
change in sales would not be as large (i.e. only
0.67 as large)
In the laptop example, the elasticity of supply 𝜀𝑠
was approximately 1.11, which means that the
supply of laptops is elastic. Specifically, if price
changed by a certain percentage, the producer
would respond with a 1.11 times larger
percentage change.
Example 2: the demand function for cigarettes is
(with quantity in millions):
𝑃𝑑 = −40𝑄𝑑 − 400
and the price of cigarettes is $80/carton. Find and
interpret the elasticity of demand for cigarettes.
Plan: use the elasticity equation
𝜀𝑑 = 𝛥𝑄
𝛥𝑃 ×
𝑃
𝑄
and the demand function to find the elasticity.
Solution: 𝑄 needs to be isolated in the demand
equation so the gradient can be determined:
𝑃 + 400 = −40𝑄𝑑
𝑃 + 400
−40= 𝑄𝑑
𝑄𝑑 = −1
40𝑃 + 100
The gradient observed is −1/40. Substitute this
into the elasticity formula:
𝜀𝑑 = −1
40 ×
𝑃
𝑄
46
Given 𝑃 = 80, to get 𝑄 substitute this into the
rearranged demand function:
𝑄𝑑(80) = −1
40(80) + 100
𝑄𝑑 80 = 98
So 98million cartons are sold. The elasticity is:
𝜀𝑑 = −1
40 ×
80
98
𝜀𝑑 = −80
3920= −
1
49≈ −0.02
Taking the absolute value, |𝜀𝑑 | = 0.02 which
means it is (highly) inelastic. For any price changes,
the change in quantity will be significantly less
(only 0.02 of the change in price).
A highly inelastic product like cigarettes means
prices can be changed by a large percentage
without the loss of many sales. In reality, this is the
case; smoking is addictive so when price increases,
people still need to smoke, but maybe won’t buy
as many cartons.
Example 3: a company selling beer conducted
research and found that elasticity of demand was
−1.2. Interpret this number. If the company
increased price by 15%, how much would quantity
demanded change?
Plan: interpret the elasticity from the theory. Use
the original definition of elasticity to work out by
what percentage quantity will change:
𝜀𝑑 =%𝛥𝑄
%𝛥𝑃
Solution: the absolute value of 𝜀𝑑 is 1.2, and since
1.2 > 1, then beer is an elastic good. This means
that if price changes by a certain percentage,
quantity demanded will change by a larger
amount.
Now, using the original definition of elasticity, and
the fact that 𝜀𝑑 = −1.2 and %𝛥𝑃 = 15%,
substitute into the original definition:
−1.2 =%𝛥𝑄
15%
Rearrange to isolate %𝛥𝑄:
−1.2 × 15% = %𝛥𝑄
%∆𝑄 = −18%
So the percentage change in 𝑄 will be larger than
the percentage change in 𝑃, but in the opposite
direction (as it is the elasticity of demand).
Exercises: 1. Determine and interpret the elasticity for the
following functions: 𝑎) 𝑄𝑠 = 1.6𝑃𝑠 − 27 𝑎𝑡 𝑄 = 27
𝑏) 𝑃𝑑 = 49 − 0.7𝑄𝑑 𝑎𝑡 𝑄 = 40
𝑐) 𝑄𝑑 = −2.7𝑃𝑑 + 111 𝑎𝑡 𝑄 = 100
𝑑) 𝑃𝑠 = 50 + 0.1𝑄𝑠 𝑎𝑡 𝑄 = 50
𝑒) 𝑄𝑠 = 1.7𝑃𝑠 − 104 𝑎𝑡 𝑃 = 20
2. Interpret the following elasticities, and determine the effect on quantity of a 10% price decrease. 𝑎) 𝜀𝑠 = 1.19
𝑏) 𝜀𝑑 = −2.3
𝑐) 𝜀𝑠 = 0.43
𝑑) 𝜀𝑑 = 1.0
𝑒) 𝜀𝑑 = 0.97
chapter two summary
Linear equations have the general form: 𝑦 = 𝑚𝑥 + 𝑐
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑚 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛=
𝑦2 − 𝑦1
𝑥2 − 𝑥1
The gradient of a line is the same at all points
along it. 1. steeper lines have larger (absolute) gradients
(𝑚 values). 2. lines cutting the 𝑦 − 𝑎𝑥𝑖𝑠 higher up have
larger 𝑐 value; the 𝑐 value shifts the line up or down.
Coordinates are written as (𝑥, 𝑦). To be able to draw a line, you need at least two points (sets of coordinates).
To find: 1. the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, replace all values of 𝑦 with
zero, then solve for 𝑥. 2. the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡, replace all values of 𝑥 with
zero, then solve for 𝑦.
Linear equations with different gradient intersect only once and there is only one set of coordinates that is similar on both lines.
47
DEMAND: this is what you do as a consumer. You demand goods and services. A demand curve slopes downwards. SUPPLY: this is what a firm does. A supply curve slopes upwards. Demand and supply can only exist when quantity and price are non-negative.
Percentage changes are found:
%∆𝑃 =∆𝑃
𝑃× 100 =
𝑐𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑖𝑐𝑒
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑝𝑟𝑖𝑐𝑒× 100
Elasticity of demand is the responsiveness of quantity demanded to a change in price.
𝜀𝑑 =%𝛥𝑄
%𝛥𝑃=
𝛥𝑄
𝛥𝑃 ×
𝑃
𝑄
The 𝛥𝑄 and 𝛥𝑃 are symbols which are different from 𝑄 and 𝑃. The 𝛥𝑄 and 𝛥𝑃 mean change in 𝑄 and change in 𝑃, respectively.
The negative sign of the elasticity of demand means that as prices changes in one direction (either a positive or negative change), then the quantity demanded will move in the opposite direction. The positive value of the elasticity of supply means that as prices change, the quantity will change in the same direction as the price change.
To interpret elasticity: take the absolute value of the elasticity, then apply the following rules:
ε < 1 : INELASTIC
ε = 1 : UNITARY ELASTIC
ε > 1 : ELASTIC
chapter two questions
1. Rearrange the following linear equations into the general form: 𝑎) 𝑦 − 4 = 𝑥 − 5 𝑏) 2𝑦 + 𝑥 = 3
𝑐) 2𝑦 − 𝑥 = 𝑥 + 3 − 1 𝑑) 2𝑥 − 𝑦 = 4
𝑒) 14𝑥 −2𝑦
5= 𝑦 − 3 𝑓)
2𝑦 + 1
4= 𝑥
𝑔) 𝑦
−5+
𝑥
3= 4 )
𝑦 − 𝑥
𝑥 − 3= 4
𝑖) 3𝑦 − 4𝑥
3 − 𝑥= 14 − 5
2. For the following graph:
Determine: a) The order of the gradients 𝑚 from lowest to
highest. b) The order of the steepness (the absolute value
of the gradients) from shallowest to steepest. c) The order of the values of 𝑐 from lowest to
highest. 3. Determine the gradients of each of the following
linear functions, then plot them: 𝑎) 𝑦 = 4𝑥 + 1
𝑏) 2𝑦 = 6𝑥 − 4
𝑐) 3𝑦 − 1 = 4𝑥 + 1
𝑑) 𝑦 − 𝑥
3= 2
𝑒) 2𝑥 − 3𝑦 =3
2
𝑓) 12𝑦 − 18𝑥 − 16 = 4 + 2𝑥
𝑔) 𝑦 + 3𝑥
2𝑦 − 3= 4
) 𝑦 = 4
4. For each of the following graphs, determine the equation of the line: 𝑎)
𝑏)
5. Determine the equation of the line passing through
each of the two points: 𝑎) 1,3 𝑎𝑛𝑑 3,15 𝑏) 2,1 𝑎𝑛𝑑 5,16 𝑐) 0,0 𝑎𝑛𝑑 3,4 𝑑) −1,7 𝑎𝑛𝑑 7, −1 𝑒) −5, −2 𝑎𝑛𝑑 3, −12
6. Determine the equation of the line passing through the point indicated, with the corresponding gradient: 𝑎) 2,3 𝑚 = 2
𝑏) −1, −1 𝑚 = 3
𝑐) 7,7 𝑚 = 1
𝑑) −3, −6 𝑚 = −2
𝑒) 1.4, −3.1 𝑚 = 2.5
(−2,3)
2, −9
𝑥
𝑦
(0, −6)
5,0
𝑥 𝑦
𝐵
𝐴
𝐶
𝑥
𝑦 𝐷 ??? ???
𝐹
48
7. Determine the coordinates where the following lines intersect: 𝑎) 𝑦 = 3𝑥 + 9 𝑦 = 2𝑥 + 3
𝑏) 𝑦 = 4𝑥 − 9 𝑦 = 3𝑥 − 3
𝑐) 𝑦 = −5𝑥 + 7 𝑦 = 2𝑥 + 14
𝑑) 2𝑦 + 3𝑥 − 26 = 0 5𝑦 + 4𝑥 = −51
𝑒) 𝑦 + 4.5𝑥 = 44.75 5.5𝑥 + 2𝑦 = 58
8. Given the following supply and demand functions, determine the equilibrium quantity and price.
𝑃𝑑 = −0.5𝑄𝑑 + 89 𝑃𝑠 = 2𝑄𝑠 + 44
Plot the two functions, and include the axis intercepts and the equilibrium.
9. Given the demand equation: 𝑃𝑑 = −0.4𝑄𝑑 + 24
And the supply equation: 𝑃𝑠 = 3.1𝑄𝑠 + 6.5
Determine the equilibrium price and quantity, and plot the market on a set of axes.
10. For the demand equation: 𝑃𝑑 = −0.6𝑄𝑑 + 89
Determine the elasticity at 𝑄 = 45, and interpret its meaning.
11. Determine the elasticity of supply at 𝑄 = 12 for the supply equation:
𝑃𝑠 = 1.75𝑄𝑠 + 4.8 12. A car manufacturer estimates its demand curve to
be (in hundreds of thousands): 𝑃𝑑 = −0.2𝑄𝑑 + 23.06
And its supply curve is estimated to be: 𝑃𝑠 = 3.1𝑄𝑠 + 15.58
Determine: a) The equilibrium quantity and price. b) The elasticity of demand at the equilibrium,
and its meaning. c) The elasticity of supply at the equilibrium, and
its meaning. 13. The market for steel has an estimated demand and
supply curve (in millions of tonnes) to be: 𝑃𝑑 = −1.7𝑄𝑑 + 26.48
𝑃𝑠 = −1
3𝑄𝑠 + 21.6
Determine: a) The equilibrium quantity and price. b) The elasticity of demand at the equilibrium,
and its meaning. c) The elasticity of supply at the equilibrium, and
its meaning.
49
Chapter 3
Simultaneous Equations and Matrices Solving the intersection of lines using different methods
3.1 Simultaneous Equations 50
3.2 Two Simultaneous Equations 50
3.3 Three Simultaneous Equations 52
3.4 The Matrix 54
3.5 Solving Two Equation Matrices 55
3.6 Solving Three Equation Matrices 58
3.7 Notes on Solutions to Matrices 61
3.8 Applications 62
3.9 The Determinant of a 2 × 2 Matrix 65
3.10 The Determinant of a 3 × 3 Matrix 66
3.11 Using the Jacobian Determinant 68
Chapter Three Summary 70
Chapter Three Questions 71
50
3.1 simultaneous equations
In Chapter 2, the intersection of two lines was
found mathematically. Essentially, simultaneous
equations were used to do this.
Theory: simultaneous equations are used to solve
for the variables when there are two or more
equations. It involves finding the common
intersection of multiple lines.
Given the two equations:
𝑦 = 2𝑥 + 5
𝑦 = −3𝑥 + 6
Simultaneous equations can be used to solve for 𝑥
and 𝑦. The and are notes to distinguish the
two different equations.
Theory: to be able to solve simultaneous
equations, there needs to be at least as many
equations as there are variables (this is not the
only condition).
Remember that 𝑥 and 𝑦 are two different
variables, so there is no reason why more variables
cannot be introduced. One such equation could
be:
𝑧 = 4𝑥 + 3𝑦 − 1
Here, z is the dependent variable, and x and y are
the independent variables.
Can the following equations be solved?
𝑧 = 4𝑥 + 3𝑦 − 1
𝑧 = −2𝑥 + 2𝑦 + 4
The answer is no, as according to the theory
above, there are more variables (3) than there are
equations (2), so there will be no unique solution.
In the following case, there are three equations
(, and ) and three variables (𝑥, 𝑦 and 𝑧).
𝑧 = 4𝑥 + 3𝑦 − 1
𝑧 = −2𝑥 + 2𝑦 + 4
𝑧 = 3𝑥 − 3𝑦 − 3
Thus a solution is possible (Note: see Section 3.7).
Simultaneous equations are simply a certain
number of equations to help solve for a certain
number of variables.
However, sometimes a set of lines do not all cross
at a single point, and in such a case, there is no
unique solution. Similarly, sometimes sets of
equations cross at an infinite number of points, so
there are an infinite number of solutions. This is
explained in detail in Section 3.7.
Exercises: 1. Determine if the following sets of equations can be
solved: 𝑎) 𝑦 = 2𝑥 + 3
𝑦 = −3𝑥 − 14
𝑏) 𝑦 = 𝑥 + 𝑧 − 3
𝑦 = −2 − 𝑥 − 𝑧
𝑦 − 𝑧 = 15
𝑐) 𝑦 = 𝑧 + 𝑥 − 1
𝑦 = 12𝑥 − 3
𝑦 = 𝑧 + 𝑥 + 𝑎 − 3
3.2 two simultaneous equations
Theory: to solve simultaneous equations:
1. set the 𝑦’s equal to each other.
2. substitute the functions of 𝑥 in and solve for 𝑥.
3. substitute the solutions of 𝑥 back into either of
the original equations to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of
the solution.
Note: you could also set the 𝑥’s equal to each in
step 1, and substitute for the functions of 𝑦. This
will give the same solution.
Example 1: solve for 𝑥 and 𝑦 in
𝑦1 = 2𝑥 + 1
𝑦2 = −𝑥 + 4
Plan: set 𝑦1 = 𝑦2 and substitute the equations,
then solve for 𝑥. Substitute back into one of the
51
original equations to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 at the
intersection.
Solution:
𝑦1 = 𝑦2
2𝑥 + 1 = −𝑥 + 4
3𝑥 = 3
𝑥 = 1
So 𝑥 = 1, but for a coordinate, a 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is still
needed. To find this 𝑦 − 𝑣𝑎𝑙𝑢𝑒, substitute the
𝑥 − 𝑣𝑎𝑙𝑢𝑒 back into either of the original
equations (both will give the same answer).
Using equation :
𝑦 = − 1 + 4
= 3
So the solution to the two equation is 𝑥 = 1, 𝑦 = 3
or in coordinate form: (1,3).
Try substituting the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 into equation to
see if the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is the same (it should be).
Theory: another method of solving simultaneous
equations is:
1. isolate one variable in one equation
2. substitute it into the other equation.
3. solve for the other variable,
4. substitute this solution back into either of the
original equations to solve for the other
variable.
Example 2: solve for 𝑥 and 𝑦 in
2𝑦 + 2𝑥 − 1 = 7
5𝑦 − 3𝑥 + 4 = 0
Plan: isolate one variable in one of the equations,
and substitute it into the other equation to solve
for the other variable. Substitute this solution into
any of the modified or original equations to solve
for the other variable.
Solution: choose as the numbers in front of the
variables are even and look easy to solve
compared to . Now, choose the variable easiest
to isolate. In the following, 𝑦 is isolated however 𝑥
is just as easy.
2𝑦 + 2𝑥 − 1 = 7
2𝑦 + 2𝑥 = 8
2𝑦 = 8 − 2𝑥
𝑦 = 4 − 𝑥 ’
This rearranged version is labelled ’. Substitute
(4 − 𝑥) in for 𝑦 in the other equation.
’ → (this means the modified equation 1
(labelled ’) is substituted into equation )
5 4 − 𝑥 − 3𝑥 + 4 = 0
Use the crab-claw from Chapter 1:
20 − 5𝑥 − 3𝑥 + 4 = 0
24 − 8𝑥 = 0
𝑥 = 3
To find the value of 𝑦, substitute 𝑥 = 3 into either
, ’ or .
’ is easiest as 𝑦 is already isolated:
𝑦 = −𝑥 + 4
𝑦 = − 3 + 4
𝑦 = 1
Thus the solution is 𝑥 = 3, 𝑦 = 1, or in coordinates
(3,1).
Go back to the beginning of this example, and try
to isolate 𝑥 in . Also, try to isolate 𝑥 or 𝑦 in . All
these should give the same solution.
The hardest part using of this method is knowing
which equation is easier to use and this will take
practice. However, even if you choose the harder
one, the answer will still be the same. It is also a
good idea to substitute the solution you get back
into both original equations, to double check that
you have the correct solution.
Exercises: 1. Solve the following sets of equations 𝑎) 7𝑦 = −5𝑥 + 9 𝐴𝑁𝐷 5𝑦 = −4𝑥 + 15
52
𝑏) 2𝑦 = −3𝑥 + 8 𝐴𝑁𝐷 4𝑥 + 𝑦 = 9
𝑐) 5𝑥 + 9𝑦 = 15 𝐴𝑁𝐷 6𝑦 + 4𝑥 = 15
𝑑) −6𝑥 + 7𝑦 = 15 𝐴𝑁𝐷 6𝑦 = 3𝑥 + 12
𝑒) 2𝑥 = −7𝑦 + 13 𝐴𝑁𝐷 − 4𝑥 − 5𝑦 − 17 = 0
3.3 three simultaneous equations
A similar theory applies to three equation systems.
A simple example will show you the technique
before the theory is introduced.
Example 1: Solve for 𝑥, 𝑦 and 𝑧 in
𝑧 = 2𝑥 + 3𝑦 + 6
𝑧 = 𝑥 + 𝑦 + 4
𝑧 = −2𝑥 + 5𝑦 + 8
Plan: use the substitution method to solve one
variable at a time.
Solution: To solve for three equations; first, take
one of the equations, and substitute it into the
other two equations (below, is put into both
and ):
→:
𝑧1 = 𝑧2
2𝑥 + 3𝑦 + 6 = 𝑥 + 𝑦 + 4
𝑥 + 2𝑦 = −2 ’
→:
𝑧1 = 𝑧3
2𝑥 + 3𝑦 + 6 = −2𝑥 + 5𝑦 + 8
4𝑥 − 2𝑦 = 2 ’
The two modified equations ’ and ’ are now
two simultaneous equations with two variables
(like in the last section). is no longer used.
Isolate 𝑥 in ’:
𝑥 + 2𝑦 = −2 ’
𝑥 = −2𝑦 − 2 ’’
Substitute ’’→’:
4 −2𝑦 − 2 − 2𝑦 = 2
−8𝑦 − 8 − 2𝑦 = 2
−10𝑦 = 10
𝑦 = −1
This is the solution to one variable. Substitute this
value of 𝑦 into an equation with only 𝑥 and 𝑦; use
either ’ or ’’ (’’ is easier as 𝑥 is already
isolated):
𝑥 = −2𝑦 − 2 ’’
𝑥 = −2(−1) − 2
𝑥 = 0
With two variables solved, the last variable can
now be solved (i.e. 𝑧) using :
𝑧 = 2𝑥 + 3𝑦 + 6
𝑧 = 2(0) + 3(−1) + 6
𝑧 = 3
So the solution is 𝑥 = 0, 𝑦 = −1, 𝑧 = 3, or in three
dimensional coordinates 𝑥, 𝑦, 𝑧 = 0, −1,3 .
Theory: to solve three simultaneous equations:
1. take one equation and substitute it into the
other two equations. This will make two new
equations with only two variables.
2. substitute one of those equations into the
other to solve for one of the unknowns.
3. use this solution to solve for one of the other
variables.
4. lastly, use these two solutions to solve for the
third variable.
Note: the hardest part is organising and
manipulating the three equations. The solution is
also written in coordinate form 𝑥, 𝑦, 𝑧 .
Example 2: Solve for 𝑥, 𝑦 and 𝑧 in
𝑧 − 3𝑥 + 8 = 4𝑦
4𝑧 = −𝑥 + 7𝑦 − 1
4𝑥 − 2𝑦 = 3𝑧 − 9
Plan: substitute one equation into the other two,
then use those two modified equation to solve for
one of the variables. Work backwards to solve for
the other variables.
53
Solution: 𝑧 in looks easiest to isolate
𝑧 − 3𝑥 + 8 = 4𝑦
𝑧 = 4𝑦 + 3𝑥 − 8 ’
Correctly label this as ’ as it has been slightly
modified. Then put ’ into both and :
’→:
4(4𝑦 + 3𝑥 − 8) = −𝑥 + 7𝑦 − 1
16𝑦 + 12𝑥 − 32 = −𝑥 + 7𝑦 − 1
9𝑦 + 13𝑥 = 31 ’
Again, this equation is labelled as ’ as it has been
modified.
’→:
4𝑥 − 2𝑦 = 3(4𝑦 + 3𝑥 − 8) − 9
4𝑥 − 2𝑦 = 12𝑦 + 9𝑥 − 24 − 9
−5𝑥 − 14𝑦 = −33 ’
Again, notice the labelling.
Working with only ’ and ’ isolate any variable
in either ’ or ’. Both are difficult, so use either.
Equation ’ has been used below:
−5𝑥 − 14𝑦 = −33
−5𝑥 = −33 + 14𝑦
𝑥 =−33 + 14𝑦
−5
𝑥 =33
5−
14
5𝑦 ’’
Again, this equation is labelled appropriately.
This equation (’’) is substituted into the other
equation (i.e. ’ ).
’’→’:
9𝑦 + 13(33
5−
14
5𝑦) = 31
9𝑦 +429
5−
182
5𝑦 = 31
−137
5𝑦 = −
274
5
𝑦 = −
2745
−137
5
= 2
This solution for 𝑦 is used to find the other two
unknowns. Substitute 𝑦 = 2 into either of the
equations with only two variables (i.e. ’’ or ’).
Substitute 𝑦 = 2 →’’:
𝑥 =33
5−
14
5 2
𝑥 =33
5−
28
5
𝑥 =5
5= 1
Finally, use any of the equations with all three
variables to solve for 𝑧. The easiest is ’ as 𝑧 is
already isolated.
Substitute 𝑥 = 1, 𝑦 = 2→’
𝑧 = 4𝑦 + 3𝑥 − 8
𝑧 = 4(2) + 3(1) − 8
𝑧 = 3
The solution is 𝑥 = 1, 𝑦 = 2, 𝑧 = 3, or 1,2,3 .
Solutions will not always be nice round numbers,
so it is a good habit to always work with fractions,
as then you will always have an exact answer.
Exercises: 1. Solve the following sets of equations: 𝑎) 𝑧 = 4𝑥 + 𝑦 − 6
2𝑧 = −𝑥 + 2𝑦 + 6
𝑧 = −𝑥 + 2𝑦 + 2
𝑏) 5𝑥 − 𝑦 − 𝑧 = 13
2𝑥 − 2𝑦 + 4𝑧 = 8
𝑥 − 3𝑦 + 2𝑧 = 2
𝑐) −𝑧 + 2𝑦 = −3𝑥 + 12
2𝑥 = 5𝑦 + 2𝑧 + 7
−𝑥 + 2𝑦 − 5𝑧 − 20 = 0
𝑑) 2𝑦 + 5𝑧 = 4 + 2𝑥
4𝑥 + 4𝑦 + 6𝑧 = 10
4𝑦 = 2𝑥 + 2𝑧 + 8
𝑒) 5𝑧 = −7𝑦 − 6𝑥 + 8
5𝑦 = −2𝑥 − 8𝑧 + 12
2𝑥 + 2𝑦 + 2𝑧 = 16
54
3.4 the matrix
The substitution method for solving equations was
quite easy for two equations, but was a lot harder
for three equations. There is a short–cut method
for solving simultaneous equations and it involves
using matrices.
To understand what a matrix is and how it is
created, equations need to be in a particular order.
Theory: every equation has to be in the general
matrix form:
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑
Where 𝑎, 𝑏, 𝑐 and 𝑑 are constants. Essentially, this
form has all the variables on the left in a consistent
order (e.g. alphabetical), and any numbers without
variables (i.e. solitary constants) on the right side.
For the equation
𝑧 − 3𝑦 + 4 = 5𝑥
To get it into the general matrix form, rearrange it:
−5𝑥 − 3𝑦 + 𝑧 = −4
Here, 𝑎 = −5, 𝑏 = −3, 𝑐 = 1 and 𝑑 = −4. The
order of the variables is alphabetical order, as it is
easy to remember.
Example 1: rearrange the following equations into
the general matrix form
𝑧 = 2𝑥 − 3𝑦 + 9
3𝑧 − 4 = −4𝑥 + 2𝑦
−𝑦 + 𝑧 = 𝑥 − 2
Plan: rearrange to get variables in alphabetical
order on the left, and any numbers without
variables (constants) on the right.
Solution: use alphabetical order
: 𝑧 = 2𝑥 − 3𝑦 + 9
−2𝑥 + 3𝑦 + 𝑧 = 9 ’
: 3𝑧 − 4 = 4𝑥 + 2𝑦
−4𝑥 − 2𝑦 + 3𝑧 = 4 ’
: −𝑦 + 𝑧 = 𝑥 − 2
−𝑥 − 𝑦 + 𝑧 = −2 ’
Rewrite the original equations into a new set:
−2𝑥 + 3𝑦 + 𝑧 = 9 ’
−4𝑥 − 2𝑦 + 3𝑧 = 4 ’
−𝑥 − 𝑦 + 𝑧 = −2 ’
It is from this new set that a matrix can be
constructed.
Theory: a matrix is a set of square brackets
summarising equations in a simpler form.
To construct a matrix:
1. take the numbers in front of the variables and
put them into square brackets.
2. draw a new set of brackets and vertically write
the order of the variables you have chosen
(e.g. alphabetical).
3. after writing an equals sign, the constants on
the right side are written vertically in square
brackets.
Example 2: write the following equations in matrix
form
−2𝑥 + 3𝑦 + 𝑧 = 9 ’
−4𝑥 − 2𝑦 + 3𝑧 = 4 ’
−𝑥 − 𝑦 + 𝑧 = −2 ’
Solution:
−2 3 1−4 −2 3−1 −1 1
𝑥𝑦𝑧 =
94
−2
Look at the left matrix. The first column is the 𝑥
column, the middle column is the 𝑦 column and
the right column is the 𝑧 column. The next matrix
has the 𝑥, 𝑦 and 𝑧 so that the variables are
defined. Then there is the equals sign, with the
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 from the equations as another matrix.
To understand how this system works, take your
left index finger and put it onto the −2 on the top
left of the left matrix, then put your right index
55
finger on the 𝑥 in the middle matrix. What you are
pointing to is the number in front of the 𝑥 in the
first equation (look back up at ’, it is −2𝑥). Now
move your left hand across one entry, and your
right hand down one entry. You should be on 3
with your left hand, and 𝑦 with your right, just like
in ’). Move your left finger across one more
place and your right finger down one place. You
should have the last entry of 1𝑧 (or just 𝑧). Going
across these three numbers gives the left side of
’, and this is equal to 9, which is the top entry of
the last matrix.
If you go back to the beginning and put your left
index finger at the start of the second row (i.e. the
number 4), your right index finger back on 𝑥 and
repeat the process, you will get equation ’.
Example 3: find the original equations for the
matrix
−1 −1 13 7 3
−3 0 2
𝑥𝑦𝑧 =
1152
Plan: go across each row of the left matrix and
multiply by each entry of the variables column.
Add these together and make this equal to the
entry in the respective row of the right matrix.
Solution:
−𝑥 − 𝑦 + 𝑧 = 11
3𝑥 + 7𝑦 + 3𝑧 = 5
−3𝑥 + 2𝑧 = 2
This is how matrices work. Once you are
comfortable with moving between equations and
matrices, you will find there is an easier way of
writing all the information in a matrix:
−2 3 1 ⋮4 −2 3 ⋮
−1 −1 1 ⋮
94
−2
In this form, the first column is the 𝑥 column, the
second column the 𝑦 column and the third column
the 𝑧 column. The 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 are separated from
𝑥, 𝑦 and 𝑧 by a dotted line. A good habit is to write
what the columns represent with a different
colour:
𝑥 𝑦 𝑧 𝑐𝑜𝑛𝑠𝑡
−2 3 1 ⋮4 −2 3 ⋮
−1 −1 1 ⋮
94
−2
You have to understand how a matrix relates to a
set of equations before you can solve that system.
Exercises: 1. Put the following sets of equations into a matrix: 𝑎) −3𝑥 + 5𝑦 − 12𝑧 − 12 = 4
5𝑥 + 2𝑦 − 3 = 15
−2𝑥 + 5𝑦 − 3𝑧 = 3𝑧 + 1
𝑏) 12𝑥 − 11𝑦 + 10𝑧 = 9
7𝑧 − 3𝑥 + 2𝑦 = 7𝑦 − 1
2𝑥 + 3𝑥 − 4𝑥 + 𝑦 = 𝑥 − 14
𝑐) −2𝑥 + 4𝑦 = 14
7𝑥 + 7𝑧 = −7
9𝑧 − 𝑦 + 5 = 0
𝑑) 𝑧 − 12𝑥 − 5𝑦 = 3
𝑥 = 3𝑥 − 2𝑦 + 4
13𝑥 = 12𝑦 − 4
2. Extract the equations from the following matrices.
𝑎) −2 1 −35 2 64 7 −1
⋮ ⋮ ⋮
51411
𝑐) 6 7 153 −2 02 12 2
⋮ ⋮ ⋮
−2−35
𝑏) 4 8 3
−5 12 39 −4 0
⋮ ⋮ ⋮
5−70
𝑑) 5 8 0
−5 0 −41 2 1
⋮ ⋮ ⋮
522
3.5 solving two equation matrices
Having a set of equations in matrix form allows
you to solve for 𝑥, 𝑦 and 𝑧 in an easier way than
the substitution method.
Intro example 1: solve the following equations
using matrices
2𝑦 = 3𝑥 + 5
0 = −𝑥 + 10 − 𝑦
First, these equations must be rearranged into the
general matrix form:
: 2𝑦 = 3𝑥 + 5
−3𝑥 + 2𝑦 = 5 ’
: 0 = −𝑥 + 10 − 𝑦
56
𝑥 + 𝑦 = 10 ’
Equations ’ and ’ allow construction of the
matrix:
The idea now is to change the second row to get
the number zero in the bottom left corner (circled
in blue). The reason is that an equation made from
that row, would be of the form:
0𝑥 + 𝑏𝑦 = 𝑐
Where 𝑏 and 𝑐 are constants. Then 0𝑥 is just zero,
leaving:
𝑏𝑦 = 𝑐
Which can then be solved for 𝑦.
However, the circled number cannot simply be
erased and replaced by 0, as this would change the
equation. What must be done is use something
called row operations to manipulate that bottom
row to get a zero in the bottom left corner.
Theory: A row operation is a process where a
multiple of one ROW is added/subtracted from
another ROW.
Before row operations are done on a matrix, it is
best to understand what is being done. For the
two equations below to be added to one another:
−3𝑥 + 2𝑦 = 5 ’
𝑥 + 𝑦 = 10 ’
The two left sides must be added, and the two
right sides must be added.
’ +’:
−3𝑥 + 2𝑦 + 𝑥 + 𝑦 = 5 + 10
Simplifying this gives:
−2𝑥 + 3𝑦 = 15
Back to the matrix, if the columns are added
together:
−31
21
⋮ ⋮
5
10
−2 3 15
The results are in red, which is the same as adding
the two equations together.
Theory: ROW operations involve adding/
subtracting the COLUMNS of a matrix.
Using this theory, to get a zero in the bottom left
corner, and knowing row operations can be used
without changing the equalities, how many
ROW2’s will need to be added/subtracted to how
many ROW1’s?
Look only at the first column (the 𝑥 column). If 3
times ROW2 were added to ROW1, the answer
would be zero (as 3 × 1 + (−3) = 0). This is
similar to finding a common denominator is
Chapter 1.
This ratio cannot be only applied to one column;
this same ratio must be applied to all columns. The
easiest method of doing this is to come up with a
short sentence once you have figured out the
ratio. In the above example, the sentence would
be:
“three lots of bottom plus top”
−3 2 ⋮ 51 1 ⋮ 10
For the first column:
“three lots of one plus negative three” is
3 × 1 + (−3) = 0
For the second column:
“three lots of one plus two” is
3 × 1 + 2 = 5
For the third column:
“three lots of ten plus five” is
3 × 10 + 5 = 35
The way this is represented on a matrix is:
−3 2 ⋮ 51 1 ⋮ 10
𝑅2’ = 3𝑅2 + 𝑅1
−3 2 ⋮ 50 5 ⋮ 35
−3 2 ⋮ 51 1 ⋮ 10
57
This is pronounced: “the new ROW2 is three times
the old ROW2 plus ROW1”.
Once the zero is in the bottom left, the equations
can be extracted from this matrix:
−3𝑥 + 2𝑦 = 5
0𝑥 + 5𝑦 = 35
Then working from the bottom equation up, solve
for the unknowns:
5𝑦 = 35
𝑦 = 7
Move up to the next equation, and substitute in
𝑦 = 7:
−3𝑥 + 2(7) = 5
−3𝑥 + 14 = 5
−3𝑥 = −9
𝑥 = 3
The hardest part when using matrices is finding the
ratio to be able to get zero in the bottom left
corner.
Theory: use ROW operations to find the ratio of
ROW 1 added/subtracted from ROW 2, which will
give a zero in the bottom left corner. Apply this
ratio to all columns, then extract the equations
from the modified matrix. Solve for the unknowns,
working from the bottom equation up.
Example 1: solve for 𝑥 and 𝑦 in
−1 6 ⋮ 7−4 2 ⋮ −5
Plan: use row operations to get a zero in the
bottom left corner.
Then extract the equations from this matrix, and
solve for 𝑥 and 𝑦 from the bottom up.
Solution: how many lots of −1 's need to be
added/subtracted to how many lots of −4 's to
get zero in the bottom left? Four of the top are
required to make the bottom, but if you are not
sure whether to add or subtract, then write it out:
4 × −1 ∎ −4 = 0
−4∎ −4 = 0
If you have an addition, it will make −8 so the
black space must be a subtract sign. This is where
many students stuff up, so don’t be one of them.
Write out your saying:
“four lots of top minus bottom”
For the first column:
“four lots of negative one subtract negative four”
4 × −1 − (−4) = 0
For the second column:
“four lots of six subtract two”
4 × 6 − 2 = 22
For the third column:
“four lots of seven subtract negative five”
4 × 7 − (−5) = 33
In matrix form, this whole process would be
written:
−1 6 ⋮ 7−4 2 ⋮ −5
𝑅2’ = 4𝑅1 − 𝑅2
−1 6 ⋮0 22 ⋮
7 33
There is now a zero in the bottom left corner, so
the equations can be extracted:
−𝑥 + 6𝑦 = 7
0𝑥 + 22𝑦 = 33
Solve from the bottom up:
22𝑦 = 33
𝑦 =33
22= 1.5
Up to the next equation:
−𝑥 + 6(1.5) = 7
−𝑥 + 9 = 7
𝑥 = 2
So the solution is 𝑥 = 2, 𝑦 = 1.5 or 2,1.5 .
−1 6 ⋮ 7−4 2 ⋮ −5
58
Check if the solution is correct by substituting it
into either of the original equations (the bottom
row of the matrix, just for consistency):
−𝑥 + 6𝑦 = 7
− 2 + 6 1.5 = 7
−2 + 9 = 7
This means the solution is correct.
Example 2: use matrices to solve for 𝑥 and 𝑦
−4𝑦 = −3𝑥 + 17
6 + 9𝑦 = −2𝑥
Plan: construct a matrix from the equations, then
use row operations to get a zero in the bottom left
corner. Extract the equations, then solve for the
unknowns, one at a time.
Solution: put and into general matrix form:
3𝑥 − 4𝑦 = 17 ’
2𝑥 + 9𝑦 = −6 ’
Construct the matrix:
3 −4 ⋮2 9 ⋮
17−6
Find the correct ratio from the first column
“two lots of top minus three lots of bottom”
Apply this to all columns:
3 −4 ⋮2 9 ⋮
17−6
𝑅2’ = 2𝑅1 − 3𝑅2
3 −4 ⋮0 −35 ⋮
1752
Write the equations from this matrix:
3𝑥 − 4𝑦 = 17
0𝑥 − 35𝑦 = 52
Working up, solve for 𝑦:
−35𝑦 = 52
𝑦 = −52
35
Then solve for 𝑥:
3𝑥 − 4 −52
35 = 17
3𝑥 +208
35= 17
3𝑥 =595
35−
208
35
3𝑥 =387
35
𝑥 =129
35
The solution is 𝑥 =129
35, 𝑦 = −
52
35
Then check that the solution is correct:
−4𝑦 = −3𝑥 + 17
−4 −52
35 = −3
129
35 + 17
208
35=
208
35
Exercises: 1. Solve the following matrices for the unknowns:
𝑎) 2 13 4
⋮ ⋮
128
𝑐) 1 −12 1
⋮ ⋮
99
𝑏) 3 −24 2
⋮ ⋮
−70
𝑑) 5 −1
−1 7 ⋮ ⋮
939
2. Solve the following equations using matrices. 𝑎) 2𝑥 − 𝑦 = −7 𝑐) 3𝑥 + 7𝑦 = 13
−4𝑥 + 4𝑦 = 8 5𝑥 + 9𝑦 = 11
𝑏) 𝑦 = −𝑥 + 8 𝑑) 9 − 3𝑦 = 5𝑥
4𝑥 = −2𝑦 − 3 −𝑥 = −3𝑦 + 13
3. A company initially sets a price for a good at 𝑃 = 15, and sells a quantity of 𝑄 = 30. When they increase price to 𝑃 = 20, the company sells 25 units. Find the equation of the demand function, then determine the equilibrium price and quantity given that supply is 𝑄𝑠 = 0.5𝑃𝑠 − 1.5.
3.6 solving 3 equation matrices
Solving three equation matrices uses row
operations, but there are some differences.
Theory: to solve a three equation system, zeros
are needed in the bottom left triangular area:
From this matrix, the bottom equation will be
0𝑥 + 0𝑦 + 𝑎𝑧 = 𝑏
This will have some multiple of 𝑧 equal to a
constant. The middle row will have 𝑧 and 𝑦, so 𝑦
can be solved as 𝑧 will already be known, and the
top row will have all three variables, so 𝑥 can be
found. This is the reason why zeros are needed in
𝑋 𝑋 𝑋 ⋮0 𝑋 𝑋 ⋮0 0 𝑋 ⋮
𝑋 𝑋 𝑋
59
the bottom left triangular area; working from the
bottom up, we can solve for 𝑥, 𝑦 and 𝑧.
However, these zeros need to be found in a
particular order.
Theory: the two bottom entries of the first column
must be changed to zero first, and then the
bottom entry of the second column can be
changed to zero.
The reason for this will soon emerge.
Intro example 1: solve for the unknowns in the
matrix
Firstly, both the 6 and 2 in the blue box must be
changed to zeros. For now, they will be done one
at a time, but later you can do them at the same
time. To get zeros in the blue box, ROW1 must be
used.
To change the 6 into a zero, work with ROW1 and
ROW2:
Find the ratio by asking: for the new ROW2, how
many ROW1’s will need to be added/subtracted
from how many ROW2’s? The short sentence is:
“two tops minus bottom”
3 −1 16 2 32 2 3
⋮⋮⋮
8 10 18
𝑅2′ = 2𝑅1 − 𝑅2
Remember that the “two tops minus bottom”
applies to every entry in the second row.
To change the number 2 to a zero in the third row
(i.e. the bottom left corner), ROW1 and ROW3
must be used, ignoring ROW2. How many ROW1’s
will need to be added/subtracted from how many
ROW3’s? The short sentence is:
“two tops minus three bottoms”
3 −1 10 −4 −12 2 3
⋮ ⋮ ⋮
86
18
𝑅3′ = 2𝑅1 − 3𝑅3
Now that the circled entries are zeros, the next
requirement is to change the −8 (highlighted) into
a zero. If the top row is used, it would ruin the zero
in the bottom left corner, as the number 3 from
the first row would be introduced again. However,
if ROW3 is changed using ROW2, then the blue
zeros will remain unchanged (as any ratio of two
zeros added or subtracted is still zero!).
To get a zero in place of the highlighted −8 , ask:
how many ROW2’s will you have to add/subtract
from ROW3? Two lots of ROW2 are needed, but
because of the negative signs, you may not be sure
whether to add or subtract. So write it out:
2 −4 ∎ −8 = 0
−8∎ −8 = 0
The sign can only be a subtraction, so:
“two tops minus bottom”
In matrix form:
3 −1 10 −4 −10 −8 −7
⋮ ⋮ ⋮
86
−38
𝑅3′ = 2𝑅2 − 𝑅3
3 −1 10 −4 −10 0 5
⋮ ⋮ ⋮
86
50
The matrix is reduced into the form that is
required, so 𝑥, 𝑦 and 𝑧 can now be solved. Extract
the equations from this matrix:
3𝑥 − 𝑦 + 𝑧 = 8
−4𝑦 − 𝑧 = 6
5𝑧 = 50
Work from the bottom equation up:
𝑧 =50
5= 10
Substitute into the next equation up:
−4𝑦 − 10 = 6
3 −1 10 −4 −10 −8 −7
⋮ ⋮ ⋮
86
−38
3 −1 10 −4 −12 2 3
⋮ ⋮ ⋮
86
18
3 −1 1 ⋮6 2 3 ⋮2 2 3 ⋮
8 10 18
60
−4𝑦 = 16
𝑦 = −4
Next equation up:
3𝑥 − −4 + (10) = 8
3𝑥 + 14 = 8
𝑥 = −2
The solution is 𝑥 = −2, 𝑦 = −4 and 𝑧 = 10, or
−2, −4,10 .
Finally, to check if the solution is correct,
substitute it back into the bottom row of the
original matrix:
2𝑥 + 2𝑦 + 3𝑧 = 18
2 −2 + 2(−4) + 3(10) = 18
−4 − 8 + 30 = 18
Theory: to solve a system of three equations, row
operations need to be used to get zeros in the
bottom left triangle corner of the variables matrix,
getting zeros in the left-most columns first.
Example 1: solve for 𝑥, 𝑦 and 𝑧 in the matrix
2 −1 −13 −1 45 −3 −5
⋮ ⋮ ⋮
17−11
9
Plan: use row operations to get zeros in the first
column where 3 and 5 are located using
ROW1. Then using the modified ROW2, get a zero
in the bottom entry of the second column.
Rewrite the equations from the final matrix then
work up to solve for 𝑥, 𝑦 and 𝑧.
Solution:
For ROW2: “three lots of top minus two lots of
bottom”
For ROW3: “five lots of top minus two lots of
bottom”
2 −1 −13 −1 45 −3 −5
⋮ ⋮ ⋮
17−11
9 𝑅2’ = 3𝑅1 − 2𝑅2
𝑅3′ = 5𝑅1 − 2𝑅3
2 −1 −10 −1 −110 1 5
⋮ ⋮ ⋮
177367
To get a zero in the bottom entry of the second
column:
“one lot of top plus bottom”
2 −1 −10 −1 −110 1 5
⋮ ⋮ ⋮
177367
𝑅3’ = 𝑅2 + 𝑅1
2 −1 −10 −1 −110 0 −6
⋮ ⋮ ⋮
1773
140
Rewrite the equation from this matrix:
2𝑥 − 𝑦 − 𝑧 = 17
−𝑦 − 11𝑧 = 73
−6𝑧 = 140
Solve for 𝑧 using the bottom equation:
𝑧 = −140
6= −
70
3
Solve for 𝑦 using the next equation up:
−𝑦 − 11 −70
3 = 73
−𝑦 = 73 −770
3
−𝑦 =219
3−
770
3
𝑦 =551
3
Solve for 𝑥 using the next equation up:
2𝑥 − 551
3 − −
70
3 = 17
2𝑥 −481
3= 17
2𝑥 =51
3+
481
3
2𝑥 =532
3
𝑥 =266
3
The solution is: 𝑥 =266
3, 𝑦 =
551
3, 𝑧 = −
70
3 or
266
3,
551
3, −
70
3 .
Check this solution is correct:
5𝑥 − 3𝑦 − 5𝑧 = 9
5 266
3 − 3
551
3 − 5 −
70
3 = 9
61
1330
3−
1653
3+
350
3= 9
27
3= 9
Note: the notes next to any matrix, such as
𝑅3’ = 𝑅2 + 𝑅1 refer to the what is being done to
that matrix, and the solution is shown in the
following matrix. The 𝑅1, 𝑅2 etc. do not refer to
the original matrix, just the matrix the note is
written next to.
These notes are also a good way of finding any
mistakes, if the final check is found to be incorrect.
Exercises: 1. Solve the following matrices.
𝑎) 5 2 −92 2 31 3 −4
⋮ ⋮ ⋮
912
−13 𝑐)
4 −1 −27 3 43 3 −1
⋮ ⋮ ⋮
8−1025
𝑏) 2 −3 42 1 72 −3 7
⋮ ⋮ ⋮
815 𝑑)
1 −5 12 5 63 −3 14
⋮ ⋮ ⋮
15122
2. Solve the following sets of equations using matrices. 𝑎) −2𝑥 + 6𝑦 − 3𝑧 = 6
𝑥 + 3𝑦 − 𝑧 = 17
4𝑥 + 2𝑦 + 4𝑧 = 4
𝑏) 6𝑥 + 4𝑦 + 6𝑧 = 2
−𝑥 + 2𝑦 − 𝑧 = 3
2𝑥 − 3𝑦 + 3𝑧 = 6
𝑐) 3𝑦 − 6 = −𝑧 + 4𝑥
2𝑧 + 𝑥 = 2𝑦 + 𝑥 + 12
6𝑥 + 𝑦 + 𝑧 + 4 = 0
𝑑) 4𝑥 = −7𝑦 − 𝑧 + 5
4𝑦 = 𝑥 + 𝑧 + 15
−2𝑥 − 2𝑧 − 10 = −2𝑦
3. A computing company makes computers at to capacity in three identical factories in three different states. Each factory uses labour 𝐿 , metal 𝑀 and silicone chips 𝑆 . Due to differences in union powers, taxation laws and property rights, the costs of 𝐿, 𝑀 and 𝑆 in these three states are not the same:
State 1: $7.20/𝐿, $18.50/𝑀 and $31.00/𝑆
State 2: $7.50/𝐿, $21.00/𝑀 and $29.00/𝑆
State 3: $9.00/𝐿, $16.50/𝑀 and $30.00/𝑆 At full capacity, the total cost for each of the three factories is:
State 1: $3373.00
State 2: $3589.00
State 3: $3564.00 Determine: a) Three expressions equating units of each of the
three inputs to total cost.
b) The amount of 𝐿, 𝑀 and 𝑆 used in each of the three identical factories.
3.7 notes on solutions to matrices
When the intersection of two lines was found back
in Chapter 2, if the two lines had the same
gradient 𝑚 but different constants 𝑐, then there
would be no intersection. Similarly, if two lines had
the same gradient 𝑚 and the same constants 𝑐
(that is, two equations for the same line), then
there would be infinite intersections (every point
on the two lines is an intersection).
Theory: since matrices are just a simpler form of
writing out a number of equations, then there can
be:
A single solution (what we have been doing in
the last two sections)
No solutions (if there is no common
intersection of all lines)
Infinite solutions (if we do not have enough
equations and/or if some of the equations are
the same)
Only after you have tried getting the zeros in the
bottom left triangle can you determine if there is
infinite, a unique, or no solutions.
The following is a matrix where there are no
solutions (notice that there are zeros in the
bottom left triangle):
2 −1 4 ⋮0 2 3 ⋮0 0 0 ⋮
−9 5 4
If the bottom row is written in equation form:
0𝑥 + 0𝑦 + 0𝑧 = 4
0 = 4
But it is impossible for 4 to be equal to zero, thus
there are no solutions.
62
An example with an infinite amount of solutions
(again, notice that the zeros in the bottom left
triangle):
−5 1 5 ⋮0 1 3 ⋮0 0 0 ⋮
9 6 0
This matrix has the same number of equations as
there are variables, however one of these
equations is all zeros:
0𝑥 + 0𝑦 + 0𝑧 = 0
This gives no information, because any values of 𝑥,
𝑦 and 𝑧 will give 0 = 0, which is already known. If
the other equations are extracted:
−5𝑥 + 𝑦 + 5𝑥 = 9
𝑦 + 3𝑧 = 6
The result is two equations and three variables, so
this cannot be solved for a unique solution, thus
there are infinitely many solutions.
Exercises: 1. Determine the number of solutions for the
following matrices:
𝑎) 6 2 46 6 43 4 2
⋮ ⋮ ⋮
53
20 𝑐)
7 5 −321 3 714 34 −38
⋮ ⋮ ⋮
−175
−146
𝑏) 12 2 66 3 7
18 19 41
⋮ ⋮ ⋮
55
12 𝑑)
5 2 46 6 43 4 2
⋮ ⋮ ⋮
532
2. Determine if the following sets of equations can be solved, then solve them if possible.
𝑎) −12𝑥 − 5𝑧 = −7𝑥 + 19
𝑧 = −8𝑦 − 4𝑥 + 8
−6𝑥 + 34.5𝑦 − 4.5𝑧 − 52.5 = 0
𝑏) −8𝑥 − 3𝑦 = 14 − 𝑧
−5𝑧 − 12 = −6𝑥 − 4𝑦
4𝑥 + 5𝑦 + 2𝑧 = 2
𝑐) 12𝑧 + 4𝑥 = −9𝑦 + 16
2𝑥 + 9.5𝑦 + 3𝑧 = 15
−6𝑥 − 3.5𝑦 = 24𝑧 − 10
3. Find the value(s) of 𝑎 where there are zero solutions, infinite solutions and a unique solution.
6 4 −33 8 31 1 𝑎
⋮ ⋮ ⋮
1283
3.8 applications
An application you will surely come across is the
macroeconomic model of an economy.
Economic theory: the macroeconomic model of an
economy determines the income of all the people
in an economy, and is written:
𝑌 = 𝐶 + 𝐼 + 𝐺
Where 𝑌 is the total income of an economy, 𝐶 is
the total consumption expenditure in an economy,
𝐼 is the sum of all investment expenditure in an
economy, and 𝐺 the government spending.
To be able to solve for 𝑌, 𝐶, 𝐼 or 𝐺, other
independent equations need to be found. That is,
above is one equation with four variables, so it
cannot be solved.
After earning income from working, the
government takes a proportion of it through taxes,
so that income is reduced. If the government taxes
at a rate 𝑡 (e.g. 20%: 𝑡 = 0.20), then the income
actually obtained is called disposable income
(denoted 𝑌𝑑 ):
𝑌𝑑 = 1 − 𝑡 𝑌
Most people do not spend everything they earn,
but rather save a proportion of it. The assumption
is made that individuals save at a rate “𝑠” of their
disposable income (i.e. the income they physically
receive).
𝑆 = 𝑠𝑌𝑑
Where “𝑆” is the total savings of all consumers.
Whatever is not saved, must be spent on
consumption goods:
𝐶 = 1 − 𝑠 𝑌𝑑
From before, 𝑌𝑑 = 1 − 𝑡 𝑌 so this allows us to
get rid of the 𝑌𝑑 in the above two equations:
𝑆 = 𝑠 1 − 𝑡 𝑌
𝐶 = 1 − 𝑠 1 − 𝑡 𝑌
Lastly, for anyone in an economy to invest money
(say in constructing an office block), that money
needs to come from somewhere; it comes from
savings. Thus 𝑆 = 𝐼, so from the above equation:
63
𝐼 = 𝑠 1 − 𝑡 𝑌
Where the right side is the equivalent of “𝑆”.
This gives three equations, and six variables:
𝑌 = 𝐶 + 𝐼 + 𝐺
𝐼 = 𝑠 1 − 𝑡 𝑌
𝐶 = 1 − 𝑠 1 − 𝑡 𝑌
However, assumptions are made about 𝐺, 𝑠 and 𝑡.
Government spending (𝐺) is constant regardless of
the level of income (as even in a recession, the
government spends approximately the same
amount); the savings rate 𝑠 takes a long time to
change, so is assumed to be constant; the
government sets the tax rate 𝑡 and is usually the
same for at least a year, and if it is changed, it is
not changed much.
Working through the following example will show
you how to solve this set of equations.
Example 1: if the government of an economy
spends $300billion and taxes at a rate of 30%,
with the citizens, on average, saving 20% of their
income, what is the consumption, investment and
total income of the economy?
Plan: use the model of a closed economy
𝑌 = 𝐶 + 𝐼 + 𝐺
𝐶 = 1 − 𝑠 1 − 𝑡 𝑌
𝐼 = 𝑠 1 − 𝑡 𝑌
Substitute all the values that are known, then
rearrange into the general matrix form, use row
operations to get the zeros in the bottom left
triangle, then solve for 𝐶, 𝐼 and 𝑌.
Solution:
𝑌 = 𝐶 + 𝐼 + 300
𝐶 = 1 − 0.2 1 − 0.3 𝑌
𝐼 = 0.2 1 − 0.3 𝑌
The variables will be arranged in alphabetical
order: 𝐶, 𝐼 and 𝑌. Rearranging the equations into
the general matrix form:
’: −𝐶 − 𝐼 + 𝑌 = 300
’: 𝐶 − 0.56𝑌 = 0
’: 𝐼 − 0.14𝑌 = 0
Construct the matrix:
−1 −1 11 0 −0.560 1 −0.14
⋮ ⋮ ⋮
30000
Use row operations to get zeros in the bottom two
entries of the first column:
−1 −1 11 0 −0.560 1 −0.14
⋮ ⋮ ⋮
30000
𝑅2′ = 𝑅1 + 𝑅2
−1 −1 10 −1 0.440 1 −0.14
⋮ ⋮ ⋮
300300
0
Get a zero in the bottom entry of the second
column:
−1 −1 10 −1 0.440 1 −0.14
⋮ ⋮ ⋮
300300
0
𝑅3′ = 𝑅2 + 𝑅3
−1 −1 10 −1 0.440 0 0.3
⋮ ⋮ ⋮
300300300
Extract the equations from this last matrix:
−𝐶 − 𝐼 + 𝑌 = 300
−𝐼 + 0.44𝑌 = 300
0.3𝑌 = 300
Work from the bottom up:
𝑌 =300
0.3= 1000
Substitute this into the next equation up:
−𝐼 + 0.44 1000 = 300
𝐼 = 140
Lastly:
−𝐶 − 𝐼 + 𝑌 = 300
−𝐶 − 140 + 1000 = 300
𝐶 = 560
64
This economy has consumption of $560billion,
invests $140billion and has a total income of
$1,000billion (or $1trillion).
Example 2: If a government has a tax rate of 40%
and spends $600billion, and the people in the
economy save, on average, 20% of their income,
what is the consumption, investment and total
income of the economy?
Plan: use the model for a closed economy
𝑌 = 𝐶 + 𝐼 + 𝐺
𝐼 = 𝑠 1 − 𝑡 𝑌
𝐶 = 1 − 𝑠 1 − 𝑡 𝑌
Substitute in the known information (𝑠 = 0.2, 𝑡 =
0.4, 𝐺 = 600), then rearrange to get into the
general matrix form. Use row operations to get
zeros in the bottom left triangle, then solve for
𝐶, 𝐼 and 𝑌.
Solution: substitute in all the known information:
𝑌 = 𝐶 + 𝐼 + 600
𝐼 = 0.2 1 − 0.4 𝑌 = 0.12𝑌
𝐶 = 1 − 0.2 1 − 0.4 𝑌 = 0.48𝑌
Rearrange into the general matrix form:
−𝐶 − 𝐼 + 𝑌 = 600 ’
𝐼 − 0.12𝑌 = 0 ’
𝐶 − 0.48𝑌 = 0 ’
Construct the matrix:
−1 −1 10 1 −0.121 0 −0.48
⋮ ⋮ ⋮
60000
𝑅3′ = 𝑅1 + 𝑅3
−1 −1 10 1 −0.120 −1 0.52
⋮ ⋮ ⋮
6000
600
𝑅3′ = 𝑅2 + 𝑅3
−1 −1 10 1 −0.120 0 0.4
⋮ ⋮ ⋮
6000
600
Extract the equations from this last matrix:
−𝐶 − 𝐼 + 𝑌 = 600
𝐼 − 0.12𝑌 = 0
0.4𝑌 = 600
Work upwards:
𝑌 =600
0.4= 1500
Then up one equation:
𝐼 − 0.12 1500 = 0
𝐼 = 180
Up one more equation:
−𝐶 − 180 + 1500 = 600
𝐶 = 720
Thus consumption is $720billion, investment is
$180billion and total income is $1,500billion.
If you think it would be easier to solve the model
using the substitution method, you would be
correct. Unless you are not given enough
information to solve for actual numbers.
Example 3: Given the following closed economy
𝑌 = 𝐶 + 𝐼 + 𝐺
𝐶 = 1 − 𝑠 1 − 𝑡 𝑌
𝐼 = 𝑠(1 − 𝑡)𝑌
and assuming the tax rate is 15% and government
spending is $600billion, what is the solution to this
system of equation in terms of the savings rate 𝑠?
Plan: substitute the numbers that are known, and
then rearrange to get into the general matrix form,
assuming 𝑠 is just a number. Solve for 𝐶, 𝐼 and 𝑌 in
terms of 𝑠.
Solution: substitute in the known numbers
𝑌 = 𝐶 + 𝐼 + 600
𝐶 = 1 − 𝑠 1 − 0.15 𝑌
𝐶 = 0.85𝑌 − 0.85𝑠𝑌
𝐼 = 𝑠 1 − 0.15 𝑌 = 0.85𝑠𝑌
Rearrange to get into the general matrix form
−𝐶 − 𝐼 + 𝑌 = 600 ’
𝐶 − (0.85 − 0.85𝑠)𝑌 = 0 ’
𝐼 − 0.85𝑠𝑌 = 0 ’
65
Remember that 𝑠 is the savings rate and is a
constant. Construct the relevant matrix:
−1 −1 11 0 −0.85 + 0.85𝑠0 1 −0.85𝑠
⋮ ⋮ ⋮
60000
𝑅2′ = 𝑅1 + 𝑅2
Manipulate this to get the zero in the middle entry
of the first column:
−1 −1 10 −1 0.15 + 0.85𝑠0 1 −0.85𝑠
⋮ ⋮ ⋮
600600
0 𝑅3
′ = 𝑅2 + 𝑅3
Then get a zero in the bottom entry of the second
column:
−1 −1 10 −1 0.15 + 0.85𝑠0 0 0.15
⋮ ⋮ ⋮
600600600
Rewrite the equations:
−𝐶 − 𝐼 + 𝑌 = 600
−𝐼 + 0.15 + 0.85𝑠 𝑌 = 600
0.15𝑌 = 600
Work from the bottom up:
𝑌 = 4000
Up one more equation:
−𝐼 + 0.15 + 0.85𝑠 4000 = 600
−𝐼 + 600 + 3400𝑠 = 600
𝐼 = 3400𝑠
Then:
−𝐶 − 3400𝑠 + 4000 = 600
𝐶 = 3400 − 3400𝑠
The answer in terms of the savings rate 𝑠 is:
Consumption is $(3400 − 3400𝑠)billion,
Investment is $(3400𝑠)billion, and
Total income of the economy is $4000billion.
Matrices make easy work of a large number of
equations.
Exercises: 1. Given an economy with the following facts, solve
for 𝐶, 𝐼 and 𝑌 using matrices. Government spending = 400, tax rate = 20% and savings rate = 15%.
2. Given an economy with the following facts, solve for 𝐶, 𝐼 and 𝑌 using matrices.
Government spending = 700, tax rate = 10% and savings rate = 25%.
3. Given an economy with the following facts, solve for 𝐶, 𝐼 and 𝑌 using matrices in terms of the tax rate 𝑡.
Government spending = 500, savings rate = 18%.
3.9 the determinant of a 𝟐 × 𝟐 matrix
For a small number of equations, using matrices
seems pointless, but with a large number of
equations, matrices simplify the mathematics.
This is why you should know a few different ways
of solving matrices.
Theory: the determinant of a 2 × 2 matrix is found
by multiplying the numbers in the right hand
diagonal, and then subtracting the multiplication
of the numbers in the left hand diagonal.
For the matrix:
𝐹 = 𝑎 𝑐𝑏 𝑑
The determinant of matrix 𝐹 is:
𝐹 = 𝑎𝑑 − 𝑏𝑐
Not to be confused with absolute values, the
determinant is written with two vertical lines
around either the whole matrix, or the letter
defining a matrix.
For the 2 × 2 matrix:
2 34 7
The determinant is:
2 × 7 − (4 × 3) = 14 − 12 = 2
An easy way of remembering this (especially if
you’re right handed) is right hand diagonal take
away left hand diagonal (actually do the karate
chops!).
To show that the determinant is being found,
straight lines are drawn either side of the matrix:
3 −45 −8
Example 1: find the determinant of the matrix
3 −45 −8
66
Solution: the determinant is the multiplication of
the right diagonal minus the multiplication of the
left diagonal which gives:
3 × −8 — 4 × 5
= −24 − (−20)
= −24 + 20
= −4
Exercises: 1. Find the determinant of the following matrices:
𝑎) 2 −12 3
𝑑) 4 47 5
𝑔) −1 −22 −6
𝑏) −3 2−29 5
𝑒) 2 43 6
) 8 −3
−16 6
𝑐) 6 128 11
𝑓) −3 5−5 −3
𝑖) −2 47 0
3.10 the determinant of a 𝟑 × 𝟑 matrix
This is a harder section, so make sure you are very
comfortable with all the material in Section 3.9.
Intro example 1: find the determinant of the 3 × 3
matrix:
2 5 −3
−1 −1 −15 −3 −2
Solution: Circle the first entry (2), then cross out
all the numbers in that row and column, but leave
the circled number alone:
With the remaining numbers, draw a matrix:
−1 −1−3 −2
Find the determinant of this matrix using the right
hand/left hand method.
𝐷𝑒𝑡 = −1 −2 − −3 −1 = −1
Go on to the next number along in the original
3 × 3 matrix; circle it and cross out the vertical and
horizontal numbers:
Draw the matrix from the remaining numbers, and
find the determinant:
−1 −15 −2
𝐷𝑒𝑡 = −1 −2 − 5 −1 = 7
Go on to the last number:
Finding the determinant of the remaining
numbers:
−1 −15 −3
𝐷𝑒𝑡 = −1 −3 − 5 −1 = 8
Then bring it all together.
Theory: The determinant of a 3 × 3 matrix is:
𝐷𝑒𝑡 = + 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
× det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥
− 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
× det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥
+ 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
× 𝑑𝑒𝑡 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥
And the “circled numbers” are the numbers in the
top row.
Notice also the highlighted signs and how they
alternate (+, −, +) which we will come back to
soon.
The determinant of the original 3 × 3 matrix is:
𝐷𝑒𝑡 = + 2 −1 − 5 7 + −3 8
= −61
The +, −, + pattern does NOT come from the sign
of the numbers in the original matrix, but it is a
pattern unrelated to the original matrix. Many
students stuff up here, so don’t be one of them.
This pattern works for the top row, but as you will
soon see, sometimes you will not want to use the
top row.
2 5 −3
−1 −1 −15 −3 −2
2 5 −3
−1 −1 −15 −3 −2
2 5 −3
−1 −1 −15 −3 −2
67
Theory: when finding the determinant using any
row or column, the pattern for the determinant
equation is given by the matrix:
+ − +− + −+ − +
An easy way of remembering this matrix is that the
top left is an addition sign, and then the signs
alternate across every row and down every
column.
Remember, this does NOT mean that the numbers
in the original 3 × 3 matrix are positive or
negative, but rather that when you come to find
the determinant, this is the pattern of signs you
put in front of the “𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 ×
𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡” in the formula.
In the introductory example, the top row was
used, but any row or any column can be used.
There is a good reason why the +, −, + matrix
should be remembered.
Theory: when asked to find the determinant of a
matrix and one of the rows or columns has one (or
more) zeros in it, use that row or column.
It makes life a lot easier, as the following example
demonstrates.
Example 1: find the determinant of the matrix
𝐴 = −2 3 1−1 0 44 −5 −3
Plan: use the middle column (because it has a
zero): circle the top number and cross out the
other numbers in that row and column. Then find
the determinant of the remaining numbers.
Repeat for the other two numbers in the middle
column.
Use the “sign” matrix to get the correct pattern:
+ − +− + −+ − +
Since the middle column is being used, the signs in
the determinant equation will be −, +, −:
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 =
− 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟𝑚𝑎𝑡𝑟𝑖𝑥
+ 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟𝑚𝑎𝑡𝑟𝑖𝑥
− 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥
Solution: Circle the first number:
Find the determinant of the remaining matrix:
𝐷𝑒𝑡 = −1 44 −3
= −13
Circle the next number:
𝐷𝑒𝑡 = −2 14 −3
= 2
For the last number:
𝐷𝑒𝑡 = −2 1−1 4
= −7
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 =
− 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟𝑚𝑎𝑡𝑟𝑖𝑥
+ 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟𝑚𝑎𝑡𝑟𝑖𝑥
− 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 × 𝑑𝑒𝑡 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥
𝐷𝑒𝑡 𝐴 = 𝐴
= − 3 −13 + 0 2 — 5 −7
= 4
But the middle number is (0)(2) = 0, so the
𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 × 𝑧𝑒𝑟𝑜 will always be zero. This is
why columns/rows with lots of zeros are chosen;
because the determinant of a circled zero does not
need to be found.
Try the above example using the middle row. You
should get the same answer.
−2 3 1−1 0 44 −5 −3
−2 3 1−1 0 44 −5 −3
−2 3 1−1 0 44 −5 −3
68
Finding determinants of 3 × 3 matrices is not too
difficult, but it requires a lot of work. The more
you practice, the easier it will be.
Exercises: 1. Find the determinants of the following 3 × 3
matrices:
𝑎) 2 5 105 6 31 −2 8
𝑑) −1 4 −14 5 4
−1 4 −1
𝑏) 3 7 −15 8 41 9 9
𝑒) 3 2 −80 7 51 2 1
𝑐) 11 12 15
−11 19 183 23 27
𝑓) 0 5 3
11 −6 25 −6 0
3.11 using the Jacobian determinant
Previously, the values of 𝑥, 𝑦 and 𝑧 were found by
finding the zeros in the bottom left triangle. The
following is a different way of finding the solution
to a set of equations.
Intro example 1: to solve the matrix
2 5 −3
−1 −1 −15 −3 −2
⋮ ⋮ ⋮
190
−10
find the determinant of the 3 × 3 “variables”
matrix (left of the dotted line) which was done in
the previous section. The determinant was −61.
This is denoted by |𝐴| and it is the determinant of
the original 3 × 3 “variables” matrix.
Next, the column of the variable to be solved is
REPLACED by the column of constants. That is, to
solve for 𝑥, replace the 𝑥 column (i.e. first column)
in the matrix with the constants column (i.e. the
numbers on the right of the dotted line):
19 5 −30 −1 −1
−10 −3 −2 = 𝐴𝑥
Find the determinant of this new matrix using the
method from the last section.
Use the middle row as it has a zero (remembering
the “signs” matrix):
𝐴𝑥 = − 0 5 −3
−3 −2 + −1
19 −3−10 −2
− (−1) 19 5
−10 −3
𝐴𝑥 = −1 −68 − −1 −7 = 61
Then to find the actual value of 𝑥, the following
fraction is used:
𝑥 = 𝐴𝑥
𝐴 =
61
−61= −1
Theory: to solve for a variable using the Jacobian
Method, firstly find the determinant of the
variables matrix (|𝐴|). Then replace the
appropriate column (for the variable to be solved)
with the constants column, and find the
determinant of this (|𝐴𝑏 |). The solution is this
second determinant divided by the determinant of
the variables matrix:
𝑏 = 𝐴𝑏
𝐴
Example 1: solve for 𝑦 and 𝑧 using the Jacobian
Method in the matrix that was started above
2 5 −3
−1 −1 −15 −3 −2
⋮ ⋮ ⋮
190
−10
Plan: use the Jacobian method; find the
determinant of the variables matrix. Then replace
the constants column into the column of the
variable being solved, and find the determinant of
this new 3 × 3 matrix. Divide this determinant by
the determinant of the original variables
determinant to solve for the unknown variable.
Solution: to solve for 𝑦, replace the 𝑦 column with
the constants column
2 19 −3
−1 0 −15 −10 −2
= 𝐴𝑦
Find the determinant of this matrix using the
middle row:
69
𝐴𝑦 = − −1 19 −3
−10 −2 + 0
2 −35 −2
− (−1) 2 195 −10
𝐴𝑦 = 1 −68 − −1 −115 = −183
Solve for 𝑦:
𝑦 = 𝐴𝑦
𝐴 =
−183
−61= 3
To solve for 𝑧:
2 5 19
−1 −1 05 −3 −10
= 𝐴𝑧
Find the determinant (yourself!) to make sure it is
𝐴𝑧 = 122, which then gives:
𝑧 = 𝐴𝑧
𝐴 =
122
−61= −2
Thus 𝑥 = −1, 𝑦 = 3, 𝑧 = −2, or −1, 3, −2 .
To make sure this answer is correct, substitute it
back into the bottom equation of the original
matrix:
5𝑥 − 3𝑦 − 2𝑧 = −10
5 −1 − 3 3 − 2 −2 = −10
−10 = −10
This is a complicated method, but the more you
practice the easier it becomes. When you have a
zero in a matrix, use it, and remember that you do
not need to find the determinant of the smaller
matrix when that zero is circled.
Example 2: solve for 𝑥, 𝑦 and 𝑧 in the following
matrix using the Jacobian Method
2 4 41 −2 10 1 −3
⋮ ⋮ ⋮
05
10
Plan: find the determinant of the variables matrix,
using the 3 × 3 method. Call this 𝐴 . Replace the
𝑥 column with the constants column. Find the
determinant of this matrix, and call it 𝐴𝑥 . Solve
for 𝑥 by:
𝑥 = 𝐴𝑥
𝐴
Repeat for 𝑦 and 𝑧 with the appropriate columns.
Also, when finding the determinant, remember the
“signs” matrix:
+ − +− + −+ − +
Solution: determinant of the variables matrix:
𝐴 = 2 4 41 −2 10 1 −3
Use the bottom row:
𝐴 = + 0 4 4
−2 1 − 1
2 41 1
+ (−3) 2 41 −2
𝐴 = −1 −2 + −3 −8 = 26
To find 𝐴𝑥 :
𝐴𝑥 = 0 4 45 −2 1
10 1 −3
Use the top row (leaving the zero out from now
on):
𝐴𝑥 = − 4 5 1
10 −3 + (4)
5 −210 1
𝐴𝑥 = −4 −25 + 4 25 = 200
To find 𝐴𝑦 replace the middle column with the
constants, and use the top row:
𝐴𝑦 = 2 0 41 5 10 10 −3
𝐴𝑦 = + 2 5 1
10 −3 + (4)
1 50 10
𝐴𝑦 = 2 −25 + 4 10 = −10
To find 𝐴𝑧 , replace the middle column with the
constants, and use the top row:
𝐴𝑧 = 2 4 01 −2 50 1 10
𝐴𝑧 = + 2 −2 51 10
− (4) 1 50 10
𝐴𝑧 = 2 −25 − 4 10 = −90
Finally, for a solution:
𝑥 = 𝐴𝑥
𝐴 =
200
26=
100
13
70
𝑦 = 𝐴𝑦
𝐴 =
−10
26=
−5
13
𝑧 = 𝐴𝑧
𝐴 =
−90
26=
−45
13
The solution is 𝑥 = 100/13, 𝑦 = −5/13,
𝑧 = −45/13 or 100
13, −
5
13, −
45
13 .
Check if the answer is correct by substituting it
back into one of the rows of the original matrix. To
be consistent, use the bottom equation:
𝑦 − 3𝑧 = 10
−5
13− 3 −
45
13 = 10
−5
13+
135
13= 10
=−5 + 135
13=
130
13= 10
The solutions are fractions, so leave them as
fractions!
Exercises: 1. Solve the following matrices.
𝑎) 8 −1 2
−2 −2 −74 1 3
⋮ ⋮ ⋮
1145
𝑐) 1 −1 −32 −1 −4
−8 5 −3
⋮ ⋮ ⋮
777
𝑏) 3 6 8
−1 1 2−3 −2 1
⋮ ⋮ ⋮
121311
𝑑) 4 −6 2
−2 −1 −51 −7 −1
⋮ ⋮ ⋮
1065
2. Solve the following sets of equations using Jacobian determinants. 𝑎) 6𝑥 + 3𝑦 + 3𝑧 = 18
2𝑥 − 𝑦 − 𝑧 = 24
−2𝑥 + 2𝑦 − 4𝑧 = 15
𝑏) 4𝑥 + 5𝑦 = 2𝑧 + 6
5𝑥 = 𝑦 + 5𝑧 + 2
−3𝑥 − 𝑧 = −5𝑦 + 6
𝑐) 6𝑥 + 4𝑦 + 6𝑧 = 3
𝑧 − 𝑦 + 4𝑥 − 9 = 0
2𝑦 − 2𝑥 − 3𝑧 = 8
𝑑) 𝑥 − 9𝑦 + 3𝑧 = 2
−𝑥 − 2𝑦 + 2𝑧 = 6
4𝑥 − 2𝑦 + 2𝑧 = 8
3. A toy manufacturing company uses metal (𝑚) plastic (𝑝) and labour (𝑙). The company owns three identical factories in a country where the costs of each of the three inputs is different:
City 1: metal costs $2/𝑢𝑛𝑖𝑡, plastic costs $8/𝑢𝑛𝑖𝑡, labour costs $7/𝑢𝑛𝑖𝑡.
City 2: metal costs $5/𝑢𝑛𝑖𝑡, plastic costs $6/𝑢𝑛𝑖𝑡, labour costs $3/𝑢𝑛𝑖𝑡.
City 1: metal costs $7/𝑢𝑛𝑖𝑡, plastic costs $3/𝑢𝑛𝑖𝑡, labour costs $9/𝑢𝑛𝑖𝑡.
If all factories are at full capacity, the costs for each of the factories is as follows:
City 1 factory costs are $342
City 2 factory costs are $255
City 3 factory costs are $375 Determine: a) The equation in terms of 𝑚, 𝑝 and 𝑙 for each of
the three cities. b) The amount of labour, plastic and metal used
in each of the three identical factories.
chapter three summary
Method 1 to solve two equation systems: 1. set the 𝑦’s equal to each other. 2. substitute the functions of 𝑥 in and solve for 𝑥. 3. Substitute the solutions of 𝑥 back into either of the
original equations to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the solution.
Note: setting the 𝑥’s equal to each other initially will give the same answer.
Method 2 to solve two equation systems: 1. isolate one variable in one equation 2. substitute it into the other equation. 3. solve for the other variable, 4. then substitute this solution back into any one of
the original equations to solve for the other variable.
To solve three simultaneous equations: 1. take one equation and substitute it into the other
two equations. This will make two new equations with only two variables.
2. substitute one of those equations into the other to solve for one of the unknowns.
3. use this solution to solve for one of the other variables.
4. lastly, use these two solutions to solve for the third variable.
The solution is also written in coordinate form 𝑥, 𝑦, 𝑧 . The matrix ready form of an equation is:
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑
To construct a matrix: 1. take the numbers in front of the variables and put
them into square brackets. 2. draw a new set of brackets and vertically write the
order of the variables you have chosen (e.g. alphabetical).
3. after writing an equals sign, the constants on the right side are written vertically in square brackets.
Manipulating matrices involve ROW operations; this is the process of adding/subtracting the COLUMNS of a matrix.
For two equation systems, a zero is needed in the bottom left corner. For three equation systems: zeros are needed in the bottom two entries of the first column (this must be
71
done first), as well as the bottom entry of the second column.
Matrices can have:
A single solution
No solutions (if there is no common intersection of all lines)
Infinite solutions (if there are not enough equations and/or if some of the equations are the same)
The macroeconomic model of an economy is: 𝑌 = 𝐶 + 𝐼 + 𝐺 Where 𝑌 is the total income of an economy, 𝐶 is the total consumption expenditure in an economy, 𝐼 is the sum of all investment expenditure in an economy, and 𝐺 is government spending. Consumption is determined by what is not saved (at a savings rate 𝑠) and not taxed (at the tax rate 𝑡): 𝐶 = 1 − 𝑠 1 − 𝑡 𝑌 Investments comes from savings, 𝑆 = 𝐼: 𝐼 = 𝑠 1 − 𝑡 𝑌 Government spending (𝐺) is constant regardless of the level of income; the savings rate 𝑠 takes a long time to change, so is assumed to be constant; the government sets the tax rate 𝑡 and is usually the same for at least a year.
For the matrix:
𝐹 = 𝑎 𝑐𝑏 𝑑
The determinant is: 𝐹 = 𝑎𝑑 − 𝑏𝑐 The determinant of a 3 × 3 matrix is: 𝐷𝑒𝑡 = + 1𝑠𝑡 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
× det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 − 2𝑛𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
× det 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 + 3𝑟𝑑 𝑐𝑖𝑟𝑐𝑙𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟
× 𝑑𝑒𝑡 𝑜𝑓𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑚𝑎𝑡𝑟𝑖𝑥 When finding the determinant using any row or column, the pattern for the determinant equation is given by the matrix:
+ − +− + −+ − +
It is easiest to find the determinant of a matrix using the row/column with the most zeros.
Using the Jacobian Method to solve a system of equations:
𝑏 = 𝐴𝑏
𝐴
Where 𝐴 is the determinant of the “variables” matrix, and 𝐴𝑏 is the determinant of the “variables” matrix with the 𝑏 column replaced with the solitary constants matrix.
chapter three questions
1. Convert the following sets of equations into matrix form: 𝑎) 𝑦 = 3𝑥 − 1 𝑏) − 3 = −𝑦 − 𝑥 𝑦 = −2𝑥 − 17 15 + 𝑦 = −3𝑥
𝑐) 17𝑦 − 13 = 𝑥 + 1 𝑑) 11𝑥 − 3 = 1.4𝑦 2𝑦 − 3𝑥 − 1 = 0 11𝑥 − 3 = 1.5𝑦 𝑒) 3𝑥 = 3𝑦 − 4 + 𝑧 𝑓) 𝑎 + 𝑐 = 𝑏 − 4
4𝑦 − 15𝑥 = 13 − 𝑏 = 𝑐 − 𝑎 − 1 1.4𝑧 − 𝑦 = 𝑥 3𝑎 − 3𝑏 − 𝑐 = 1
2. Solve the following sets of equations without using matrices: 𝑎) 5𝑥 + 𝑦 = 14 2𝑥 + 8𝑦 + 2 = 0
𝑏) 4𝑦 = 𝑥 + 6 7𝑥 + 3𝑦 = 20
𝑐) 3𝑥 = −7𝑦 + 30.5 4𝑥 + 𝑦 = 17.75
𝑑) 3𝑥 = −2𝑦 − 4𝑥 + 9 𝑥 + 𝑦 = −3𝑧 + 5 𝑥 + 𝑦 = −4𝑧 + 6
𝑒) 2𝑥 + 2𝑦 + 2𝑧 = 16 3𝑥 + 3𝑧 = 𝑦 + 20 3𝑦 + 𝑧 = 𝑥 + 6
3. Solve the following sets of equations using matrices: 𝑎) 12𝑥 + 7𝑦 = 19 𝑏) 𝑥 = 𝑦 + 1 8𝑥 + 13𝑦 = 21 3𝑥 + 3𝑦 − 9 = 0
𝑐) 6𝑦 = −8𝑥 + 6 𝑑) 12𝑦 + 22 = 5𝑥 5𝑦 + 6 = −3𝑥 3𝑥 + 8𝑦 + 2 = 0
𝑒) 7𝑥 + 12𝑦 = 319 𝑓) 2𝑥 + 13𝑦 = 140 9𝑦 = 18𝑥 − 63 2𝑥 + 8𝑦 = 80
𝑔) 𝑦 = 3𝑥 + 7 ) 4𝑥 = 𝑦 − 5.75 8𝑥 + 2𝑦 = −10.5 2𝑥 + 𝑦 = 25.25
4. If possible, solve the following matrices for a unique solution:
𝑎) 2 1 ⋮
−2 2 ⋮ 8
−2 𝑏)
1 4 ⋮ 5 3 ⋮
2926
𝑐) −1 1 ⋮ 2 −2 ⋮
3−6
𝑑) 1 3 ⋮ 3 1 ⋮
106
𝑒) −2 −4 ⋮ 3 2 ⋮
14−9
𝑓) 4 5 ⋮ 3 6 ⋮
10084
𝑔) 1 5 ⋮ 2 −1 ⋮
42−9.5
) 3 3 ⋮
−2 2 ⋮ 2418
𝑖) 12 13 ⋮ 1 2 ⋮
104.516.5
𝑗) −3 12 ⋮ −18 7 ⋮
25.523
𝑘) 1 8 ⋮ 1 9 ⋮
3640.2
𝑙) −1 3 ⋮ −1 5 ⋮
18.433.6
𝑚) −9 1 ⋮ 4 1 ⋮
−65.826.5
5. Use matrices to solve for the following demand and supply equations: 𝑎) 𝑃𝑑 = −2𝑄𝑑 + 62.5 𝑃𝑠 = 2𝑄𝑠 + 28.5
𝑏) 𝑃𝑑 = −1.5𝑄𝑑 + 59.75 𝑃𝑠 = 1.8𝑄𝑠 + 30.5
𝑐) 𝑃𝑑 = −1.4𝑄𝑑 + 72.6 𝑃𝑠 = 2.2𝑄𝑠 + 4.2
𝑑) 𝑃𝑑 = −4.3𝑄𝑑 + 349/30 𝑃𝑠 = 4.1𝑄𝑠 + 97/30
𝑒) 𝑃𝑑 = −2𝑄𝑑 + 24.07 𝑃𝑠 = 0.41𝑄𝑠 + 3.7537
6. Solve the following sets of equations using matrices: 𝑎) 4𝑥 + 𝑦 + 8𝑧 = 21 5𝑥 + 2𝑦 − 𝑧 = −6 6𝑥 + 3𝑦 + 2𝑧 = 3
𝑏) − 2𝑥 + 𝑦 + 2 = 0 2𝑥 + 𝑦 − 3𝑧 + 11 = 0
72
4𝑦 + 𝑥 = 𝑧 + 12
𝑐) 2.5𝑥 + 3𝑦 + 7𝑧 = 92 1.5𝑥 + 2𝑦 + 𝑧 = 17 3𝑥 − 2𝑦 + 𝑧 = 16
𝑑) 5𝑥 + 2𝑦 + 𝑧 + 4 = 0 1𝑥 + 5𝑦 + 3𝑧 − 5 = −5 3𝑥 + 𝑦 + 4𝑧 + 13 = 0
7. If possible, solve the following matrices:
𝑎) −3 2 12 3 31 −1 −2
⋮ ⋮ ⋮
4−14
𝑏) −2 8 12−1 7 04 1 2
⋮ ⋮ ⋮
4923
13.5
𝑐) 4 7 −71 −12 −114 5 11
⋮ ⋮ ⋮
−7−4949
𝑑) 1.2 12.2 1.73.3 −8.3 84.5 7.1 −2.9
⋮ ⋮ ⋮
67.4320.9128.08
𝑒)
4/3 2 27/9 −11/9 9
−1/9 1/2 1
⋮ ⋮ ⋮
26/3254/943/36
𝑓) 1.3 1.75 2.3
−1.3 2.1 −1.42.5 3.1 4.1
⋮ ⋮ ⋮
11.21−2.0720.51
𝑔) 4.1 4.2 0.23.6 4.4 03.8 2.6 1.1
⋮ ⋮ ⋮
−18.31−17.24−16.23
) −1.1 2.34 −1.21−1.22 4.11 −0.98
1 0.23 1.31
⋮ ⋮ ⋮
−0.18.12
10.78
8. Determine the number of solutions (none, one or infinite) for the following matrices. Justify your answer.
𝑎) 2 1 83 1 15 −2 −3
⋮ ⋮ ⋮
191221
𝑏) 2 1 3
−2 4 103 −1 −2
⋮ ⋮ ⋮
15205
𝑐) 2 1 58 −3 2
−10 −1.5 −14
⋮ ⋮ ⋮
74
−18 𝑑)
3 2 7−8 −3 45 2 1
⋮ ⋮ ⋮
021
𝑒) 1 5 −33 −21 185 1 3
⋮ ⋮ ⋮
8−918
𝑓) 5 −2 −1
−1 −4 59 −8 3
⋮ ⋮ ⋮
−37
20
9. Find the determinant of the following 2 × 2 matrices.
𝑎) 4 71 −1
𝑏) 18 12 3
𝑐) −2 −2−2 3
𝑑) 3 11 3
𝑒) 3 31 1
𝑓) −1 1.53.5 1
𝑔) 2.2 1.58 7
) 1/3 5/72/3 1/7
𝑖) 8/7 2.22/9 1/3
𝑗) 1.1 22 1.1
𝑘) 2 22 2
𝑙) 0 01 1
10. Find the determinant of the following 3 × 3 matrices:
𝑎) 5 2 61 3 21 1 2
𝑏) −1 5 71 1 23 −3 2
𝑐) 4 2 41 −2 21 0 −3
𝑑) −1 4 30 8 0
−2 1 −1
𝑒) 7 6 43 −1 6
−1 8 0 𝑓)
19 23 1420 41 180 37 0
11. Find the solution to the following matrices using the Jacobian Method:
𝑎) −2 4 1−1 1 1−2 1 9
⋮ ⋮ ⋮
62
16
𝑏) 7.5 5.5 91.5 3.5 −2.52.5 1 1.5
⋮ ⋮ ⋮
952119
𝑐) −4.5 2.25 8.254.5 1.75 −3.25
3.75 −0.5 1
⋮ ⋮ ⋮
60−139.25
𝑑)
1/3 1/2 1/91/6 1/4 2/95/6 3/4 1/3
⋮ ⋮ ⋮
1/38/302/3
12. A mobile phone manufacturer has three identical factories in Malaysia, Indonesia and Taiwan, all running at capacity. To make mobile phones, plastic (𝑃), silicone chips (𝑆) and labour (𝐿) is required. These three inputs all have different cost structures in the three countries (e.g. due to unions) which are: Malaysia: 𝑃 = $7.00, 𝑆 = $19.00, 𝐿 = $14.00 Indonesia: 𝑃 = $9.00, 𝑆 = $21.00, 𝐿 = $11.00 Taiwan: 𝑃 = $4.00, 𝑆 = $20.00, 𝐿 = $25.00 If the total costs of the three factories is: Malaysia: 𝑇𝐶 = $2853.00 Indonesia: 𝑇𝐶 = $2873.00 Taiwan: 𝑇𝐶 = $3594.00 Determine: a) Equations relating inputs and costs for the three
factories. b) The amount of labour, plastic and silicone chips
used in each of the three identical factories. 13. For the macroeconomic model of an economy with
the following facts, solve for 𝐶, 𝐼 and 𝑌 using matrices.
Government spending = 600, tax rate = 10% and savings rate = 10%.
14. For the macroeconomic model of an economy with the following facts, solve for 𝐶, 𝐼 and 𝑌 using matrices.
Government spending = 540, tax rate = 23% and savings rate = 9%.
15. For the macroeconomic model of an economy with the following facts, solve for 𝐶, 𝐼 and 𝑌 in terms of the savings rate 𝑠.
Government spending = 200, tax rate = 20%.
73
Chapter 4
Non-linear functions Moving away from lines towards curves
4.1 Defining Non-Linear Functions 74
4.2 Defining a Quadratic Function 75
4.3 Quadratic Graphs 76
4.4 Sketching Quadratic Functions 77
4.5 The Cubic Function 80
4.6 The Exponential Function 82
4.7 The Logarithmic Function 84
4.8 Logarithmic Graphs 88
4.9 The Natural Number 𝑒 89
4.10 The Hyperbolic Function 90
4.11 Economic Applications 91
Chapter Summary 93
Chapter Four Questions 94
74
4.1 defining non-linear functions
Theory: a function is defined as having only one
𝑦 − 𝑣𝑎𝑙𝑢𝑒 for every 𝑥 − 𝑣𝑎𝑙𝑢𝑒. That is, for every
value on the 𝑥 − 𝑎𝑥𝑖𝑠, there is only one
𝑦 − 𝑣𝑎𝑙𝑢𝑒 plotted by the curve. If this occurs for
all 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 on a given curve, then the curve is a
function. This does not mean that two different
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 cannot give the same 𝑦 − 𝑣𝑎𝑙𝑢𝑒. A
function is still a function if multiple 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠
plot to a sinlge 𝑦 − 𝑣𝑎𝑙𝑢𝑒.
The curve above is a function as all 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠
plot to only one 𝑦 − 𝑣𝑎𝑙𝑢𝑒: 𝐴→𝐵, 𝐶→𝐷. This is
the only thing that makes it a function. Two
different 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 (such as 𝐹 and 𝐺) can plot to
a single 𝑦 − 𝑣𝑎𝑙𝑢𝑒.
Theory: to find if a curve is a function, apply the
vertical line test:
𝐴 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑓 𝑎𝑙𝑙 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑙𝑖𝑛𝑒𝑠 𝑐𝑟𝑜𝑠𝑠
𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑜𝑛𝑙𝑦 𝑜𝑛𝑐𝑒
The curve above passes the vertical line test, so it
is a function, but the following curve does not pass
the vertical line test, and therefore a single
𝑥 − 𝑣𝑎𝑙𝑢𝑒 (e.g. 𝐻) plots two different 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠
(𝐽 and 𝐾). Part of the curve passes the vertical line
test, but unless the whole thing passes, it is not a
function.
Different people denote functions differently. For
example, 𝑦 has been used when referring to linear
functions (lines), but there are many ways of
defining functions: 𝑓(𝑥), 𝑔(𝑥), 𝐷(𝑥), 𝑚, 𝑟 etc. All
these mean the same thing: they are all
“𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑥”. Graphically, the letters denote
the vertical axis, and 𝑥 is the horizontal axis.
Sometimes the letters might mean something: if
referring to revenue from selling a certain quantity
of computers, the function may be defined as:
𝑅 𝑄 = 50𝑄 − 0.6𝑄2
Where 𝑄 is the quantity of computers sold (the
𝑥 − 𝑎𝑥𝑖𝑠) and 𝑅 𝑄 means
“𝑟𝑒𝑣𝑒𝑛𝑢𝑒 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑠𝑜𝑙𝑑”, (the
𝑦 − 𝑎𝑥𝑖𝑠).
Common non-linear functions include:
Look closely at the graphs above and you will see
that for each 𝑥 − 𝑣𝑎𝑙𝑢𝑒 (on the horizontal axis),
there is only one value for each function (colour).
All these curves pass the vertical line test (try it!),
therefore, they are all functions.
Cubic
Hyperbolic
Quadratic
Logarithmic
Exponential
𝑥
𝑦
H
J
K
L
M
𝑥
𝑦
𝐴 𝐶
𝐵
𝐷
𝐹 𝐺
𝐸
𝑥
𝑦
75
Exercises: 1. Determine which of the following curves are
functions.
4.2 defining a quadratic function
The quadratic is by far the most common non–
linear function you will come across. It involves the
squaring of a variable.
Theory: the general form of a quadratic is:
𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐
Where 𝑎, 𝑏 and 𝑐 are constants.
Example 1: find 𝑎, 𝑏 and 𝑐 for
𝑦 = 2𝑥2 + 3𝑥 − 5
Solution: 𝑎 = 2, 𝑏 = 3 and 𝑐 = −5.
Example 2: find 𝑎, 𝑏 and 𝑐 for
𝑓 𝑥 = −3𝑥2 − 2𝑥 + 4
Solution: 𝑎 = −3, 𝑏 = −2 and 𝑐 = 4.
It is easier to have the quadratic in the general
form to get the constants 𝑎, 𝑏 and 𝑐.
Example 3: find 𝑎, 𝑏 and 𝑐 for
𝑔 𝑥 = 2 − 7𝑥2 + 3𝑥
Solution: rearrange: 𝑔(𝑥) = −7𝑥2 + 3𝑥 + 2
So 𝑎 = −7, 𝑏 = 3 and 𝑐 = 2.
Example 4: find 𝑎, 𝑏 and 𝑐 in:
𝑥 = 2 − 18𝑥2
Solution:
𝑥 = −18𝑥2 + 2
where 𝑎 = −18, 𝑏 = 0 and 𝑐 = 2.
Note that 𝑏 or 𝑐 can equal zero, but 𝑎 cannot,
otherwise it would be a linear function.
The general form of the quadratic is not the only
quadratic form, however it is the easiest to work
with. When presented with other forms, rearrange
them into the general form using the crab-claw
method from Chapter 1.
Example 5: find the values of 𝑎, 𝑏 and 𝑐 for
𝑦 = 𝑥 + 2 (𝑥 − 3)
Plan: simplify using the crab-claw method.
Solution: draw the ‘crab-claw’, and then follow the
lines:
𝑦 = 𝑥2 + 2𝑥 − 3𝑥 − 6
𝑦 = 𝑥2 − 𝑥 − 6
So 𝑎 = 1, 𝑏 = −1 and 𝑐 = −6.
Example 6: find the general form of
3𝑦
(3𝑥 − 1) = 2 𝑥 + 1
Plan: simplify and rearrange using the crab-claw
Solution: Multiply both sides by (3𝑥 − 1) to get rid
of it on the left and have it as a multiplication on
the right.
3𝑦 = 2 𝑥 + 1 3𝑥 − 1
Then apply the crab-claw on the right side:
3𝑦 = 2 3𝑥2 + 3𝑥 − 𝑥 − 1
The 2 out front of the square brackets goes into
every term in the square brackets, after the inside
of the square brackets have been simplified:
3𝑦 = 6𝑥2 + 4𝑥 − 2
Isolate 𝑦 by dividing both sides by 3:
𝑦 =6𝑥2 + 4𝑥 − 2
3= 2𝑥2 +
4
3𝑥 −
2
3
3𝑦 = 2 𝑥 + 1 3𝑥 − 1
𝑦 = 𝑥 + 2 (𝑥 − 3)
𝑥
𝑦
𝐴
𝐵
𝐶
𝐷
𝐸
76
Example 7: find the general form of
𝑦 = 𝑥 + 2 2 − 3
Plan: simplify and rearrange using the crab-claw
Solution: this is solved using the crab-claw
method, but the 𝑥 + 2 2 must first be rewritten
as two separate parts:
𝑦 = 𝑥2 + 2𝑥 + 2𝑥 + 4 − 3
Simplify:
𝑦 = 𝑥2 + 4𝑥 + 1
Note: remember from Chapter 1:
𝑥 + 2 2 ≠ 𝑥2 + 22
You cannot simply put the squared into every term
inside the brackets. This is where many students
stuff up, so don’t be one of them!
Exercises: 1. Determine the value of 𝑎, 𝑏 and 𝑐 in the following
quadratics: 𝑎) 𝑦 = 6𝑥2 − 5𝑥 − 1
𝑏) 𝑦 = 1 − 2𝑥 + 𝑥2
𝑐) 𝑦 − 6 − 𝑥2 = 2𝑥
𝑑) 𝑦 =2𝑥2
3+
1
2−
1
3𝑥
𝑒) 𝑦 = 8 − 8𝑥 − 3𝑥2
𝑓) 𝑦 − 2
𝑥 + 1= 2𝑥 − 1
𝑔) 𝑦 = 𝑥 + 1 2 − 8𝑥 + 23
) 𝑦 − −𝑥 + 2 2 = 2𝑥 − 1 2
4.3 quadratic graphs
Quadratics have a distinctive shape which you
must recognise.
Three different quadratics are drawn above in
general form with 𝑏 = 0 and 𝑐 = 0 (i.e. 𝑦 = 𝑎𝑥2).
The difference between the three graphs is the
value of 𝑎.
Theory: The larger the value of 𝑎, the steeper the
curve (black); similarly, the smaller the value of 𝑎,
the shallower the curve (blue).
The following functions are of the form:
𝑦 = 𝑥2 + 𝑐
That is, 𝑏 is set to zero.
The shape of the graphs are identical, except for
their vertical position (vertical displacement).
Theory: the value of 𝑐 in the quadratic equation
determines the vertical displacement of the graph.
Look at all the graphs so far; the turning points
(minimum points in the cases so far) have been on
the 𝑦 − 𝑎𝑥𝑖𝑠. However, if the value of 𝑏 is not
zero, there is a vertical and horizontal shift.
𝑦 = 𝑥2 + 3
𝑦 = 𝑥2 − 3
𝑦 = 𝑥2 − 1
𝑥
𝑦
𝑦 = 2𝑥2
𝑦 = 𝑥2
𝑦 = 0.5𝑥2 𝑥
𝑦
𝑦 = 𝑥 + 2 (𝑥 + 2) − 3
77
Notice how the turning point has moved away
from the 𝑦 − 𝑎𝑥𝑖𝑠 and there has also been a
vertical shift.
Theory: changing the value of 𝑏 in the general
quadratic form shifts the graph both horizontally
and vertically.
Finally, a note about the sign of 𝑎.
Theory: a positive value of 𝑎 will result in
quadratics with a shape of a ‘smiley’ face, whereas
when the value of 𝑎 is negative, the shape
becomes inverted; a ‘sad’ face.
The values of 𝑎, 𝑏 and 𝑐 still change the graph in a
similar way, but the sign of 𝑎 determines if it is a
‘smiley’ or ‘sad’ face.
Exercises: 1. Match the graphs with the equations.
𝑦 = −3𝑥2 𝑦 = 𝑥2 + 2 𝑦 = 𝑥2 + 10𝑥 + 25 𝑦 = −𝑥2 + 8𝑥
2. Match the graphs to the equations
𝑦 = − 3𝑥2 + 2 −𝑦 + 5 = 𝑥2 𝑦 = 𝑥 − 3 2 + 3 𝑦 = −𝑥2 + 2𝑥 𝑦 = 3𝑥2 + 10𝑥 + 28
4.4 sketching quadratic functions
All quadratics can be described as having: a
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡; a turning point; and either two,
one or no 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠.
To find these three main features of quadratics,
theory from Chapter 2 must be applied.
Theory: To find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 of any function,
set all the 𝑥’s in the equation equal to zero.
Example 1: Find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 of
𝑦 = −2𝑥2 + 8𝑥 − 6
Plan: set all the 𝑥’s equal to zero.
Solution:
𝑦 = −2(0)2 + 8 0 − 6 = −6
So the point 0, −6 is the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡.
The 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is also simply the value of the
constant 𝑐.
To find the 𝑥– 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(𝑠), set all the 𝑦’s equal
to zero. In the example above, this would be:
0 = −2𝑥2 + 8𝑥 − 6
Rearranging this to isolate 𝑥 is very difficult, so a
formula is used to solve for 𝑥. The formula is called
the Quadratic Formula (QF) which you must
memorise.
𝑥
𝑦
𝑥
𝑦
𝑦 = −𝑥2 + 4𝑥
𝑦 = −𝑥2 − 2𝑥
𝑥
𝑦
𝑦 = 𝑥2 + 4𝑥
𝑦 = 𝑥2 − 2𝑥
𝑥
𝑦
78
Theory: the Quadratic Formula is:
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
This is a very important formula, and it may be
easier to memorise the words “negative 𝑏 plus or
minus the square root of 𝑏 squared minus four 𝑎𝑐,
all over two 𝑎”. This formula is finding where a
quadratic function crosses the 𝑥 − 𝑎𝑥𝑖𝑠 and these
point(s) are called roots.
Example 2: determine the roots of
0 = −2𝑥2 + 8𝑥 − 6
Plan: use the quadratic formula
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
Solution:
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 =−8 ± 82 − 4 −2 (−6)
2(−2)
=−8 ± 64 − 48
−4
=−8 ± 16
−4
=−8 ± 4
−4→ 𝑥 = 1 𝑂𝑅 𝑥 = 3
The last step is split into two. Instead of having the
± sign, it is separated out into a + and a − to get
two answers:
𝑥 =−8 + 4
−4=
−4
−4= 1 𝑂𝑅
−8 − 4
−4=
−12
−4= 3
So there are two 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠: (1,0) and (3,0).
The only point that remains to be found to
determine the nature of a quadratic is the turning
point (the point where the function “turns”). If you
memorised the quadratic formula, it makes finding
the turning point (TP) easier. Remembering that a
point has an 𝑥 and 𝑦 value, so once we find the
𝑥 − 𝑣𝑎𝑙𝑢𝑒, the corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒 needs to
be found.
Theory: to find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point
(TP), rewrite the Quadratic Function and only look
at the things not under the square root sign (i.e.
only look at the circled part):
Then to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the turning point,
substitute this 𝑥 − 𝑣𝑎𝑙𝑢𝑒 into the original
function.
Example 3: find the coordinates of the turning
point of the quadratic
𝑦 = −2𝑥2 + 8𝑥 − 6
Plan: use the Quadratic Formula ( but ignore
everything under the square root sign) to find the
𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point, then substitute
this value into the original equation to find the
𝑦 − 𝑣𝑎𝑙𝑢𝑒.
Solution:
𝑇𝑃 𝑥 − 𝑣𝑎𝑙𝑢𝑒 =−𝑏
2𝑎=
−8
2 −2 = 2
Substitute 𝑥 = 2 into the original equation to find
the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the turning point:
𝑦 2 = −2 2 2 + 8 2 − 6
= −8 + 16 − 6
= 2
Giving the turning point (2,2).
A different short-cut is knowing that the
𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point will always be half
way between the two 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠.
Example 4: Plot the results from Examples 1 − 3.
Plan: plot and label the coordinates, then join up
all the dots with a nice curve.
𝑇𝑃 𝑥 − 𝑣𝑎𝑙𝑢𝑒 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎=
−𝑏
2𝑎
79
Solution:
The dotted black line is what the graph should be,
and the blue line is a free–hand sketch. As long as
it approximately looks like a quadratic, and goes
through the labelled coordinates, then the curve
will be fine. Notice that this graph is a ‘sad’ face,
which from the last section, the value of 𝑎 must be
negative (𝑎 = −2). Also notice that the turning
point is half way between the two 𝑥 −
𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠.
On a related topic, on the following graphs notice
that the blue function cuts the 𝑥 − 𝑎𝑥𝑖𝑠 twice, the
red function only just touches the 𝑥 − 𝑎𝑥𝑖𝑠 (cuts it
once), and the black function does not cross the
𝑥 − 𝑎𝑥𝑖𝑠 at all (cuts it zero times).
Theory: From the Quadratic Formula, it is possible
to determine how many times a quadratic crosses
the 𝑥 − 𝑎𝑥𝑖𝑠. The value underneath the square-
root sign determines this:
𝑏2 − 4𝑎𝑐
If the value of 𝑏2 − 4𝑎𝑐 is positive, then the
curve will cross the 𝑥 − 𝑎𝑥𝑖𝑠 twice (as the
square root of a positive number will have two
solutions – a positive and negative number).
If the value of 𝑏2 − 4𝑎𝑐 equals zero, then the
curve crosses (touches) the 𝑥 − 𝑎𝑥𝑖𝑠 only once
(the square root of zero is zero).
If the value of 𝑏2 − 4𝑎𝑐 is negative, then the
curve does not cross the 𝑥 − 𝑎𝑥𝑖𝑠 (as the
square root of a negative number does not
exist).
Example 5: using the Quadratic Formula,
determine how many times the following equation
crosses the 𝑥 − 𝑎𝑥𝑖𝑠
𝑦 = 2𝑥2 − 16𝑥 + 32
Plan: find the value of the part under the square
root sign in the Quadratic Formula: 𝑏2 − 4𝑎𝑐.
If it is positive → 2 intercepts, if zero → 1 intercept,
and if negative → 0 intercepts.
Solution:
𝑏2 − 4𝑎𝑐 = −16 2 − 4 2 32
= 256 − 256
= 0 ∴ only one x − intercept
Example 6: Sketch 𝑦 − 8𝑥2 + 26𝑥 = −7
Plan: rearrange to get the general form, then find
the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 by setting all 𝑥’s to zero. Find
the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 by setting 𝑦 = 0 and then
applying the Quadratic Formula:
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
Use the part of the QF not under the square root
sign to find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the TP, then put back
into original equation to get the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the
TP.
Solution: rearrange into the general form
𝑦 = 8𝑥2 − 26𝑥 − 7
𝑦 − 𝑖𝑛𝑡: 𝑦 0 = 8 0 2 − 26 0 − 7
𝑦 = 𝑥2 + 4𝑥 + 8
𝑦 = 𝑥2 − 5
𝑥
𝑦 𝑦 = 𝑥2 + 2𝑥 + 1
(2,2)
(3,0)
(1,0)
(0, −6)
𝑦
𝑥
80
= −7
∴ 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑎𝑡 (0, −7)
𝑥 − 𝑖𝑛𝑡: 0 = 8𝑥2 − 26𝑥 − 7
𝑥 − 𝑖𝑛𝑡 𝑠 =−(−26) ± (−26)2 − 4 8 (−7)
2(8)
=26 ± 676 + 224
16
=26 ± 900
16
=26 ± 30
16→ 𝑥 = 3.5 𝑂𝑅 𝑥 = −0.25
Coordinates of 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 are: (3.5,0) and
−0.25,0 .
𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡:
𝑇𝑃 𝑥 − 𝑣𝑎𝑙𝑢𝑒 =−𝑏
2𝑎=
−(−26)
2(8)=
26
16= 1.625
𝑇𝑃 𝑦 − 𝑣𝑎𝑙𝑢𝑒 :
𝑦(1.625) = 8 1.625 2 − 26(1.625) − 7
= −28.125 ∴ 𝑇𝑃 𝑎𝑡 (1.625, −28.125)
Example 7: Sketch 3 𝑥 − 3 = 𝑥2 − 𝑦
Plan: set all 𝑥’s to zero to find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡,
then rearrange to get into general form, then use
the QF to find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 and 𝑇𝑃.
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
Solution: rearrange into the general form
3𝑥 − 9 = 𝑥2 − 𝑦
𝑦 = 𝑥2 − 3𝑥 + 9
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡:
𝑦 = (0)2 − 3 0 + 9 = 9
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 coordinate: (0,9)
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠:
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 =−(−3) ± (−3)2 − 4 1 (9)
2(1)
=3 ± 9 − 36
2=
3 ± −27
2
Since there is a negative number under the square
root sign, it means there are no 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠.
𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑃𝑜𝑖𝑛𝑡:
𝑥 − 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑇𝑃 = −𝑏
2𝑎= −
−3
2 1 =
3
2= 1.5
𝑦 − 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑇𝑃: 𝑦(1.5) = 1.5 2 − 3(1.5) + 9
= 𝑦 1.5 = 2.25 − 4.5 + 9 = 6.75
TP coordinate: (1.5,6.75)
Plot all the points, and remember, there are no
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 and it is a “smiley” face as 𝑎 is
positive:
Exercises 1. Find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 (if any) for the following
quadratics: 𝑎) 𝑦 = 𝑥2 + 2𝑥 − 15
𝑏) 𝑦 = 2𝑥2 − 20𝑥 + 42
𝑐) 𝑦 = 3𝑥2 + 18𝑥 + 27
𝑑) 𝑦 = 𝑥2 + 2𝑥 + 9
2. Sketch the following quadratics. 𝑎) 𝑦 = −𝑥2 − 5𝑥 − 6
𝑏) 𝑦 = 𝑥 + 2 2 − 1
𝑐) 𝑦 = 𝑥 − 1 2 + 𝑥 + 2 2
𝑑) 𝑦 = − −𝑥 + 1 1 − 𝑥 + 1
𝑒) 𝑦 = 𝑥2 + 𝑥 − 1 2 − 3
𝑓) 𝑦 = − 𝑥 − 2 2 − 2
4.5 the cubic function
Theory: The cubic function is a function where the
highest power of 𝑥 is 3 (i.e. 𝑥3).
For example:
𝑦 = 2𝑥3 − 4𝑥2 + 3𝑥 − 17
(1.5,6.75)
(0,9)
𝑦
𝑥
(1.625, −28.125)
(3.5,0) (−0.25,0)
(0, −7)
𝑦
𝑥
81
Notice that there is an 𝑥2, but the highest power is
3 (i.e. the 2𝑥3) so this is a cubic function, not a
quadratic function.
Theory: The general form of a cubic function is:
𝑦 = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑
Where 𝑎, 𝑏, 𝑐 and 𝑑 are constants.
The simplest cubic is:
𝑦 = 𝑎𝑥3
where 𝑏, 𝑐 and 𝑑 are all zero. These simple cubic
functions have an ‘𝑆’ shape:
Here, there is only one 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 (through the
origin), and unlike the quadratic functions, the
‘ends’ of the graph go in different directions. That
is, as 𝑥 becomes more positive (i.e. to the right of
the origin on the 𝑥 − 𝑎𝑥𝑖𝑠), the value on the
𝑦 − 𝑎𝑥𝑖𝑠 also becomes more positive; similarly,
when the value of 𝑥 becomes more negative (to
the left of the origin on the 𝑥 − 𝑎𝑥𝑖𝑠), the value on
the 𝑦 − 𝑎𝑥𝑖𝑠 also becomes more negative.
Compare this to a quadratic where both ends of
the graph go in the same direction (either both
ends go up, or both ends go down).
More complex cubic functions look like:
These functions still have the general ‘𝑆’ shape; it
is the values of 𝑎, 𝑏, 𝑐 and 𝑑 that determine the
sharpness of this ‘𝑆’ shape. The blue graph is much
‘sharper’ than the black one, which is ‘sharper’
than the red one.
Also, count how many times each colour crosses
the 𝑥 − 𝑎𝑥𝑖𝑠. The blue one crosses three times,
the black one crosses twice (touching the 𝑥 − 𝑎𝑥𝑖𝑠
counts as a cross), and the grey curve cuts the
𝑥 − 𝑎𝑥𝑖𝑠 only once.
Theory: depending on the values of 𝑎, 𝑏, 𝑐 and 𝑑, a
cubic can cross the 𝑥 − 𝑎𝑥𝑖𝑠 once, twice or three
times.
Just like in quadratics, the sign of 𝑎 in a cubic
function has an interpretation.
Theory: when 𝑎 is positive, the ends of the graph
extend to the top right and bottom left ↙↗ , but
when 𝑎 is negative, the ends of the graph extend
to the top left and bottom right ↖↘ .
The reason why 𝑎 is important is that as 𝑥
becomes very large in a positive direction or in a
negative direction, the cubed term dominates over
all other terms.
The number of times a cubic function crosses the
𝑥 − 𝑎𝑥𝑖𝑠 is determined by the values of 𝑎, 𝑏, 𝑐 and
𝑑, with the sign of 𝑎 determining the direction the
ends of the graph extends. Notice the sign of 𝑎 in
𝑦 = 𝑥3 + 2
𝑦 = 𝑥3 − 𝑥2 − 21𝑥 + 45
𝑦
𝑥
𝑦 = 𝑥3 − 𝑥2 − 20𝑥
𝑦 = 2𝑥3
𝑦 = 𝑥3
𝑦 = 0.5𝑥3
𝑦
𝑥
82
the following graph and the direction the ends
extend.
Exercises 1. Match the following graphs to the equations:
𝑎) 𝑦 = 𝑥 + 4 𝑥 + 2 2 + 𝑥 𝑏) 𝑦 = 𝑥3 + 4 𝑐) 𝑦 = −3𝑥3 + 2.5 𝑑) 𝑦 = 𝑥 − 3 𝑥 + 1 −3 − 𝑥
Hint: expand all the brackets before deciding.
4.6 the exponential function
Before defining the mathematical exponential
function, an intuitive understanding is needed. You
are given $100 to go to the casino and have fun.
You play the coin-toss, and (somehow) can predict
the outcome every single time. After the first
round, you win $100. You then have $200, and
bet it again. You win, and now you have $400. You
bet it all again, and win again. Now you have $800.
You bet it once more and win, and end up with
$1600. Every time you play, you double your
earnings, but you don’t win the same amount
every time. The first round you won $100, and the
last round you won $800, which are not the same,
but every round, you doubled your earnings. This is
what exponential growth means. After every
round (or time-period), the final amount is a
certain ratio of the amount of the previous time-
period.
In the example above, the amount of money (𝑚)
after 𝑡 turns (assuming you keep winning) can be
defined by:
𝑚 = 100 2𝑡
The 2 comes from the fact that earnings are
doubled every time-period, and the 100 from the
initial amount.
Theory: an exponential function is one which has a
variable in the index. The general form of an
exponential function is
𝑦 = 𝑏(𝑎𝑥) + 𝑐
Where 𝑎, 𝑏 and 𝑐 are constants.
𝑎 is a number that determines the growth rate
(e.g. doubling;𝑎 = 2, tripling;𝑎 = 3 etc.)
𝑏 is the initial amount. It determines the shape
of the graph and where it intersects the
𝑦 − 𝑎𝑥𝑖𝑠.
𝑐 a constant that shifts the graph up or down,
but doesn’t change the shape of the graph. It is
also the value of the asymptote (the value the
function will approach but will never reach).
The simplest exponential is where 𝑏 = 1 and 𝑐 =
0:
𝑦 = 𝑎𝑥
This exponential function cuts the 𝑦 − 𝑎𝑥𝑖𝑠 at
0,1 as any number to the power of 0 equals 1.
𝑦
𝑥
𝑦 = −𝑥3 + 5
𝑦 = −𝑥3 − 𝑥2 + 21𝑥 + 45 𝑦
𝑥
83
The shape of an exponential function:
The black line is the simplest exponential function,
and crosses the 𝑦 − 𝑎𝑥𝑖𝑠 at 𝑦 = 1, as 20 = 1. It is
also the value of 𝑏 (= 1). The asymptote is at
𝑥 = 0 as the further left the curve extends, the
closer it gets to the 𝑥 − 𝑎𝑥𝑖𝑠 (i.e. 𝑦 = 0).
The red line has the 3 out front, meaning that it
crosses the 𝑦 − 𝑎𝑥𝑖𝑠 at 𝑦 = 3 but also that the
general shape has been changed to be steeper.
Similar to the black line, the asymptote is at 𝑥 = 0.
The blue line is the same as the black line, except
that it has 3 subtracted from it. This shifts the
graph down 3 units and also changes the
asymptote to 𝑦 = −3. The graph will get closer
and closer to 𝑦 = −3 but will never reach it. The
other two functions have 𝑐 = 0 which means the
asymptote for those two functions is the 𝑥 − 𝑎𝑥𝑖𝑠.
Theory: to sketch an exponential function, the
𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 and the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the
asymptote are required (remember, an asymptote
is a value that the function approaches but never
reaches).
The 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is found by setting all 𝑥’s
equal to zero.
The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the asymptote is any
number added onto the end of the
exponential function (i.e. the number 𝑐).
Example 1: sketch the two exponential functions
on a single set of axes
1. 𝑦 = 3 2𝑥 + 4
2. 𝑦 = 0.5 2−𝑥 − 2
Plan: to find the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠, set the
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 to zero. The asymptote is the number
added to the end of the exponential.
Solution: for the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠
1. 𝑦 = 3 20 + 4 = 3 + 4 = 7
2. 𝑦 = 0.5 20 − 2 = −1.5
For the asymptotes
1. 𝑦 = 4
2. 𝑦 = −2
Notice that for the first function above, there is a 3
in front which makes the graph much steeper.
Conversely, the 0.5 in the second function makes it
shallower. There is also a negative sign in front of
𝑥 in the second function above meaning it is
flipped along the 𝑦 − 𝑎𝑥𝑖𝑠.
A concept similar to growth is that of decay; the
reverse of growth. For growth, the more time
passes, the faster the thing grows. In decay, the
more time passes, the slower things decay.
Theory: the general form of the decay function is
very similar to that of growth, except there is a
negative in front of the index x:
𝑦 = 𝑏(𝑎−𝑥) + 𝑐
𝑦 = 3 2𝑥 + 4
𝑦 = 0.5 2−𝑥 − 2
𝑦
𝑥
𝑦 = 4
𝑦 = −2
𝑦 = 3(2𝑥)
𝑦 = 2𝑥
𝑦 = 2𝑥 − 3
𝑦
𝑥
0,1
0,3
0, −2
84
The graph which was shown in the last exercise is
one of decay, as are the following:
Exercises: 1. Match the following graphs with the equations:
𝑎) 𝑦 = 3 2𝑥 𝑏) 𝑦 = 3 2−𝑥 𝑐) 𝑦 = 5 3𝑥 𝑑) 𝑦 = 4 2−𝑥 + 5
2. Sketch the following equations on a single set of axes (they must be correct, relative to one another!):
𝑎) 𝑦 = 5 2𝑥 − 4 𝑏) 𝑦 = 2−𝑥 + 3 𝑐) 𝑦 = −2 3𝑥 + 1 𝑑) 𝑦 = 4 2−𝑥 + 5
4.7 the logarithmic function
For an exponential function such as:
𝑦 = 𝑎𝑥
𝑦 is already isolated, but sometimes 𝑥 needs to be
isolated. It is impossible to isolate 𝑥 without using
logarithms. A logarithm (log for short) is a
mathematical method used to solve exponents.
Intro Example 1: use common sense to solve:
10𝑥 = 10
Obviously 𝑥 = 1. Now how about:
10𝑥 = 100
The answer is 𝑥 = 2, as 102 = 100. These are
easy, but how about:
10𝑥 = 50
The answer is not 𝑥 = 1.5 because 101.5 ≈ 31.6. It
is exactly this sort of situation where logs are used.
Theory: the definition of a logarithm is:
For 𝑎𝑏 = 𝑐
𝑏 =log 𝑐
log 𝑎= log𝑎 𝑐
Where 𝑏 is called the index, and 𝑎 is called the
base. However, with the following log rules, this
definition can be easily found.
There are some log rules you will have to learn to
be able to solve these problems. Without knowing
these rules, you will make lots of mistakes. Back to
the previous problem:
10𝑥 = 50
Theory: RULE 1: taking logs of the whole of both
sides separately allows the index to be brought
down in front of the log.
Example 1: solve for 𝑥 in
10𝑥 = 50
Solution: log the whole of both sides
log 10𝑥 = log 50
Then bring down any exponents
𝑥 log(10) = log(50)
This is read as “𝑥 × log(10) is equal to log(50)”.
log(10) and log(50) are just numbers (the values
can be found using a calculator). Treat it as you
would any other number. To isolate 𝑥 in:
5𝑥 = 15
Divide both sides by 5. It is the same for
𝑥 log(10) = log(50)
log 10𝑥 = log 50
log 𝑎𝑏 = 𝑏 log 𝑎
𝑦
𝑥
𝑦 = 3 2−𝑥 + 1
𝑦 = 2−𝑥 𝑦
𝑥
𝑦 = 1
85
Divide both sides by log(10):
𝑥 = log(50)
log(10)
That is the answer.
Always give this exact answer, but you can also
give a decimal answer; put it into your calculator
and you should get 1.69897 (5𝑑. 𝑝. ).
The place where most students fail is not realising
that something has to be logged. The logarithm
changes one thing to another. If you log nothing,
then how can it be changed? When travelling in
Europe, dollars need to be exchanged for Euros,
however if you give the bank nothing, nothing will
be exchanged! However, if you give the bank
$100, then that $100 will be changed to Euros.
The same applies to logs; you have to log
something!
Example 2: solve for 𝑥
73 3 𝑥 = 4
Solution: You could divide both sides by 73 and
then apply RULE 1. But if both sides were not
divided by 73 to begin with, and instead logged:
log[ 73 3 𝑥 ] = log[4]
The square brackets on the left is 73 × 3𝑥 , so the
index of 𝑥 cannot be brought down front. It is not
73 × 3 𝑥 , but rather two separate parts being
multiplied together (see Chapter 1).
Theory: RULE 2: the log of a multiplication of two
terms is equal to the addition of the logs of the
separate terms.
log(𝑎 ∙ 𝑏) = log 𝑎 + log 𝑏
Example 2 (cont): solving for 𝑥 in
log[ 73 3 𝑥 ] = log[4]
Solution: by RULE 2, 73 and 3𝑥 can be separated
into two logs, with an addition sign between:
log 73 + log 3𝑥 = log 4
Move the log 73 onto the other side by subtracting
it from both sides to get:
log 3𝑥 = log 4 − log 73
RULE 1 can now be applied to bring 𝑥 out front on
the left side:
𝑥 log 3 = log 4 − log 73
To solve for 𝑥, divide the whole of both sides by
log 3:
𝑥 =log 4 − log 73
log 3≈ −2.6435 (4𝑑. 𝑝. )
Finally, the last rule is similar to RULE 2, but it is for
division instead of multiplication:
Theory: RULE 3: the log of a division of two terms
is equal to the subtraction of the separate logs.
log 𝑎
𝑏 = log 𝑎 − log 𝑏
Example 3: solve for 𝑥 without simplifying first
4𝑥
56= 12
Solution: log the whole of both sides:
log 4𝑥
56 = log 12
Use RULE 3 to separate the 4𝑥 and the 56 by
subtracting their logs:
log 4𝑥 − log 56 = log 12
Use RULE 1 to bring 𝑥 out front and rearrange:
𝑥 log 4 = log 12 + log 56
Use RULE 2 in reverse on the right side:
𝑥 log 4 = log 12 × 56
Rearrange and simplfy:
𝑥 =log 12 × 56
log 4 =
log 672
log 4 ≈ 4.6962 (4𝑑. 𝑝. )
Theory: all rules can be used forwards or
backwards. It all depends on what is being found.
Think about it; if RULE 2 can be used to separate
the log of a multiplication into two log additions,
then there is no reason why it cannot be used to
86
bring two log additions into a single log
multiplication. Practice with your calculator to
make sure. The same applies to both RULE 1 and
RULE 3.
Note: when given:
log 672
log 4
many students think they can bring everything
inside of a single log:
log 672
log 4 = log
672
4
This is not correct! Never write this! When two
logs are divided, they cannot be simplified further.
Similarly, if two logs are multiplied, they cannot be
simplified further.
One last rule is:
Theory:
RULE 4a: log(1) = 0, which is like saying 𝑦0 = 1.
RULE 4b: log 10 = 1 (ln 𝑒 = 1; this will be
explained later in the chapter)
Remember that you can use these rules backwards
or forwards. It all depends on what is being
isolated and where it is.
Example 3: a country’s Gross Domestic Product
(𝐺𝐷𝑃) has the following form (in $billions):
𝐺𝐷𝑃 = 4.3 1.1𝑡
When will 𝐺𝐷𝑃 reach $7billion?
Plan: substitute 7 for the left side in the above
equation (as 𝐺𝐷𝑃 is set to $7b), then solve using
log rules.
Solution:
7 = 4.3 1.1𝑡
This could be solved in a number of ways, but the
following is the harder way. Try find the short-cut.
Log the whole of both sides:
log 7 = log 4.3 1.1𝑡
Separate the right side into two logs using RULE 2:
log 7 = log 4.3 + log(1.1𝑡)
Move log 4.3 to the other side, and bring 𝑡 down
using RULE 1:
log 7 − log 4.3 = 𝑡 log(1.1)
Using RULE 3, bring the left side under a single log:
log 7
4.3 = 𝑡 log(1.1)
Then divide both sides by log 1.1:
𝑡 =log
74.3
log 1.1≈ 5.113 𝑦𝑒𝑎𝑟𝑠 (3𝑑. 𝑝. )
The short-cut is to simplify first, then use RULE 1.
However, there are times where you have to use
all the rules, so learn them.
Example 4: inflation is the growth in prices. The
estimated inflation rate, relative to the base year
0, is determined by the following function:
𝑝 𝑡 = 1.01 𝑡 1.02 𝑡−1 +1
51
The government wants to find out when the price
level will double.
Plan: when the price level has doubled, 𝑝 𝑡 = 2.
Substitute this into the equation above, and use
log rules to solve for 𝑡.
Solution:
2 = 1.01 𝑡 1.02 𝑡−1 +1
51
Rearrange before applying any log rules:
2 −1
51= 1.01 𝑡 1.02 𝑡−1
Log the whole of both sides:
log 2 −1
51 = log[ 1.01 𝑡 1.02 𝑡−1]
Using RULE 2, separate the square bracket into
two separate logs with an addition sign in
between:
log 2 −1
51 = log 1.01 𝑡 + log 1.02 𝑡−1
Use RULE 1 to bring down the two exponents in
brackets:
87
log 2 −1
51 = 𝑡 log 1.01 + (𝑡 − 1) log 1.02
Half crab-claw the log 1.02 into the bracket:
log 2 −1
51 = 𝑡 log 1.01 + 𝑡 log 1.02 − log 1.02
Move the log 1.02 to the other side:
log 2 −1
51 + log 1.02 = 𝑡 log 1.01 + 𝑡 log 1.02
Factorise out the 𝑡 on the right side:
log 2 −1
51 + log 1.02 = 𝑡 log 1.01 + log 1.02
Divide both sides by the blue part:
log 2 −
151
+ log 1.02
log 1.01 + log 1.02 = 𝑡
Simplify the numerator (top) with RULE 2, and the
denominator (bottom) also with RULE 2:
𝑡 =log 1.02 2 −
151
log[ 1.01 1.02 ]≈ 23.63 𝑦𝑒𝑎𝑟𝑠 (2𝑑. 𝑝. )
This seems very complex, but only use rules when
you are trying to do something specific. That is,
isolating 𝑡 requires RULE 1, separating out two
parts which are logged requires RULE 2. Always
think about what you are trying to do as well as
the final outcome, and then decide which rules are
needed.
Theory: the logarithmic function log is usually used
to the base 10. There are many bases, but the
most commonly used bases are 10 and 𝑒:
log10 𝑋
log𝑒 𝑋
When a logarithm to the base 𝑒 is used, it is called
ln (the natural logarithm). All the same log rules
apply, but it is easier to use ln when working with
exponential functions involving 𝑒. NOTE: see
Section 4.9 below for a definition of 𝑒. For now,
treat it as just another number.
Example 5: An economist has come up with an
equation that tells how many cars (in thousands)
Nissan will be producing at a given year into the
future
𝐶 𝑡 = 48𝑒0.03𝑡 + 2
The head of Nissan wants you to find out when:
a) Output will break the 80,000 cars/year level.
b) Output will double.
Plan:
a) replace 𝐶(𝑡) with 80, and solve using log rules,
but instead of log, use ln (as the base is 𝑒).
b) Find out the number of cars at 𝑡 = 0 (i.e. the
starting point), then find 𝑡 when this will
double.
Solution:
a) Replace 𝐶 𝑡 with 80 and move the 2 to the
other side:
80 − 2 = 48𝑒0.03𝑡
Take the natural logarithm (ln) of both sides:
ln 78 = ln 48𝑒0.03𝑡
Since the square brackets have two things being
multiplied, RULE 2 will need to be used:
ln 78 = ln 48 + ln 𝑒0.03𝑡
Move ln 48 to the left side, and using RULE 1, bring
the 0.03𝑡 down in front:
ln 78 − ln 48 = 0.03𝑡 ln 𝑒
Use RULE 4b, ln 𝑒 = 1, so:
ln 78 − ln 48 = 0.03𝑡
Use RULE 3 to bring the two logs on the left into
one log, and then divide both sides by 0.03:
ln
7848
0.03= t ≈ 16.18 𝑦𝑒𝑎𝑟𝑠 2𝑑. 𝑝.
Determine how many cars are produced currently:
𝐶 0 = 48𝑒0.03 0 + 2 = 50
So 50 doubled is 100:
100 = 48𝑒0.03𝑡 + 2
98 = 48𝑒0.03𝑡
88
98
48= 𝑒0.03𝑡
ln 98
48 = 0.03𝑡 ln 𝑒
ln
9848
0.03= 𝑡 ≈ 23.79𝑦𝑒𝑎𝑟𝑠 (2𝑑. 𝑝. )
In a problem with 𝑒, ln is used because ln 𝑒 can be
cancelled off (as ln 𝑒 = 1). If instead, both sides
were logged, the result would be log 𝑒 and this is
NOT equal to 1. If you used log instead, you can
still get the same answer, however a lot more
work is involved. As a general rule, use ln as it
allows manipulation using all the same log rules.
Remember to be consistent; if you use a log, keep
using log in that question; if you use ln for a given
problem, keep using ln throughout that problem.
Exercises: 1. Solve for 𝑥 leaving answers in the simplest form:
𝑎) 15𝑥 = 10
𝑏) 15𝑥+1 = 10
𝑐) 8 2𝑥 = 3𝑥−1
𝑑) 25 5𝑥 = 9 1𝑥5𝑥+1
𝑒) 2𝑥16𝑥+1 = 31 3𝑥
𝑓) ln(53) − ln(25) = 𝑥 ln 5
𝑔) 3𝑥2+2𝑥 = 9 3𝑥
) 8𝑥2−2𝑥 = 323−𝑥
2. Solve for 𝑥 in terms of 𝑦 in the following: 𝑎) 𝑦 = 15𝑥 + 18
𝑏) 𝑦 = 2𝑥𝑦3𝑥
𝑐) 𝑦 = 13𝑥2𝑥+132𝑥−1
𝑑) 𝑦 = 2𝑥232
𝑒) 3𝑦 + 42 = 243𝑥+1
3. If a country’s GDP (in $billion) grows according to the following equation:
𝐺𝐷𝑃𝑡 = 700 1.2 𝑡 Determine when GDP will reach $1200𝑏𝑖𝑙𝑙𝑖𝑜𝑛.
4. The government has employed you to study the inflation within a particular city, and provide future forecasts. You determine that the historic inflation rate can be accurately estimated using:
𝑝 𝑡 = 1.035 𝑡−1 +7
207
a) Forecast the price level in 4 years. b) Determine how long (in years) it will take for
prices to double. c) Determine how long (in years) it will take
prices to triple. 5. A manufacturer of laptop computers forecasts
costs will rise according to the equation: 𝐶 𝑡 = 1.03𝑒0.02𝑡 − 0.03
a) Determine the cost level in 5 years. b) Determine when costs will double.
4.8 logarithmic graphs
A log function has the general shape:
Theory: the key aspects of this graph are the
general shape of a lower case ‘r’, and that it is
asymptotic (approaches but never touches) to a
particular 𝑥 − 𝑣𝑎𝑙𝑢𝑒.
The graph comes from the equation of the general
form:
𝑦 = 𝑎 log(𝑥 + 𝑏) + 𝑐
Where 𝑎 determines the sharpness of the ‘r’
shape, 𝑏 shifts the whole graph left or right, and 𝑐
shifts of the whole graph up or down.
As you can see above, the grey line is exactly the
same as the black line in shape, except it has been
shifted right. The red line is also exactly the same
as the black one, except it has shifted vertically.
The blue line shows how 𝑎 determines the
“sharpness” of the curve.
Sketching log graphs is not as important as
knowing the four log rules and their applications,
so it will not be covered any further.
𝑦
𝑥
𝑦 = log 𝑥
𝑦 = log 𝑥 − 4
𝑦 = 5log 𝑥
𝑦 = log 𝑥 + 4
𝑦
𝑥
89
Exercises: 1. Match the graphs to the equations.
𝑎) 𝑦 = 4 log(𝑥 − 2)
𝑏) 𝑦 = log(𝑥 − 4)
𝑐) 𝑦 = log 𝑥 + 4
𝑑) 𝑦 = 2 log(𝑥)
2. Sketch the following functions on the same set of axes (make sure they are correct relative to each other):
𝑎) 𝑦 = log(𝑥) − 3
𝑏) 𝑦 = log(𝑥 + 2) + 2
𝑐) 𝑦 = 3 log 𝑥
𝑑) 𝑦 = 2.5 log(𝑥) + 1
4.9 the natural number 𝒆
This section is all about a special number. This
number is called 𝑒 and is approximately equal to
2.7183 (4d.p.). It is like the number pi (𝜋), which is
used for circles, but 𝑒 is used to describe things
that are continuous (i.e. things that happen all the
time).
For example, a bank offering continuous
compounding of any savings will have to use the
number 𝑒, as say $1000 in an account must be
compounded all the time.
That is one situation of growth, but 𝑒 works just as
well in decay; business examples of continuous
decay are not as common as growth and usually
very complex.
Theory: any time 𝑒 is used, the variable in question
must be changing continuously. In very large
populations, 𝑒 can be a good approximation of
growth and decay.
A simple example will demonstrate this theory: if a
colony of ants doubles every year, which of the
following is occurring:
1. The population of the colony is constant
throughout the year, and then on December
31st, it suddenly doubles.
2. The population of the colony constantly
increases throughout the year, and at
December 31st, it just reaches the doubling
point.
The colony of ants obviously grows throughout the
year, and it is such a large colony that we can
assume that it grows continuously. It is in this sort
of situation where the number 𝑒 is used.
Theory: the growth of anything continuous has the
general form:
𝐴𝑡 = 𝐴0𝑒𝑟𝑡
In words; the amount available at some time 𝑡 in
the future 𝐴𝑡 is equal to the initial amount (𝐴0)
multiplied by the continuous growth base (𝑒) to
the power of the rate of growth per time period
multiplied by the number of time periods (𝑟𝑡).
This equation is very similar to the exponential
growth rate equation.
Example 1: the 𝐺𝐷𝑃 of England is estimated to be
growing at 2.3% per year. Currently, the 𝐺𝐷𝑃 is
£4trillion. Estimate the 𝐺𝐷𝑃 in 10 years time.
Plan: 𝐺𝐷𝑃 can be assumed to grow continuously
as the citizens of England are constantly earning
money throughout the year, so the 𝑒 growth
equation can be used. Substitute the growth rate,
time and initial amount into:
𝑦
𝑥
90
𝐴𝑡 = 𝐴0𝑒𝑟 𝑡
then solve for 𝐴𝑡 , which is the 𝐺𝐷𝑃 in ten years.
Solution: replace the known variables in the
exponential growth rate equation:
𝐴10 = 4𝑒 0.023 10
The trillion has been left out, but will be brought
back at the end. Also, 0.023 has been used instead
of 2.3%. The reason is that for use in most
equations, the growth rate must be in decimal
form, not percentage form.
𝐴10 = 4𝑒0.23
𝐴10 = £5.03 𝑡𝑟𝑖𝑙𝑙𝑖𝑜𝑛 (2. 𝑑. 𝑝. )
Example 2: Woolworths, a large company, has a
growth in customers across it’s 400 stores in
Australia at a rate of 2.6% per year. If they
currently have 3.67million customers,
approximately how many years will it take for the
customer base to grow to 5million?
Plan: since 400 stores is a large number of stores,
and the growth of a population is accurately
approximated by continuous growth, the 𝑒 growth
formula can be used:
𝐴𝑡 = 𝐴0𝑒𝑟 𝑡
Substitute in all known information, then
rearrange to solve for 𝑡.
Solution: 𝐴0 = 3.67, 𝑟 = 0.026 and 𝐴𝑡 = 5, so the
equation is:
5 = 3.67𝑒0.026𝑡
Divide both sides by 3.67:
5
3.67= 𝑒0.026𝑡
Take the natural log (ln) of the whole of both
sides:
ln 5
3.67 = ln 𝑒0.026𝑡
ln 5
3.67 = 0.026𝑡
𝑡 =ln
53.67
0.026≈ 11.89𝑦𝑒𝑎𝑟𝑠 2𝑑. 𝑝.
Exercises: 1. Solve for 𝑥 in the following:
𝑎) 15 = 𝑒𝑥 + 1
𝑏) 15 = 𝑒𝑥+1
𝑐) 170 = 12𝑒0.1+𝑥
𝑑) 13 = 6.5𝑒0.03𝑥
𝑒) 21 = 10.5𝑒0.027𝑥 − 1
2. If 𝐺𝐷𝑃 of Fiji grows at a rate of 4.1%p.a. and is currently $50𝑏𝑖𝑙𝑙𝑖𝑜𝑛, determine:
a) The GDP in 5 years. b) How long it takes for the GDP to double. c) How long it takes for the GDP to triple.
3. A dishonest bank lends you $5000 to purchase a car, at a rate of 4%p.a. The fine-print stated that the amount is compounded every second. Determine the amount you will have to pay back in 3 years.
4. An investment portfolio is estimated to grow continuously for 7 years at 6.5% p.a. and then at 8.3% p.a. every year after that. If the initial investment is $1,000, determine: a) when the investment will double in value. b) how many years until the portfolio reaches a
value of $5,000. c) The value of the portfolio after 8 years, 312
days.
4.10 the hyperbolic function
Despite hyperbolic functions not being very
common, the basics are still important. You have
probably seen the function:
𝑦 = 1
𝑥
This is the simplest hyperbola. What makes
hyperbolas special is that they have two
asymptotes. Remember that an asymptote is a line
to which the function approaches but never
reaches. The following is the graph of this simplest
hyperbola:
𝑦
𝑥
𝑦 =1
𝑥
91
In this case, the hyperbola (the two blue lines
constitute a single function) has two asymptotes;
the 𝑥 − 𝑎𝑥𝑖𝑠 and the 𝑦 − 𝑎𝑥𝑖𝑠. For the 𝑦 − 𝑎𝑥𝑖𝑠
asymptote, the closer the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 get to zero
(trace it with your fingers by going towards the
origin), the greater the value of the function on the
𝑦 − 𝑎𝑥𝑖𝑠 becomes (but it never touches the
𝑦 − 𝑎𝑥𝑖𝑠). Similarly, the second asymptote is the
𝑥 − 𝑎𝑥𝑖𝑠 and the greater the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 become
(both in a positive or negative direction) the closer
the function gets to the 𝑥 − 𝑎𝑥𝑖𝑠, but never
touches it.
Theory: the general form of a hyperbola is:
𝑦 = 𝑑
𝑎𝑥 + 𝑏+ c
The denominator (𝑎 and 𝑏) determines the
locations of the vertical asymptote. That is,
setting the whole denominator (𝑎𝑥 + 𝑏) equal
to zero and solving for 𝑥 is how the vertical
asymptote is found.
The value of 𝑐 determines the location of the
horizontal asymptote.
The value of 𝑑 determines how close the curve
gets to the intersection of the two asymptotes.
To make this last point clear, the following graph
shows the same hyperbolic function, with different
values of 𝑑.
The larger the value of 𝑑 (red), the closer to the
intersection of the asymptotes the graph will lie.
The smaller 𝑑 is (blue), the further from the
intersection of the asymptotes the graph will lie.
Example 1: sketch the following function
𝑦 = 1
2𝑥 + 3+ 5
Plan: for the vertical asymptote, set the
denominator equal to zero and solve for 𝑥. The
horizontal asymptote is the number added on to
the fraction.
Solution: for the vertical asymptote:
2𝑥 + 3 = 0
2𝑥 = −3
𝑥 = −1.5
For the horizontal asymptote, 𝑦 = 5. Lightly draw
the vertical and horizontal asymptotes, then draw
in the general shape of a hyperbola.
The dotted vertical line crosses the 𝑥 − 𝑎𝑥𝑖𝑠 at
𝑥 = −1.5, and the horizontal line 𝑐 = 5.
Exercises:
1. Match the following graphs to the equations:
𝑦 =2
2𝑥 + 4 𝑦 = −5 −
1
−6 + 𝑥
𝑦
𝑥
𝑦
𝑥
𝑦 = 5
𝑥 = −1.5
𝑦 =1
2𝑥 + 3+ 5
𝑦 =3
𝑥
𝑦 =1
𝑥
𝑦
𝑥
92
𝑦 =3
3𝑥 − 9+ 3 𝑦 = 3 +
3
−3 + 𝑥
2. Sketch the following functions:
𝑎) 𝑦 =1
2𝑥− 1
𝑏) 𝑦 =2
𝑥 − 1+ 2
𝑐) 𝑦 =1
𝑥
𝑑) 𝑦 = −1
𝑥
4.11 applications
Economic theory: 𝐺𝐷𝑃 per capita is a measure of
the mean income per person in a particular
country. It is defined as:
𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 =𝐺𝐷𝑃
𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛
Example 1: a developing country like Papua New
Guinea (PNG) has a 𝐺𝐷𝑃 of 𝑈𝑆$60billion with a
growth of 4% per year. PNG has a population of
7million, which grows at 1.9%. When will 𝐺𝐷𝑃 per
capita reach 𝑈𝑆$15,000? Answer in years and
days.
Plan: 𝐺𝐷𝑃 and populations are both very large so
can be assumed to grow at an exponential rate:
𝐴𝑡 = 𝐴0𝑒𝑟 𝑡
Two equations are needed; one for 𝐺𝐷𝑃 growth
and one for population growth, as both grow at
different rates. These are then combined to get
𝐺𝐷𝑃 per capita.
Solution: the growth rate for 𝐺𝐷𝑃 is:
𝐺𝐷𝑃𝑡 = 𝐺𝐷𝑃0𝑒𝑟1𝑡
Substitute what is known:
𝐺𝐷𝑃𝑡 = 60 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑒0.04𝑡
The population growth rate is:
𝑃𝑜𝑝𝑡 = 𝑃𝑜𝑝0𝑒𝑟2𝑡
𝑃𝑜𝑝𝑡 = 26 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑒0.019𝑡
Here, 𝑟1and 𝑟2 have been used as 𝐺𝐷𝑃 and
population grow at different rates. Combining the
two formulas, 𝐺𝐷𝑃 per capita is:
𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 =𝐺𝐷𝑃𝑡
𝑃𝑜𝑝𝑡
𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 =60 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 𝑒0.04𝑡
26 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑒0.019𝑡
Simplify the 60billion/26million
(billion/million=thousand):
𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 =60,000 𝑒0.04𝑡
26𝑒0.019𝑡
Simplify further using index rules (Chapter 1):
𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 =60,000 𝑒0.04𝑡−0.019𝑡
26
=60,000 𝑒0.21𝑡
26
This is the 𝐺𝐷𝑃 per capita function, but what is
required is when 𝐺𝐷𝑃 per capita will reach
𝑈𝑆$15,000. Set 𝐺𝐷𝑃 per capita to 15,000, and
solve:
15,000 =60,000 𝑒0.21𝑡
26
390,000
60,000= 𝑒0.21𝑡
6.5 = 𝑒0.21𝑡
Take logs of the whole of both sides:
ln 6.5 = ln(𝑒0.21𝑡)
Simplify:
ln 6.5 = 0.21𝑡 ln 𝑒
Remember ln 𝑒 = 1:
𝑡 =ln 6.5
0.21 ≈ 8.91𝑦𝑒𝑎𝑟 (2𝑑. 𝑝. )
Finally, to get the solution in years and days, take
the decimal (i.e. 0.91) and multiply it by the
number of days in a year (assume 365) to get
𝑡 = 8 𝑦𝑒𝑎𝑟𝑠, 333 𝑑𝑎𝑦𝑠 (to the nearest day).
Example 2: a company that takes care of its
employees spends 𝑆 dollars per year on the well-
93
being of each employee, where 𝑆 is related to
profits (in millions of dollars), by the function:
𝑆 = 2500 1 − 𝑒−0.1𝜋
Determine:
a) The spending on each employee when profits
are $5million.
b) The profit level when spending on each
employee will double from that in part a).
c) The profit level when spending on each
employee is $2,000.
Plan:
a) Set 𝜋 to 5 and solve for 𝑆.
b) Multiply the 𝑆 from a), then substitute it in for
𝑆 in the equation; solve for 𝜋.
c) Set 𝑆 = 2,000 and solve for 𝜋.
Solution:
a) Set 𝜋 to 5:
𝑆 = 2500 1 − 𝑒−0.1 5
= 2500 1 − 𝑒−0.5 ≈ 983.67 2𝑑. 𝑝.
b) Multiply the exact value determined in a) by 2:
2 × 2500 1 − 𝑒−0.5 = 5000 1 − 𝑒−0.5
Set this equal to the original equation:
5000 1 − 𝑒−0.5 = 2500 1 − 𝑒−0.1𝜋
Divide both sides by 2500:
2 1 − 𝑒−0.5 = 1 − 𝑒−0.1𝜋
Rearrange to isolate 𝜋:
𝑒−0.1𝜋 = 1 − 2 1 − 𝑒−0.5
Take the natural log of the whole of both sides:
ln 𝑒−0.1𝜋 = ln 1 − 2 1 − 𝑒−0.5
−0.1𝜋 = ln 1 − 2 1 − 𝑒−0.5
𝜋 =ln 1 − 2 1 − 𝑒−0.5
−0.1≈ 15.462million
c) Set 𝑆 = 2000 and solve:
2000 = 2500 1 − 𝑒−0.1𝜋
4
5= 1 − 𝑒−0.1𝜋
𝑒−0.1𝜋 = 0.2
ln 𝑒−0.1𝜋 = ln 0.2
−0.1𝜋 = ln 0.2
𝜋 =ln 0.2
−0.1≈ 16.094million
Exercises: 1. Spain’s economy has a 𝐺𝐷𝑃 approximating
€800billion and grows at approximately 3.7%p.a. If the population of Spain is currently 40million, and is growing at 0.7%p.a., determine: a) When the population will reach 50𝑚𝑖𝑙𝑙𝑖𝑜𝑛. b) The current GDP per capita. c) The GDP per capita in 5 years time. d) When the GDP per capita will reach €30,000.
2. A company invests 𝑆 dollars in the health of each of its employees according to the equation:
𝑆 = 750 1 − 𝑒−0.08𝜋 Where 𝜋 is profit in millions of dollars. Determine: a) The amount spent on each individual if the
company has a profit of $7million. b) The total amount spent on healthcare when
profits are $10million and there are 19 employees.
c) The profit level when the amount spent on each employee is double that in part a).
d) The number of employees when profits are $15million and total spending on healthcare is $15,200.
chapter four summary
A function is defined as having only one 𝑦 − 𝑣𝑎𝑙𝑢𝑒 for every 𝑥 − 𝑣𝑎𝑙𝑢𝑒. A curve is a function if all vertical lines cross the curve only once.
The general form of a quadratic is: 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 The value of 𝑎 determines the sharpness of the quadratic. The value of 𝑐 in the quadratic equation determines the vertical shift of the graph. Changing the value of 𝑏 in the general quadratic form shifts the graph both horizontally and vertically.
A positive value of 𝑎 will result in quadratics with a shape of a ‘smiley’ face, whereas when the value of 𝑎 is negative, the shape becomes inverted; a ‘sad’ face.
The Quadratic Formula is:
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
The 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the turning point (TP), if found in the Quadratic Formula by evaluating everything not under the square root sign. Then to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the turning point, substitute this 𝑥 − 𝑣𝑎𝑙𝑢𝑒 into the original function. 𝑏2 − 4𝑎𝑐 determines the number of solutions:
94
𝑏2 − 4𝑎𝑐 > 0 → two solutions
𝑏2 − 4𝑎𝑐 = 0 → one solution
𝑏2 − 4𝑎𝑐 < 0 → no solutions
The cubic function is the name of a function where the highest power of 𝑥 is 3. The general form of a cubic is: 𝑦 = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 A cubic can cross the 𝑥 − 𝑎𝑥𝑖𝑠 once, twice or three times. When 𝑎 is positive, the ends of the graph extend to the top right and bottom left ↙↗ , but when 𝑎 is negative, the ends of the graph extend to the top left and bottom right ↖↘ .
An exponential function has the general form: 𝑦 = 𝑏(𝑎𝑥) + 𝑐
𝑎 is a number that determines the growth rate
𝑏 is the initial amount.
𝑐 a constant that shifts the graph up or down To sketch an exponential function, determine:
The 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 is found by setting all 𝑥’s equal to zero.
The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 of the asymptote is 𝑐. The general form of the decay function is:
𝑦 = 𝑏(𝑎−𝑥) + 𝑐
The definition of a logarithm is: For 𝑎𝑏 = 𝑐
𝑏 =log 𝑐
log 𝑎= log𝑎 𝑐
Where 𝑏 is called the index, and 𝑎 is called the base.
LOG RULE 1 log 𝑎𝑏 = 𝑏 log 𝑎 LOG RULE 2: log(𝑎 ∙ 𝑏) = log 𝑎 + log 𝑏
LOG RULE 3: log 𝑎
𝑏 = log 𝑎 − log 𝑏
LOG RULE 4a: log(1) = 0 LOG RULE 4b: log 10 = 1 (ln 𝑒 = 1) All these rules can be used forwards or backwards. The logarithmic graph has the general form: 𝑦 = 𝑎 log(𝑥 + 𝑏) + 𝑐 Where 𝑎 determines the sharpness of the “r” shape, 𝑏 the shifting of the whole graph left or right, and 𝑐 the shifting of the whole graph up or down.
Any time the number 𝑒 is used, the variable in question must be changing continuously. The growth of anything continuous has the general form: 𝐴𝑡 = 𝐴0𝑒𝑟 𝑡
The general form of a hyperbola is:
𝑦 = 𝑑
𝑎𝑥 + 𝑏+ c
Setting the whole denominator (𝑎𝑥 + 𝑏) equal to zero and solving for 𝑥 is how the vertical asymptote is found.
The value of 𝑐 determines the location of the horizontal asymptote.
The value of 𝑑 determines how close the curve gets to the intersection of the two asymptotes.
𝐺𝐷𝑃 per capita is a measure of the mean income per person in a particular country. It is defined as:
𝐺𝐷𝑃 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎 =𝐺𝐷𝑃
𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛
chapter four questions
1. Determine if the following curves are functions:
2. Determine the values of 𝑎, 𝑏 and 𝑐 for the following
quadratics: 𝑎) 𝑦 = 3 − 4𝑥 − 5𝑥2 𝑏) 𝑦 = 3 − 𝑥 𝑥 + 2
𝑐) 𝑦
𝑥 + 2= 𝑥 − 2
𝑑) 𝑦 − 1
𝑥 − 3= 𝑥 + 1
3. Match the following quadratics to the functions.
𝑦 = 𝑥2 − 6𝑥 𝑦 = −3𝑥2 + 27𝑥 − 54
𝑦 = 𝑥2 − 25𝑥 + 150 𝑦 = −𝑥2 + 18𝑥 − 45 4. Find the roots of the following quadratics:
𝑎) 𝑦 = 𝑥2 + 3𝑥 − 6 𝑏) 𝑦 = 4𝑥2 − 9 𝑐) 𝑦 = 𝑥 − 1 𝑥 + 3 𝑑) 𝑦 = 𝑥 + 4 2 𝑒) 𝑦 = 3𝑥 − 6 2 − 9 𝑓) 𝑦 = 𝑥2 − 2𝑥 + 12
𝑔) 𝑦 + 5
𝑥 − 3= 𝑥 + 4
5. Sketch the following quadratics: 𝑎) 𝑦 = 𝑥2 − 3𝑥 − 10 𝑏) 𝑦 = 𝑥2 − 3𝑥 − 28 𝑐) 𝑦 = 𝑥2 + 11𝑥 + 30 𝑑) 𝑦 = 𝑥 − 3 2 𝑒) 𝑦 = −4 𝑥 − 5 2 𝑓) 𝑦 = 𝑥2 − 4𝑥 − 13
𝑔) 𝑦 + 6
𝑥 − 3= 𝑥 + 4
𝑥
𝑦
𝑥
𝑦
𝐴
𝐵
𝐶
𝐷
𝐸
95
) 𝑦 𝑥 − 1 −1 =2
𝑥 + 2 −1
6. Determine the number of solutions for: 𝑎) 𝑦 = 𝑥2 + 5𝑥 + 6 𝑏) 𝑦 = −𝑥2 − 6𝑥 + 16 𝑐) 𝑦 = 4𝑥2 − 3𝑥 + 1 𝑑) 𝑦 = 𝑥2 − 18 𝑒) 𝑦 = 14𝑥2 − 2𝑥 − 2 𝑓) 𝑦 = 𝑥2 + 6𝑥 + 9
7. Match the following functions to their equations:
𝑦 = −3𝑥3 − 3 𝑦 = 2𝑥3 − 3
𝑦 = −𝑥3 + 5𝑥2 + 48𝑥 − 252 𝑦 = 𝑥3 + 12𝑥2 + 11𝑥 − 168
8. Match the following exponential functions to their equations:
𝑦 = 1.5𝑥 𝑦 = 3 2𝑥 𝑦 = 2−𝑥 + 2
9. Sketch the following exponential functions on the same set of axes. 𝑎) 𝑦 = 2𝑥 𝑏) 𝑦 = 3𝑥 + 2 𝑐) 𝑦 = 4 2𝑥 𝑑) 𝑦 = 3 3−𝑥 − 1
10. Match the following logarithmic functions to their equations:
𝑦 = 4 log 𝑥 𝑦 = log 𝑥2 𝑦 = 2 log 𝑥 − 2 + 5
11. Match the following to their equations:
𝑦 = −1
𝑥 − 2− 4 𝑦 =
1
𝑥 + 4− 6 𝑦 =
2
𝑥 − 5+ 5
12. Solve for 𝑥 in the following logarithmic functions: 𝑎) 12 = log 4𝑥 𝑏) − 13 = 2 log 3𝑥 𝑐) 7 = log 2𝑥+3 𝑑) log 2𝑥 = 4𝑥 + 2 𝑒) log 2𝑥−4 = 0 𝑓) 18 = 4 ln 2𝑥 𝑔) − log 8𝑥+1 = 182𝑥 ) 7 − log 2𝑥 = 𝑥 − 1 𝑖) 25 − ln 3 = − ln 6𝑥−1
13. Solve for 𝑥 in the following exponential functions: 𝑎) 15 = 2𝑥 𝑏) 18 = 4 3𝑥
𝑐) 9 − 23 = 4𝑥−1 𝑑) 34𝑥 9 = 3𝑥2
𝑒) 2𝑥−1 = 32−3𝑥 𝑓) 4 2𝑥−3 = 2𝑥2
𝑔) 14 2𝑥−3 = 213𝑥−3 14. The 𝐺𝐷𝑃 of Indonesia is approximately
𝑈𝑆$550billion, and is growing at 4.1% p.a. If the population is currently 120million, and is growing at 1.1% p.a. Determine: a) When the population will reach 140million. b) The 𝐺𝐷𝑃 in 10 years time. c) When 𝐺𝐷𝑃 will reach $1trillion. d) When 𝐺𝐷𝑃 per capita will reach 𝑈𝑆$10,000.
15. The price level of an economy is accurately estimated by the function:
𝑝 𝑡 = 1.03 𝑡 1.05 𝑡−1 +1
21
Determine: a) The price level in five years time. b) When the price level will reach 1.5 times that
of the current level. c) When the price level will double.
16. A company invests 𝑟 dollars in the relaxation of each of its employees, according to the amount of profit (in millions) generated by the firm:
𝑟 = 1200 1 − 𝑒−0.1𝜋 Determine: a) The amount spent on each employee when
profits are $5million. b) The total amount spent on relaxation when
profits are $10million, and there are 20 employees.
c) The profit level when spending on each employee is $1000.
d) The number of employees when profits are $12million, and total spending on relaxation is about$10,900.
𝑦
𝑥
𝑦
𝑥
𝑦
𝑥
𝑦
𝑥
96
Chapter 5
Single Variable
Differentiation Finding the slope of curves at any point along the function
5.1 What is Differentiation? 97
5.2 Differentiation by First Principles 99
5.3 Differentiation Rules: Power Rule 103
5.4 Differentiation Rules: Chain Rule 104
5.5 Differentiation Rules: Product Rule 105
5.6 Differentiation Rules: Quotient Rule 107
5.7 Differentiation Rules: 𝑒 Rule 109
5.8 Differentiation Rules: ln Rule 110
5.9 The Second Derivative 111
5.10 The Gradient Function Graph 112
5.11 Simple Applications 113
Chapter Five Summary 114
Chapter Five Questions 115
97
5.1 what is differentiation?
Engineers, economists, financial analysts,
managers and most professionals use
differentiation in everyday tasks.
Theory: differentiation is finding the rate that one
thing changes when something else is changed.
Typically, it is the rate of change of the 𝑦 − 𝑎𝑥𝑖𝑠
when there is a change in the 𝑥 − 𝑎𝑥𝑖𝑠.
A revenue function is determined by how much
output is sold. Differentiation gives the rate of
change of revenue for changes in output (or the
marginal revenue; how much extra revenue will
be obtained from selling an extra unit of
output).
A profit function is also determined by the
quantity of goods sold. Differentiating this will
give the rate of change of profit as output
changes (or marginal profit).
Theory: the rate of change of a function is found by
finding the slope of a curve at a given point. The
slope of a curve at a given point is found by
determining the slope of a line just touching that
point (i.e. tangent to that point).
Four lines have been drawn tangent to the blue
function:
The gradient at a point along a curve gives
information about the function. If the slope is
positive, the function is increasing (e.g. points 𝐴
and 𝐷). This means the rate of change of 𝑦 as 𝑥
changes is positive (from the slope of the tangent
line).
If the slope is negative, it means the function is
decreasing (e.g. point 𝐵). Interpreting this, the rate
of change of 𝑦, as 𝑥 changes, is negative (as the
point has a negative gradient).
Theory: the gradient of a line tangent to a given
point is the rate of change of 𝑦 for changes in 𝑥.
The following is a profit function:
At point 𝐸, the tangency has a positive gradient, so
the rate of change of profit (𝑦 − 𝑎𝑥𝑖𝑠) for changes
in output (𝑥 − 𝑎𝑥𝑖𝑠) is positive. If output is
increased from 𝑄1, profits will also increase beyond
𝜋1. On the other hand, for point 𝐹, the tangency
has a negative gradient meaning the rate of change
of profit for changes in output is negative; as
output increases beyond 𝑄2, profit falls below 𝜋2.
Theory: The gradient of a given point along a curve
is often called the marginal value.
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐𝑎𝑛𝑔𝑒 = 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
For the profit function, the gradient would be called
marginal profit, because the slope determines the
extra profit if output were to increase. Marginal
means “at the margin”, or “extra”.
Theory: Differentiation is the process of finding the
gradient at all points along a function.
𝐸
𝜋
𝑄
𝐹
𝜋1
𝜋2
𝑄1 𝑄2
𝐴
𝐵
𝐶
𝐷
𝑦
𝑥
98
The gradient function is a function that gives the
gradient of the original function at all the points
along the original function.
For the functions above, the gradient (or slope) of
the tangent changes when moving along the
𝑥 − 𝑎𝑥𝑖𝑠. Differentiation provides a way of finding
the gradients of any given point along a curve.
The following shows a simple use of gradients.
Example 1: if output is currently 𝑄 = 50, and the
profit function is
𝜋 𝑄 = −𝑄2 + 110𝑄 − 1000
As a manager, you would like to know if increasing
output will increase profits.
Plan: if the rate of change of profit (marginal profit)
is positive, then additional output will generate
more profit. If the rate of change of profit (marginal
profit) is negative, additional output will reduce
profits.
Production is to increase from 𝑄 = 50 to 𝑄 = 51.
Solution: The profit function is a “sad” quadratic (as
𝑎 < 0), meaning there is a maximum. If current
output is to the left of the maximum, an increase of
one unit will increase profits (as the marginal profit
is positive). If current output is to the right of the
maximum, an extra unit will reduce profits
(negative marginal profit). Look at points 𝐸 and 𝐹
of the last graph.
Find the turning point (which is a maximum) using
part of the quadratic formula:
𝑥 − 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑇𝑃 = −𝑏
2𝑎= −
110
−2=
110
2= 55
Since 𝑄 = 50 is to the left of the maximum, the
slope at 𝑄 = 50 must be positive, so the marginal
profit is positive. This means that if output were to
increase to 𝑄 = 51, total profits will increase.
Find the profit at 𝑄 = 50 and at 𝑄 = 51 to make
sure of this result:
𝜋 50 = − 50 2 + 110 50 − 1000
= 2000
𝜋 51 = −(51)2 + 110(51) − 1000
= 2009
Thus profits will increase if output is increased by
one unit.
The actual value of the gradient at 𝑄 = 50 was not
determined, but from working with quadratics, it
was known that anything to the left of the
maximum must have a positive gradient. Most of
the time, however, it will not be so simple.
The rest of this chapter is devoted to finding the
gradient function mathematically. However,
before this can be done, the gradient function
must be denoted by something that distinguishes
it from the original function.
Theory: mathematicians denote the derivative
(gradient) function in a number of ways:
𝑦 ′ is pronounced “𝑦 prime”
𝑑𝑦
𝑑𝑥 is pronounced “dee y dee x”. For a small
change in 𝑥 (the 𝑑𝑥), 𝑦 will change by a certain
small amount (𝑑𝑦). The 𝑑 means very small.
Exercises: 1. Determine if the following points have a positive or
negative gradient
2. Using your knowledge of quadratics and turning
points, for the function 𝑦 − 7 − 3𝑥 = −𝑥2
Determine if the following points have a positive or negative gradient:
𝑥 = −1, 𝑥 = 2, 𝑥 = 6 3. Order the following points from lowest (i.e. most
negative) to highest (i.e. most positive):
𝑦
𝑥
𝐴
𝐵 𝐶
𝐷
𝐸 𝐹
𝐺
𝐻
99
Hint: use a ruler and draw in the tangencies.
5.2 differentiation by first principles
Differentiation is simply finding the gradient
function of an original function. The following is
the long method, but it is necessary that you
understand it.
Intro example 1: beginning with the function
𝑓 𝑥 = 4 𝑥 − 2 2 + 2
for which the gradient at 𝑥 = 3 is to be found. On
the graph, draw a line through this point and
another point on the curve, say 1 unit up from 3
(i.e. 𝑥 = 4):
This is called a rough approximation of the
gradient at 𝑥 = 3. It is just an approximation, as it
is obvious that the gradient of this line (red) and
the tangent (blue) line are not the same. Ignore
this for the time being and determine the gradient
of this red line.
From Chapter 2, to find the gradient of a line, two
points are needed (i.e. two coordinates). The first
𝑥 − 𝑣𝑎𝑙𝑢𝑒 is 𝑥 = 3, but for a coordinate we also
need a 𝑦 − 𝑣𝑎𝑙𝑢𝑒. To find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒,
substitute 𝑥 = 3 into the original function:
𝑓 𝑥 = 4 𝑥 − 2 2 + 2
𝑓(3) = 4 3 − 2 2 + 2
𝑓 3 = 4 1 2 + 2 = 6
The first coordinate is (3,6).
The other point on the red line is 1 unit above 3,
so 𝑥 = 4. Again, the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is found using the
original function:
𝑓 𝑥 = 4 𝑥 − 2 2 + 2
𝑓(4) = 4 4 − 2 2 + 2
𝑓 4 = 4 2 2 + 2 = 18
which gives the second coordinate (4,18).
Plot these two points on a set of axes (previous
graph), then find the gradient:
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛=
𝑦2 − 𝑦1
𝑥2 − 𝑥1=
18 − 6
4 − 3=
12
1= 12
Thus 12 is the gradient of the red line. However,
this red line is not a good approximation to the
blue line. To get a better approximation, we look
back to the start of this problem and instead of
having a point 1 unit up from 3, a point a little
closer to 𝑥 = 3 is used, say 0.5 units up from 3 (i.e.
the other 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is 3.5):
This red line is a closer approximation to the actual
gradient than the black dotted line (which was the
first approximation). Despite this new
approximation not being perfect, the gradient is
still found because this approximation is better.
Two coordinates are still needed: one is still the
point (3,6). The other coordinate has the
𝑦
𝑥 3 4 3.5
𝑦
𝑥 3 4
𝑦
𝑥
𝐴
𝐵
𝐸
𝐹
𝐺
𝐶 𝐷
100
𝑥 − 𝑣𝑎𝑙𝑢𝑒 as 𝑥 = 3.5. The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 comes from
the original function:
𝑓 𝑥 = 4 𝑥 − 2 2 + 2
𝑓(3.5) = 4 3.5 − 2 2 + 2
𝑓 3.5 = 4 1.5 2 + 2 = 11
Plot the two points on a set of axes (previous
graph), then finding the gradient of the connecting
line:
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛=
𝑦2 − 𝑦1
𝑥2 − 𝑥1=
11 − 6
3.5 − 3
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =5
0.5= 10
This is a closer approximation to the actual
gradient at 𝑥 = 3, but it is still not the actual
gradient. How would we get a better
approximation? Instead of adding 0.5 to 3,
something smaller could be added, like 0.01, or
0.001 etc. This will not be done mathematically,
but the graph below shows what is meant:
As points closer and closer to 𝑥 = 3 are chosen as
the “other coordinate”, the approximations
become better and better (red arrow). So using
very small additions to 𝑥 = 3 gives approximations
very close to the tangent at 𝑥 = 3 (blue line).
Theory: to mathematically find an approximation
to the tangent at a point 𝑥 (this was 3 in the above
example), a point is chosen a very small distance
𝑠 above this 𝑥 − 𝑣𝑎𝑙𝑢𝑒. The other point will have
an 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of 𝑥 + 𝑠 (i.e. the point in question
plus a very small amount). In the above example, 𝑠
was initially 1, then 0.5. Thus the two points are:
𝑥1 = 𝑥 (this is the base point)
𝑥2 = 𝑥 + 𝑠 (an 𝑥 − 𝑣𝑎𝑙𝑢𝑒 a little more than 𝑥1)
The 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of these two 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 are
found using the original function 𝑓 𝑥 .
Since these are not numbers but rather letters, the
same process is done as if they were numbers.
For the first point 𝑥1, the function will be
𝑓(𝑥1), which is simply 𝑓(𝑥) ; the function that
is given. The coordinate is then 𝑥, 𝑓 𝑥 .
For the second point 𝑥2 = 𝑥 + 𝑠, the function
will be 𝑓 𝑥2 = 𝑓(𝑥 + 𝑠). The coordinate is
then 𝑥 + 𝑠, 𝑓 𝑥 + 𝑠 .
Having two coordinates allows the gradient of this
line to be found using rise/run:
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =𝑟𝑖𝑠𝑒
𝑟𝑢𝑛=
𝑦2 − 𝑦1
𝑥2 − 𝑥1
=𝑓 𝑥 + 𝑠 − 𝑓 𝑥
𝑥 + 𝑠 − 𝑥
=𝑓 𝑥 + 𝑠 − 𝑓(𝑥)
𝑠
This generalises what was done graphically.
However, remember that the gradient of the line is
not a good approximation of the actual gradient of
a point if the value of 𝑠 is too large. That is why
mathematicians invented something called a limit.
Theory: the limit as “𝑠 goes to zero”: lim𝑠→0
The idea is to get 𝑠 to “go to” zero (but not be
zero) because when 𝑠 is very small the
approximation becomes better and better.
Differentiation by First Principles is the formula:
𝑓 ′ = lim𝑠→0
𝑓 𝑥 + 𝑠 − 𝑓 𝑥
𝑠
Example 1: Find the gradient function of
𝑦 = 5𝑥 + 2
Plan: use the first principles formula;
𝑦
𝑥 4 3.5 3.75 3.25 3
101
𝑓 ′ = lim𝑠→0
𝑓 𝑥 + 𝑠 − 𝑓 𝑥
𝑠
Solution: the trick with this formula is that only the
𝑓(𝑥 + 𝑠) part ever needs to be found. To find this,
replace every 𝑥 in the original function with
(𝑥 + 𝑠) in brackets:
The original function is
𝑓(𝑥) = 5𝑥 + 2
Replace every 𝑥 with 𝑥 + 𝑠 in brackets:
𝑓(𝑥 + 𝑠) = 5(𝑥 + 𝑠) + 2
𝑓(𝑥 + 𝑠) = 5𝑥 + 5𝑠 + 2
Substitute the 𝑓(𝑥 + 𝑠) in the first principles
formula with what was found above:
𝑓 ′ = lim𝑠→0
[5𝑥 + 5𝑠 + 2] − 𝑓 𝑥
𝑠
Then replace the 𝑓(𝑥) part, which is simply the
original function:
𝑓 ′ = lim𝑠→0
5𝑥 + 5𝑠 + 2 − [5𝑥 + 2]
𝑠
= lim𝑠→0
5𝑥 + 5𝑠 + 2 − 5𝑥 − 2
𝑠
Simplify the top by cancelling:
𝑓 ′ = lim𝑠→0
5𝑠
𝑠
And then the 𝑠
𝑠 part can be crossed off:
Only after crossing off 𝑠
𝑠 can every other 𝑠 be
replaced with zero. But since there is no 𝑠 left:
𝑓 ′(𝑥) = lim𝑠→0
5 = 5
The lim𝑠→0
part can now be ignored as there is no
longer an 𝑠 left, giving the gradient function:
𝑓 ′ 𝑥 = 5 𝑜𝑟 𝑑𝑓 𝑥
𝑑𝑥= 5
To repeat, this is the gradient function, and it is
different from the original function, however, they
are related (see section 5.10).
Think about the answer 𝑓 ′ = 5. The original
function was a line with equation 𝑦 = 5𝑥 + 2, and
from Chapter 2, the gradient of this line is 𝑚 = 5,
which is the same answer.
A line has the same gradient all along it, but the
gradient of a quadratic changes along its curve.
Example 2: find the gradient function of
𝑦 = 3𝑥2 − 𝑥 − 6
Plan: use the first principles formula
𝑑𝑦𝑑𝑥
= lim𝑠→0
𝑓 𝑥 + 𝑠 − 𝑓 𝑥
𝑠
Solution: The only thing we need to find in this
formula is 𝑓(𝑥 + 𝑠) and this is done by replacing
every 𝑥 in the original function with (𝑥 + 𝑠) in
brackets:
𝑓(𝑥) = 3𝑥2 − 𝑥 − 6
𝑓 𝑥 + 𝑠 = 3 𝑥 + 𝑠 2 − (𝑥 + 𝑠) − 6
Write out the “squared” term as two terms:
𝑓 𝑥 + 𝑠 = 3 𝑥 + 𝑠 𝑥 + 𝑠 − 𝑥 + 𝑠 − 6
𝑓 𝑥 + 𝑠 = 3 𝑥2 + 2𝑥𝑠 + 𝑠2 − 𝑥 + 𝑠 − 6
𝑓 𝑥 + 𝑠 = 3 𝑥2 + 6𝑥𝑠 + 3𝑠2 − 𝑥 − 𝑠 − 6
This is the 𝑓(𝑥 + 𝑠) part done, so replace 𝑓(𝑥)
with the original function in the first principles
formula:
𝑑𝑦𝑑𝑥
= lim𝑠→0
𝑓 𝑥 + 𝑠 − 𝑓 𝑥
𝑠
= lim𝑠→0
3𝑥2 + 6𝑥𝑠 + 𝑠2 − 𝑥 − 𝑠 − 6 − [3𝑥2 − 𝑥 − 6]
𝑠
The tricky part is to simplify this. The easiest way is
to look at the numerator (top), and you will see a
lot of terms can be cancelled off (colour coded
below). This results in something much simpler.
The ultimate goal is to get rid of the 𝑠 on the
bottom, by cancelling it off with an 𝑠 on top.
= lim
𝑠→0
𝟑𝒙𝟐 + 6𝑥𝑠 + 𝑠2 − 𝒙 − 𝑠 − 𝟔 − 𝟑𝒙𝟐 + 𝒙 + 𝟔
𝑠
𝑓 𝑥 + 𝑠 = 3 𝑥 + 𝑠 𝑥 + 𝑠 − 𝑥 + 𝑠 − 6
𝑓 ′ = lim𝑠→0
5𝒔
𝒔
𝑓 ′ = lim𝑠→0
𝟓𝒙 + 5𝑠 + 𝟐 − 𝟓𝒙 − 2
𝑠
102
Note that the negative sign must be put into all the
terms of 𝑓(𝑥) when it is taken away from
𝑓(𝑥 + 𝑠).
𝑑𝑦
𝑑𝑥= lim
𝑠→0
6𝑥𝑠 + 𝑠2 − 𝑠
𝑠
To cancel off the 𝑠 in the denominator, an 𝑠 in the
numerator should be factorised out first. Only then
can it be cancelled off with the bottom one.
𝑑𝑦
𝑑𝑥= lim
𝑠→0
𝑠(6𝑥 + 𝑠 − 1)
𝑠
Having cancelled off the 𝑠 on the bottom, all other
𝑠’s can be replaced with zero, and the limit
notation removed:
𝑑𝑦
𝑑𝑥= 6𝑥 + 0 − 1 = 6𝑥 − 1
This is the derivative function. It is the function
that gives the derivative (gradient) at every point
on the original function. For example, the gradient
of the original function at 𝑥 = 2 would be found
by substituting 𝑥 = 2 into the gradient function
(i.e. 𝑑𝑦
𝑑𝑥 2 = 6 2 − 1 = 11).
Example 3: using first principles, find the
derivative function of
𝑦 =3
𝑥
Plan: use the first principles formula
𝑑𝑦𝑑𝑥
= lim𝑠→0
𝑓 𝑥 + 𝑠 − 𝑓 𝑥
𝑠
Solution: find 𝑓(𝑥 + 𝑠)
𝑓(𝑥) =3𝑥
𝑓(𝑥 + 𝑠) =3
𝑥 + 𝑠
This cannot be simplified right now, so substitute
everything into the first principles formula:
𝑑𝑦
𝑑𝑥= lim
𝑠→0
3
𝑥 + 𝑠 −3𝑥
𝑠
Isolate the numerator (top) of this fraction, so
simplification is easier (it will be brought back in
later):
3
𝑥 + 𝑠−
3
𝑥=
3
𝑥 + 𝑠 𝑥
𝑥 −
3
𝑥
(𝑥 + 𝑠)
(𝑥 + 𝑠)
This is finding a Common Denominator (Chapter
1). The Common Denominator is the multiplication
of the two denominators. Get rid of the brackets:
=3
𝑥 + 𝑠 𝑥
𝑥 −
3
𝑥 𝑥 + 𝑠
𝑥 + 𝑠
=3𝑥
𝑥 + 𝑠 (𝑥)−
3𝑥 + 3𝑠
𝑥 + 𝑠 (𝑥)
Having the Common Denominator present, a single
fraction can be written:
=3𝑥 − 3𝑥 + 3𝑠
𝑥 + 𝑠 𝑥 =
3𝑥 − 3𝑥 − 3𝑠
𝑥 + 𝑠 𝑥
Cancel off the 3𝑥 − 3𝑥:
=−3𝑠
𝑥 + 𝑠 (𝑥)
Bring everything back into the first principles
formula:
𝑑𝑦
𝑑𝑥= lim
𝑠→0
−3𝑠
𝑥 + 𝑠 (𝑥)
𝑠
From Chapter 1, the division of fractions can be
written as the multiplication of the inverse:
𝑑𝑦
𝑑𝑥= lim
𝑠→0
−3𝑠
𝑥 + 𝑠 𝑥 ∙
1
𝑠
Cross out the two red 𝑠:
𝑑𝑦
𝑑𝑥= lim
𝑠→0
−3
𝑥 + 𝑠 (𝑥)
Replace the remaining 𝑠 with zero and remove the
limit notation:
𝑑𝑦
𝑑𝑥=
−3
𝑥 + 0 (𝑥)=
−3
𝑥 (𝑥)= −
3
𝑥2
Differentiation by first principles requires knowing
how to simplify and factorise, so if you’re not
comfortable spend some time revising Chapter 1.
𝑑𝑦
𝑑𝑥= lim
𝑠→0
𝒔(6𝑥 + 𝑠 − 1)
𝒔
103
Exercises: 1. Using Differentiation by First Principles, determine
the gradient function given the following original functions:
𝑎) 𝑦 = 7𝑥 − 3
𝑏) 𝑦 = −9𝑥 + 1
𝑐) 𝑦 = 𝑥2 + 2𝑥
𝑑) 𝑦 = −3𝑥2 + 14𝑥 − 11
2. Using First Principles, determine the gradient function of the following:
𝑎) 𝑦 = 3𝑥2 − 𝑥3
𝑏) 𝑦 = −5𝑥2
7+
12
5𝑥 − 1
𝑐) 𝑦 =5
𝑥
𝑑) 𝑦 =3
7𝑥
𝑒) 𝑦 =1
𝑥2
5.3 differentiation rules: power rule
This is a rule which makes the last 4 pages
redundant. Sorry, but at least now you’ll know
what you are actually doing.
Theory: the power rule states that to find the
derivative function, you bring the power to the
front and then subtract 1 from the power.
For the original function: 𝑦 = 𝑎𝑥𝑛
The derivative function is: 𝑑𝑦
𝑑𝑥= 𝑛𝑎𝑥𝑛−1
Example 1: use the power rule to differentiate:
𝑦 = 𝑥2
Solution: bring the 2 out front of the 𝑥, then
subtract 1 from the power:
𝑦′ =𝑑𝑦
𝑑𝑥= 2𝑥
Example 2: find the derivative of
𝑦 = 2𝑥3
Plan: use the power rule to bring the power out
front then subtract one from the power.
Solution:
𝑦′ = (3)2𝑥2
Multiply the 2 and 3 together to give the solution:
𝑦′ = 6𝑥2
Example 3: find the gradient function of
𝑦 = 4𝑥3 − 2𝑥2 + 7𝑥 − 1
Plan: use the power rule to differentiate each
section, one at a time.
Solution: the four sections are:
Apply the power rule to each of the sections:
𝑑𝑦
𝑑𝑥= 12𝑥2 − 4𝑥 + 7
Remember: when you see 𝑥 by itself, it is really 𝑥1
and when this is differentiated, it gives 𝑥1−1 =
𝑥0 = 1, so the 𝑥 is no longer present.
Theory: the derivative of a constant is zero. It
means the gradient of a horizontal line is zero.
This is why, in the last example, the −1
differentiated to give zero.
Example 4: differentiate
𝑦 = −3
𝑥4+ 5𝑥2 − 𝑥 + 15
Plan: use the power rule, one section at a time.
Solution: you cannot simply differentiate using the
power rule if the term is not up top. The original
function must first be simplified (see Chapter 1):
𝑦 = −3𝑥−4 + 5𝑥2 − 𝑥 + 15
Then differentiate each section separately:
𝑑𝑦
𝑑𝑥= −4 −3 𝑥−4−1 + 2 5𝑥2−1 − 1 𝑥1−1 + 0
= 12𝑥−5 + 10𝑥 − 1
Watch out for the negative signs!
Exercises: 1. Differentiate the following using the power rule:
𝑎) 𝑦 = 3𝑥2
𝑏) 𝑦 = −2𝑥3 + 𝑥2
𝑐) 𝑦 = −3𝑥3
7
𝑦 = 4𝑥3 − 2𝑥2 + 7𝑥 − 1
Step 1: = 𝑥2
Step 2: = 2𝑥2−1
104
𝑑) 𝑦 =5
𝑥4−
𝑥2
5
𝑒) 𝑦 = −𝑥 −𝑥2
2−
𝑥3
3+
1
𝑥
𝑓) 𝑦 =1
𝑥+
1
𝑥2+
1
𝑥3
𝑔) 𝑦 = 33
4𝑥−3 + 𝑥2
) 𝑦 = 6 +15
−𝑥−4
𝑖) 𝑦 = 𝑥14 +1
𝑥14
𝑗) 𝑦 = −13𝑥−13 + 15 − 𝑥
𝑘) 𝑦 = 12
5.4 differentiation rules: chain rule
Intro example 1: differentiate
𝑦 = (4𝑥 + 2)3
Solution: to solve this, you might expand it using
the crab-claw, and then differentiate the
sections:
𝑦 = 64𝑥3 + 80𝑥2 + 48𝑥 + 8
Differentiate this:
𝑑𝑦
𝑑𝑥= 192𝑥2 + 160𝑥 + 48
This is a very long method (especially when you
show all working) but there is an easier method;
the chain rule.
There is the formal and informal way of
remembering the chain rule.
Theory: the informal method is: differentiate the
function as if using the power rule, then multiply
by the derivative of the inside of the brackets.
Example 1: differentiate the above example
𝑦 = (4𝑥 + 2)3
Plan: use the informal chain rule method:
differentiate the function as if using the power
rule, then multiply by the derivative of the inside
of the brackets.
Solution: Differentiate as if using the power rule:
3(4𝑥 + 2)3−1 = 3 4𝑥 + 2 2
Then multiply this by the derivative of the inside of
the brackets:
𝑑𝑦
𝑑𝑥= 3 4𝑥 + 2 2(4)
𝑑𝑦
𝑑𝑥= 12 4𝑥 + 2 2
The 12 came from multiplying the 3 and 4
together. We cannot put the 12 into the brackets
because the bracket is squared.
Example 2: Find the gradient function of
𝑦 = (𝑥2 + 3𝑥)5
Plan: use the chain rule to differentiate:
“differentiate the function as if using the power
rule, then multiply by the derivative of the inside
of the brackets”
Solution: the first underlined part of the following
is the as if part of the chain rule, and the second
underlined part is the derivative of inside the
brackets.
The 2𝑥 + 3 must be in brackets as 5 𝑥2 + 3𝑥 4
is multiplied by all of 2𝑥 + 3 . Otherwise, it would
imply that the 5 𝑥2 + 3𝑥 4 is only multiplied by
2𝑥, and then 3 is added onto the whole thing,
which is not the case!
This last example can be simplified to:
𝑑𝑦
𝑑𝑥= (10𝑥 + 15) 𝑥2 + 3𝑥 4
As the 5 can only go into every term in the bracket
which is to the power of 1.
Example 3: differentiate
𝑦 =2
3𝑥3 − 4𝑥2 + 1 −5.5
𝑑𝑦
𝑑𝑥= 5 𝑥2 + 3𝑥 4(2𝑥 + 3)
𝑦 = 4𝑥 + 2 4𝑥 + 2 4𝑥 + 2
𝑦 = 16𝑥2 + 16𝑥 + 4 4𝑥 + 2
105
Plan: move everything up top so the chain rule can
then be easily used.
Chain rule: differentiate the function as if using the
power rule, then multiply by the derivative of the
inside of the brackets.
Solution: use index rules to bring the denominator
up top and change the sign of the index:
𝑦 = 2 3𝑥3 − 4𝑥2 + 1 5.5
Apply the chain rule: the first underlined part
below is the as if part, and the second underlined
part is “the derivative of the inside of the
brackets”.
Simplify the constants (2 ∙ 5.5) into the only
brackets which are to the power of 1 (9𝑥2 − 8𝑥):
𝑦 = 99𝑥2 − 88𝑥 3𝑥3 − 4𝑥2 + 1 4.5
The formal method of the chain rule is temporarily
inventing a new function.
Theory: formally, the chain rule is:
𝑑𝑦
𝑑𝑥=
𝑑𝑦
𝑑𝑢×
𝑑𝑢
𝑑𝑥
To demonstrate this, the following function will be
used:
𝑦 = (𝑥2 + 1)7
Invent the function 𝑢 = 𝑥2 + 1, which is whatever
is inside the brackets. Then replace the brackets in
the original equation with the function 𝑢:
𝑦 = (𝑢)7
Differentiate both of these new functions:
𝑑𝑦
𝑑𝑢= 7𝑢6
And:
𝑑𝑢
𝑑𝑥= 2𝑥
If these two derivatives are multiplied together,
the wanted derivative is obtained:
𝑑𝑦
𝑑𝑢×
𝑑𝑢
𝑑𝑥=
𝑑𝑦
𝑑𝑢×
𝑑𝑢
𝑑𝑥=
𝑑𝑦
𝑑𝑥
With the equation above:
𝑑𝑦
𝑑𝑥= 7𝑢6 × 2𝑥
= 14𝑥𝑢6
= 2𝑥 7 𝑥2 + 1 6
= 14𝑥 𝑥2 + 1 6
Replace 𝑢 with what it was originally defined as
(i.e. it was defined as (𝑥2 + 1) in this question).
You get the same answer using the formal or
informal method, however always write out the
theory used.
Exercises: 1. Differentiate the following using the chain rule:
𝑎) 𝑦 = 𝑥 + 1 3
𝑏) 𝑦 = 2𝑥 − 3 4
𝑐) 𝑦 = −3𝑥 + 1 4
𝑑) 𝑦 = 𝑥2 + 2 3
𝑒) 𝑦 = 2𝑥2 − 3 4
𝑓) 𝑦 = −4𝑥2 − 𝑥 3
𝑔) 𝑦 = 12𝑥 − 𝑥3 4
) 𝑦 = 4 𝑥2 + 𝑥 + 1 5
𝑖) 𝑦 = 5 −3𝑥3
8+
1
𝑥
4
𝑗) 𝑦 = 1
𝑥+
1
𝑥2
3
𝑘) 𝑦 = 12 𝑥2 + 𝑥 + 5 +1
𝑥+
5
𝑥2
2
5.5 differentiation rules: product rule
This rule applies whenever two separate functions
are multiplied together. The following shows
examples of two separate functions:
These cannot be differentiated using the rules
learnt so far, so a new rule is required: the product
rule.
𝑦 = 𝑥 + 2 −4 𝑥4 + 2 6
𝑡 = 𝑒𝑥2(4𝑥 − 1)
𝑟 = 3𝑥
4+ 2
2
1
𝑥+ 1
−2
𝑑𝑦
𝑑𝑥= 2 5.5 3𝑥3 − 4𝑥2 + 1 5.5−1(9𝑥2 − 8𝑥)
106
Theory: given the function
where 𝑔(𝑥) and (𝑥) are separate functions. The
derivative function is then:
𝑓 ’(𝑥) = 𝑔’(𝑥) ∙ (𝑥) + 𝑔(𝑥) ∙ ’(𝑥)
Informally: “the derivative function is the
derivative of the first part multiplied by the second
part, plus the derivative of the second part
multiplied by the first part”.
𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
Example 1: find the gradient function of
𝑓(𝑥) = (15𝑥2 + 3𝑥3)(−4𝑥7 + 6)
Plan: use the product rule:
𝑓 ′ 𝑥 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
Solution: defining the two functions:
Following the product rule, do one section at a
time:
𝑑𝑒𝑟 1𝑠𝑡 = 30𝑥 + 9𝑥2
𝑑𝑒𝑟 2𝑛𝑑 = −28𝑥6
Put all the parts in brackets and substitute into the
power rule formula: 𝑓 ′ 𝑥
Try simplifying this last line using the crab claw
(twice). You should get:
𝑓 ′ 𝑥 = −120𝑥9 − 540𝑥8 + 54𝑥2 + 180𝑥
You could have crab-clawed the original function
and then applied the power rule four times. But in
harder examples, it would take too much time.
Example 2: find the gradient function of
Plan: two rules will be required: (1) the product
rule as our “base rule”, (2) the chain rule within
our base rule.
Solution: Use the product rule to find all the parts
needed (with the chain rule used within these
parts):
𝑑𝑒𝑟 1𝑠𝑡 = 3 4𝑥 2 4
= 12(4𝑥)2
𝑑𝑒𝑟 2𝑛𝑑 = 6 −3𝑥2 − 1 5 −6𝑥
= −36𝑥 −3𝑥2 − 5𝑥 5
Substitute into the product rule formula:
𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
This looks very big, but it is essentially in two parts,
separated by an addition sign. If you look carefully
in both sections, it is possible to factorise out some
of those parts (underlined below) to make it a bit
simpler:
Simplify the square brackets:
𝑑𝑦
𝑑𝑥= 4𝑥 2 −3𝑥2 − 1 5[−36𝑥2 − 12 − 144𝑥]
That is the extent of simplifying.
The product rule is not complicated, but
determining the two functions is sometimes
difficult, and the simplifications can, at times, be
complex.
Exercises: 1. Differentiate the following using the product rule:
𝑎) 𝑦 = 5𝑥 4 3𝑥2 + 1
𝑏) 𝑦 = 𝑥3 + 1 𝑥2 + 𝑥5
𝑐) 𝑦 = −3𝑥2 − 7 13𝑥4 − 𝑥
𝑑) 𝑦 = 12𝑥4 + 12𝑥5 𝑥7 2
𝑒) 𝑦 = 15𝑥3 + 7𝑥 13𝑥7 + 4
𝑓) 𝑦 = 15𝑥3 +1
𝑥
2
𝑥+ 5
𝑔) 𝑦 = 17 𝑥5 −𝑥
𝑥4
1
𝑥+
1
𝑥2
) 𝑦 = 0.25 𝑥 + 0.5𝑥 7𝑥3 + 7𝑥3
𝑑𝑦
𝑑𝑥= 4𝑥 2 −3𝑥2 − 1 5 12 −3𝑥2 − 1 + −36 4𝑥
𝑦 = (4𝑥)3 −3𝑥2 − 1 6
1𝑠𝑡 2𝑛𝑑
𝑑𝑒𝑟1𝑠𝑡 2𝑛𝑑 𝑑𝑒𝑟2𝑛𝑑 1𝑠𝑡
= (30𝑥 + 9𝑥2)(−4𝑥7 + 6) + (−28𝑥6)(15𝑥2 + 3𝑥3)
𝑓(𝑥) = (15𝑥2 + 3𝑥3)(−4𝑥7 + 6)
1𝑠𝑡 2𝑛𝑑
𝑓 𝑥 = 𝑔 𝑥 . 𝑥
1𝑠𝑡 2𝑛𝑑
= 12 4𝑥 2 −3𝑥2 − 1 6 + −36𝑥 −3𝑥2 − 1 5 4𝑥 3
107
2. Using all the differentiation rules learnt so far, differentiate the following:
𝑎) 𝑦 = 2𝑥 + 1 3 7𝑥2 − 1
𝑏) 𝑦 = 1 − 𝑥 5 2𝑥2 − 1 2
𝑐) 𝑦 = 1
𝑥+ 𝑥2
3
1
𝑥−4
2
𝑑) 𝑦 = 4 𝑥 + 𝑥7 − 𝑥 3 5
−𝑥−3− 1
5
5.6 differentiation rules: quotient rule
This rule is similar to the product rule, except now,
we have one function divided by another function.
Theory: the quotient rule applies for functions of
the form:
𝑦(𝑥) =𝑡(𝑥)
𝑏(𝑥)
For example, these can look like:
𝑦 =3𝑥2
1 + 𝑥−3
𝑦 = 𝑒2𝑥 + 𝑥
4𝑥
+ 3𝑥3
Theory: formally, the quotient rule is
𝑑𝑦
𝑑𝑥=
𝑡′ 𝑥 × 𝑏 𝑥 − 𝑏′ (𝑥) × 𝑡(𝑥)
𝑏(𝑥) 2
Informally, it is: “the derivative function is the
derivative of the top times the bottom minus the
derivative of the bottom times the top, all divided
by the bottom squared”.
For the numerator (top part of the fraction) the
order of what you differentiate first is important as
there is a minus sign.
𝑑𝑦
𝑑𝑥=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
Ask yourself, would you naturally differentiate the
top or bottom first? Most people say they would
differentiate the top first, and that is what is done.
Example 1: differentiate
𝑦 =15𝑥4 + 3𝑥2
5𝑥 + 1
Plan: use the quotient rule:
𝑑𝑦
𝑑𝑥=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
Solution: using the quotient rule, differentiate the
two parts:
𝑑𝑒𝑟 𝑇𝑂𝑃 = 60𝑥3 + 6𝑥
𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 5
Substitute each section in brackets into the
quotient rule:
𝑑𝑦
𝑑𝑥=
60𝑥3 + 6𝑥 5𝑥 + 1 − 5 (15𝑥4 + 3𝑥2)
[5𝑥 + 1]2
You don’t usually need to expand the
denominator, especially if you need to use the
crab-claw method. This is because sometimes you
might be able to cancel it off. However, always
simplify the numerator; in this last case by using
the crab-claw method. Do it yourself, as the
following is only the answer:
𝑑𝑦
𝑑𝑥=
300𝑥4 + 60𝑥3 + 30𝑥2 + 6𝑥) − 75𝑥4 + 15𝑥2
5𝑥 + 1 2
=225𝑥4 + 60𝑥3 + 15𝑥2 + 6𝑥
[5𝑥 + 1]2
3𝑥 could also be factorised:
𝑑𝑦
𝑑𝑥=
3𝑥 75𝑥3 + 20𝑥2 + 5𝑥 + 2
[5𝑥 + 1]2
If simplification is obvious, then it should be done.
Example 2: find the gradient function of
𝑓 𝑥 =3𝑥2 − 𝑥−4
2𝑥 + 1
Plan: use the quotient rule
𝑑𝑦
𝑑𝑥=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
Solution: find the derivative of each of the two
parts:
𝑑𝑒𝑟 𝑇𝑂𝑃 = 6𝑥 + 4𝑥−5
𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 2
Substitute back into the quotient rule formula
(remember the brackets):
𝑑𝑦
𝑑𝑥=
6𝑥 + 4𝑥−5 2𝑥 + 1 − 2 3𝑥2 − 𝑥−4
2𝑥 + 1 2
108
Then use the crab claw method (you should do it!)
to simplify the top:
𝑑𝑦
𝑑𝑥=
12𝑥2 + 8𝑥−4 + 6𝑥 + 4𝑥−5 − 6𝑥2 + 2𝑥−4
2𝑥 + 1 2
Bring together “like terms”:
𝑑𝑦
𝑑𝑥=
6𝑥2 + 10𝑥−4 + 6𝑥 + 4𝑥−5
2𝑥 + 1 2
Factorise out the number 2 from the top:
𝑑𝑦
𝑑𝑥=
2 3𝑥2 + 5𝑥−4 + 3𝑥 + 2𝑥−5
2𝑥 + 1 2
This next example combines the quotient rule with
other differentiation rules.
Example 3: find the derivative of
𝑦 = 4𝑥2 + 1 (−3𝑥−5)
2𝑥5
Plan: use the quotient rule as the base rule and
the product rule within the quotient rule:
𝑑𝑦
𝑑𝑥=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
Solution: find the derivative of each section of the
base rule:
𝑑𝑒𝑟 𝑇𝑂𝑃 = 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
= 8𝑥 −3𝑥−5 + 15𝑥−6 4𝑥2 + 1
= 36𝑥−4 + 15𝑥−6
𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 10𝑥4
Substitute back into the Quotient Rule:
𝑑𝑦
𝑑𝑥
= 36𝑥−4 + 15𝑥−6 2𝑥5 − 10𝑥4 4𝑥2 + 1 (−3𝑥−5)
[2𝑥5]2
Factorise out 2𝑥4 from the top, and also simplify
the denominator (which we wouldn’t usually do,
except here, we can see that part of it will cancel
off with the denominator):
𝑑𝑦
𝑑𝑥=
2𝑥4[𝑥 36𝑥−4 + 15𝑥−6 − 5 4𝑥2 + 1 −3𝑥−5 ]
4𝑥10
Cancel off the 2𝑥4 with part of the bottom:
𝑑𝑦
𝑑𝑥=
[𝑥 36𝑥−4 + 15𝑥−6 − 5 4𝑥2 + 1 −3𝑥−5 ]
2𝑥6
Crab-claw the numerator (do it yourself!) and you
should get:
𝑑𝑦
𝑑𝑥=
96𝑥−3 + 30𝑥−5
2𝑥6
A rule of thumb for simplification using the
quotient rule is: if you can’t factorise anything out
or do a simple crab-claw, then don’t bother.
Keeping track of all the separate sections as well as
simplification are the hardest parts of using the
quotient rule, but the more you practice, the
easier it will become.
Exercises: 1. Differentiate the following using the quotient rule:
𝑎) 𝑦 =𝑥 + 1
3𝑥 − 1
𝑏) 𝑦 =𝑥2
𝑥 + 1
𝑐) 𝑦 =2𝑥3 + 1
5𝑥 + 𝑥2
𝑑) 𝑦 =3𝑥2 + 4𝑥3
𝑥2
𝑒) 𝑦 =14𝑥4 + 5
13𝑥3 − 𝑥2
2. Differentiate the following using any of the rules learnt so far:
𝑎) 𝑦 = 3𝑥4 − 𝑥 3
5𝑥 − 1
𝑏) 𝑦 = 2𝑥 − 1 7
3𝑥 + 1 3
𝑐) 𝑦 = 15𝑥3 + 3𝑥 2
1 − 𝑥 2
𝑑) 𝑦 =12𝑥2𝑥8𝑥−9
1 + 𝑥 5
𝑒) 𝑦 =8 13𝑥2 + 14
18𝑥 − 3 2
𝑓) 𝑦 = −
5𝑥4
3+ 1
2
12𝑥2 − 𝑥
𝑔) 𝑦 =
1𝑥
+ 𝑥2 3
𝑥 − 1 2
) 𝑦 =
1𝑥
+1𝑥2
1𝑥
−1𝑥2
3
109
5.7 differentiation rules: 𝒆
Remembering that 𝑒 is just a number, there is no
reason why it could not have an index. In Chapter
4, an exponential function of the form 𝑦 = 2𝑥 was
discussed, but it could just as easily have been
𝑦 = 𝑒𝑥 . This is also an exponential function
because 𝑒 is a number, not a variable.
Theory: the formal 𝑒 rule is:
For the function 𝑦 = 𝑒𝑓 𝑥
The derivative is 𝑦′ = 𝑓 ′ 𝑥 𝑒𝑓 𝑥
The informal 𝑒 rule is: differentiate the exponent,
bring it out front in brackets, then multiply it by
the original function.
Example 1: differentiate the function
𝑦 = 𝑒2𝑥+1
Solution: differentiatie the exponent to get 2.
Bring this out front and then multiply it by the
original function:
𝑑𝑦
𝑑𝑥= (2)𝑒2𝑥+1
Example 2: differentiate
𝑔 𝑥 = 3𝑒𝑥2+2𝑥
Plan: use the exponential rule
“differentiate the exponent, bring it out front in
brackets, then multiply by the original equation”
Solution: differentiate the exponent:
𝑑𝑒𝑟 𝐸𝑋𝑃𝑂𝑁𝐸𝑁𝑇 = 2𝑥 + 2
Then bring it out front in brackets and multiply by
the original equation:
𝑑𝑦
𝑑𝑥= 2𝑥 + 2 (3𝑒𝑥2+2𝑥)
This can be simplified a bit, by putting the 3 into
each of the terms in the bracket:
𝑑𝑦
𝑑𝑥= (6𝑥 + 6)𝑒𝑥2+2𝑥
The 𝑒𝑥2+2𝑥 could be crab-clawed into the brackets
but in this case, there is no point.
Example 3: find the derivative of
𝑥 = 14𝑒𝑥3+11𝑥−1
Plan: use the 𝑒 rule of differentiation
𝑦 = 𝑒𝑓 𝑥 𝑑𝑦
𝑑𝑥= 𝑓 ′ 𝑥 𝑒𝑓 𝑥
Solution:
𝑑𝑒𝑟 𝐸𝑋𝑃𝑂𝑁𝐸𝑁𝑇 = 3𝑥2 + 11
Put this out front in brackets, then multiply by the
original function
𝑔′ 𝑥 = 3𝑥2 + 11 14𝑒𝑥3+11𝑥−1
Simplify the 14 into the first bracket to get:
𝑔′ 𝑥 = 42𝑥2 + 154 𝑒𝑥3+11𝑥−1
The 14 was simplified into the front bracket as
could the 𝑒𝑥3+11𝑥−1, however it looks “cleaner” if
we do not. Either way is fine.
Example 4: find the gradient function of
𝑦 = (4𝑥3 + 𝑥)𝑒−3𝑥−5
Plan: use the product rule as the base rule with
the 𝑒 rule within it:
𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
𝑦′ = 𝑓 ′ 𝑥 𝑒𝑓 𝑥
Solution: the two functions are underlined
𝑑𝑒𝑟 1𝑠𝑡 = 12𝑥2 + 1
𝑑𝑒𝑟 2𝑛𝑑 = (15𝑥−6)𝑒−3𝑥−5
Substitute into the base (product) rule:
𝑑𝑦
𝑑𝑥= 12𝑥2 + 1 𝑒−3𝑥−5
+ (15𝑥−6𝑒−3𝑥−5)(4𝑥3 + 𝑥)
𝑑𝑦
𝑑𝑥= 𝑒−3𝑥−5
(12𝑥2 + 1 + 60𝑥−3 + 15𝑥−5)
A step was deliberately missed, so you should do it
and see if you can get the same answer.
Exercises: 1. Differentiate the following using the 𝑒 rule:
𝑎) 𝑦 = 𝑒2𝑥−3
𝑏) 𝑦 = 𝑒−4𝑥−2
𝑦 = 4𝑥3 + 𝑥 𝑒−3𝑥−5
110
𝑐) 𝑦 = 𝑒𝑥2+1
𝑑) 𝑦 = 𝑒3𝑥2+𝑥
𝑒) 𝑦 = 𝑒1
𝑥+𝑥
2. Differentiate the following using any of the rules learnt so far:
𝑎) 𝑦 = 𝑒3𝑥2+1 1 + 𝑒3𝑥
𝑏) 𝑦 = 𝑒3−2𝑥2 10 − 𝑥3
𝑐) 𝑦 =15𝑥2 + 𝑥
𝑒9𝑥−2
𝑑) 𝑦 = 𝑒2𝑥−1 + 𝑥 − 2 4
𝑒) 𝑦 = 12 𝑒1𝑥 + 5𝑥
2
𝑓) 𝑦 = 1
𝑒3−2𝑥+ 𝑥
4
𝑔) 𝑦 = 1
𝑥+
1
𝑒4𝑥2 5
) 𝑦 =𝑒𝑥4−1
𝑥 + 𝑒3−3𝑥
5.8 differentiation rules: 𝐥𝐧 rule
The natural log is the log of a function with 𝑒 as
the base rather than 10. If you don’t understand
this, go back to Chapter 4, but it’s not entirely vital
for this section.
Theory: formally, the ln rule is:
For the function 𝑦 = ln 𝑓(𝑥)
The gradient function is 𝑦′ =𝑓 ′ 𝑥
𝑓 𝑥
Informally:
To differentiate the log of something in brackets:
“differentiate the inside of the brackets, then
divide by the inside of the brackets”.
If you are to find the derivative function of
𝑦 = ln(𝑥2 + 1)
Differentiate the inside of the brackets, then divide
by the inside of the brackets (NOT the whole
function, just the inside of the brackets). So in this
case, the derivative function would be:
𝑑𝑦
𝑑𝑥=
2𝑥
𝑥2 + 1
Ignore the ln when differentiating the function.
Example 1: find the gradient function of
𝑦 = ln 3
𝑥4 + 16𝑥 − 5
Plan: simplify the brackets, then use the ln rule:
“differentiate the inside of the brackets, then
divide by the inside of the brackets”
Solution: simplify the inside of the brackets:
𝑦 = ln 3𝑥−4 + 16𝑥 − 5
Differentiate using the ln rule:
𝑑𝑦
𝑑𝑥=
−12𝑥−5 + 16
3𝑥−4 + 16𝑥 − 5
That is the extent of the ln rule. It only gets more
complex when other rules also need to be used.
Example 2: find the derivative of
𝑦 = 3𝑥5 + 1 ∙ ln(𝑒𝑥4)
Plan: use the product rule (as there are two
functions being multiplied together) as the base
rule
𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
then the ln rule and the 𝑒 rule within the base
rule.
Solution: the two functions for the product rule
are underlined
𝑑𝑒𝑟 1𝑠𝑡 = 15𝑥4
In 𝑑𝑒𝑟 2𝑛𝑑 both the ln rule and the 𝑒 rule were
used. The 𝑒𝑥4 cancelled off so only 4𝑥3 was left.
Substitute into the base (product) rule:
𝑑𝑦
𝑑𝑥= 15𝑥4 ∙ ln 𝑒𝑥4
+ 4𝑥3 (3𝑥5 + 1)
Factorise out 𝑥3 and simplify inside the brackets:
𝑑𝑦
𝑑𝑥= 𝑥3 15𝑥 ∙ ln 𝑒𝑥4
+ 4 3𝑥5 + 1
𝑑𝑦
𝑑𝑥= 𝑥3 15𝑥 ∙ ln 𝑒𝑥4
+ 12𝑥5 + 4
𝑑𝑒𝑟 2𝑛𝑑 =4𝑥3𝑒𝑥4
𝑒𝑥4 = 4𝑥3
𝑦 = 3𝑥5 + 1 ∙ ln(𝑒𝑥4)
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Exercises: 1. Differentiate the following using the ln rule:
𝑎) ln 𝑥
𝑏) ln 15𝑥
𝑐) ln 𝑥2
𝑑) ln 𝑥2 + 3
𝑒) ln 3𝑥4 + 17𝑥2
𝑓) ln 𝑒𝑥+1 − 1
2. Differentiate the following using any of the rules learnt so far:
𝑎) 𝑦 = ln 𝑥2 + 1 3 − 𝑥4
𝑏) 𝑦 = 𝑒𝑥2+3𝑥 ln 𝑥2 + 3𝑥
𝑐) 𝑦 = ln −4𝑥3 + 7𝑥2 8
𝑥+ 𝑥3
𝑑) 𝑦 = ln 2
𝑥5− 𝑥 𝑒6−3𝑥
𝑒) 𝑦 = 1
𝑥+
2
𝑥2 ln 3𝑥2 − 1
𝑓) 𝑦 = ln 𝑒𝑥2+ 𝑥 4𝑥3
𝑔) 𝑦 =ln 5𝑥 − 3
𝑥2 + 1
) 𝑦 = ln ln 3𝑥
5.9 the second derivative
Theory: the second derivative of a function is
obtained by differentiating the function once, and
then differentiating the derivative.
Similar to the first derivative, there are several
different notations for a second derivative, the
common ones being:
𝑦′′ 𝑎𝑛𝑑 𝑑2𝑦
𝑑𝑥2
Example 1: find the second derivative of
𝑦 = 2𝑥2 + 1 3
Plan: differentiate the function, then differentiate
that derivative.
Solution: find the first derivative:
𝑦′ = 3 2𝑥2 + 1 2 4𝑥
𝑦′ = 12𝑥 2𝑥2 + 1 2
To find the second derivative, 𝑦′ must be
differentiated, and the product rule is required.
𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
𝑑𝑒𝑟1𝑠𝑡 = 12
𝑑𝑒𝑟2𝑛𝑑 = 2 2𝑥2 + 1 4𝑥
= 8𝑥 2𝑥2 + 1
Substitute it into the product rule:
𝑦′′ = 12 2𝑥2 + 1 2 + 8𝑥 2𝑥2 + 1 12𝑥
Factorise out 12 2𝑥2 + 1 :
𝑦′′ = 12 2𝑥2 + 1 2𝑥2 + 1 + 8𝑥2
𝑦′′ = 12 2𝑥2 + 1 10𝑥2 + 1
Example 2: find the second derivative of
𝑦 = 𝑒3𝑥−6 1 −𝑥3
−2
Plan: differentiate, then differentiate the
derivative.
Solution: simplify first:
𝑦 = 𝑒3𝑥−6 1 + 0.5𝑥3
Differentiate using the product rule:
𝑑𝑒𝑟1𝑠𝑡 = 3𝑒3𝑥−6
𝑑𝑒𝑟2𝑛𝑑 = 1.5𝑥2
Substitute into the base (product) rule:
𝑦′ = 3𝑒3𝑥−6 1 + 0.5𝑥3 + 1.5𝑥2𝑒3𝑥−6
Now to find the second derivative, you must use
the product rule twice (for the two sections shown
below). You should do this one section at a time to
avoid confusion:
𝑦′ = 3𝑒3𝑥−6 1 + 0.5𝑥3 + 1.5𝑥2𝑒3𝑥−6
Differentiate the red section first:
𝑑𝑒𝑟1𝑠𝑡 = 9𝑒3𝑥−6
𝑑𝑒𝑟2𝑛𝑑 = 1.5𝑥2
Substitute into the product rule formula:
𝑦′′ 𝑟𝑒𝑑 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 9𝑒3𝑥−6 1 + 0.5𝑥3 + 1.5𝑥23𝑒3𝑥−6
𝑦′′ 𝑟𝑒𝑑 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 9𝑒3𝑥−6 1 + 0.5𝑥3 + 4.5𝑥2𝑒3𝑥−6
Differentiate the blue section:
𝑑𝑒𝑟1𝑠𝑡 = 3𝑥
𝑑𝑒𝑟2𝑛𝑑 = 3𝑒3𝑥−6
Substitute into the product rule formula:
𝑦′′ 𝑏𝑙𝑢𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 3𝑥𝑒3𝑥−6 + 3𝑒3𝑥−61.5𝑥2
𝑦′′ 𝑏𝑙𝑢𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 = 3𝑥𝑒3𝑥−6 + 4.5𝑒3𝑥−6𝑥2
112
The second derivative is the sum of the derivative
of the red section and the blue section:
𝑦′′ = 9𝑒3𝑥−6 1 + 0.5𝑥3 + 4.5𝑥2𝑒3𝑥−6 + 3𝑥𝑒3𝑥−6
+ 4.5𝑒3𝑥−6𝑥2
Factorise out 𝑒3𝑥−6 because each term has 𝑒3𝑥−6:
𝑦′′ = 𝑒3𝑥−6 9 1 + 0.5𝑥3 + 4.5𝑥2 + 3𝑥 + 4.5𝑥2
Then simplify everything inside the square
brackets:
𝑦′′ = 𝑒3𝑥−6 4.5𝑥3 + 9 + 9𝑥2 + 3𝑥
After the first derivative was found, the 𝑒3𝑥−6
could have been factorised out, and then the
derivative found that way. Both methods will give
the same solution. Try it.
Example 3: find the second derivative of
𝑦 = ln(3𝑥4 + 𝑥)
Plan: differentiate the function, then apply
differentiation rules to differentiating the first
derivative.
Solution: the first derivative is
𝑦′ =12𝑥3 + 1
3𝑥4 + 𝑥
To differentiate this, the quotient rule is needed.
𝑑𝑒𝑟 𝑇𝑂𝑃 = 36𝑥2
𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 12𝑥3 + 1
Replacing the base (quotient) rule:
𝑦′′ =36𝑥2 3𝑥4 + 𝑥 − 12𝑥3 + 1 12𝑥3 + 1
3𝑥4 + 𝑥 2
Try simplifying this last equation to see if you get
the following:
𝑦′′ =−36𝑥6 + 12𝑥3 − 1
3𝑥4 + 𝑥 2
Exercises: 1. Find the second derivative of the following
functions: 𝑎) 𝑦 = 𝑥3 + 𝑥2 − 1
𝑏) 𝑦 =𝑥4
−4+ 𝑥2
𝑐) 𝑦 =1
𝑥2 + 𝑥
𝑑) 𝑦 = 3 5 − 3𝑥 5
𝑒) 𝑦 = 2𝑥2 1 + 3𝑥4
2. Find the second derivative of:
𝑎) 𝑦 = 𝑒2𝑥2 5𝑥2
𝑏) 𝑦 = ln 𝑥2
3+ 1
𝑐) 𝑦 = ln 4𝑥3 + 2𝑥2
𝑑) 𝑦 = 𝑒4𝑥 7 − 𝑒1𝑥
3
𝑒) 𝑦 = 𝑒2𝑥−1 ln 𝑒2𝑥2+ 1
5.10 the gradient function graph
Just like any function, a gradient function can be
graphed.
Theory: The gradient function is related to the
original function: any value of the gradient
function at a given 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is the gradient at
that same 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the original function.
That is, to find the gradient of any point on the
original function (which has not been shown
above), you go up from that point on the 𝑥 − 𝑎𝑥𝑖𝑠
to hit the gradient function, then across to the
𝑦 − 𝑎𝑥𝑖𝑠 which gives a value; this value is the
gradient of the original function.
On the graph above, the gradient of the original
function at 𝑥 = 1, is 2. At 𝑥 = 3, the gradient of
the original function is 7 (Note: the original
function is not shown).
Shown below is shown a function as well as its
gradient function, to reinforce the relationship.
𝑥
𝑦 𝑑𝑦
𝑑𝑥
1 3
2
7
113
Whenever the original function (blue) is sloping
upwards (i.e. a positive gradient), the gradient
function (red) is above the 𝑥 − 𝑎𝑥𝑖𝑠 (i.e. is
positive). For any 𝑥 − 𝑣𝑎𝑙𝑢𝑒 to the left of 𝑥 = −2,
the gradient of the original function is positive, and
so the gradient function to the left of 𝑥 = −2 is
above the 𝑥 − 𝑎𝑥𝑖𝑠. Similarly for any 𝑥 − 𝑣𝑎𝑙𝑢𝑒 to
the right of 𝑥 = 1.
Whenever the original function is sloping
downwards (i.e. negative gradient), the gradient
function is below the 𝑥 − 𝑎𝑥𝑖𝑠. That is, any
𝑥 − 𝑣𝑎𝑙𝑢𝑒 between 𝑥 = −2 and 𝑥 = 1, the slope
of the original function is negative, and so the
gradient function is below the 𝑥 − 𝑎𝑥𝑖𝑠.
Where there is a maximum or minimum on the
original function, the gradient is zero, and this is
where the gradient function crosses the 𝑥 − 𝑎𝑥𝑖𝑠.
That is, at 𝑥 = −2, 𝑥 = 1.
Theory: a positive gradient on the original function
gives a value above the 𝑥 − 𝑎𝑥𝑖𝑠 on the gradient
function. A negative gradient on the original
function gives a value below the 𝑥 − 𝑎𝑥𝑖𝑠 on the
gradient function. When the original function is at
a stationary point, the gradient function crosses
the 𝑥 − 𝑎𝑥𝑖𝑠. An inflection point on the original
function maps to a turning point on the gradient
function (i.e. either a maximum or a minimum).
The steeper the slope at a particular point on the
original function, the further the gradient function
is from the 𝑥 − 𝑎𝑥𝑖𝑠. Similarly, the smaller the
slope of the original function, the closer the
gradient function is to the 𝑥 − 𝑎𝑥𝑖𝑠 at that point.
Re-read this section before trying the following
exercises, as it takes a little time to fully
understand.
Exercises: 1. Sketch the original function (i.e. the quadratic):
𝑦 = 3 𝑥 + 3 2 − 5 (using theory from Chapter 4) then sketch the gradient function on the same set of axes.
2. Given the function graphed below, sketch in the gradient function.
3. Given the gradient function below, sketch in the
original function.
5.11 simple applications
Remember that the gradient function gives us a
method of finding the gradient of a point on the
original function.
Theory: to find the gradient of a point 𝑥 = 𝑎
where 𝑎 is a number, differentiate the original
function to get the gradient function, then put the
number 𝑎 into the gradient function. The value
𝑥
𝑦 𝒅𝒚
𝒅𝒙
𝑥
𝑦
𝒚 = 𝒇(𝒙)
𝑑𝑦
𝑑𝑥
𝑦 = 𝑓(𝑥)
𝑥
𝑦
−2 1
114
from the gradient function is the gradient of the
original function at 𝑥 = 𝑎.
Example 1: find the gradient of
𝑦 =𝑥2 − 3𝑥
𝑒4𝑥 − 1
when 𝑥 = −3 and also when 𝑥 = −1.
Plan: use the quotient rule as the base rule
𝑑𝑦
𝑑𝑥=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇 − 𝑑𝑒𝑟 𝐵𝑂𝑇 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
then use the 𝑒 rule to find the gradient function.
Substitute the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 (one at a time) into the
gradient function.
Solution:
𝑑𝑒𝑟 𝑇𝑂𝑃 = 2𝑥 − 3
𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 4 𝑒4𝑥
Substitute into the quotient rule:
𝑑𝑦
𝑑𝑥=
2𝑥 − 3 𝑒4𝑥 − 1 − 4 𝑒4𝑥 𝑥2 − 3𝑥
𝑒4𝑥 − 1 2
Simplification is not required as only the final
number is needed (i.e. the gradient at a point). To
get this number, substitute 𝑥 = −3 into the
gradient function:
𝑑𝑦
𝑑𝑥 −3
= −6 − 3 𝑒−12 − 1 − 4 𝑒−12 9 + 9
𝑒−12 − 1 2
Simplifying gives:
𝑑𝑦
𝑑𝑥 −3 =
−9𝑒−12 + 9 − 72𝑒−12
𝑒−12 − 1 2
=−81𝑒−12 + 9
𝑒−12 − 1 2≈ 8.9996 (4𝑝. 𝑑. )
Thus the gradient of the original function at
𝑥 = −3 is approximately 8.9996.
For the gradient at 𝑥 = −1, substitute 𝑥 = −1 into
the gradient function:
𝑑𝑦
𝑑𝑥 −1 =
−2 − 3 𝑒−4 − 1 − 4 𝑒−4 1 + 3
𝑒−4 − 1 2
Simplify:
𝑑𝑦
𝑑𝑥 𝑥 = −1 =
−5𝑒−4 + 5 − 16𝑒−4
𝑒−4 − 1 2
=−21𝑒−4 + 5
𝑒−4 − 1 2≈ 4.7892 (4𝑑. 𝑝. )
which is the gradient of the original function at
𝑥 = −1.
Example 2: a car manufacturer has a monthly
profit function
𝜋 = −𝑄2 + 150𝑄 − 5000
and is currently producing 70 cars per month. Use
the properties of the gradient function to
determine if it is more profitable for the firm to
increase production.
Plan: differentiate the profit function, then
substitute 𝑄 = 70 into the gradient function. If the
gradient is positive, more output will increase
profits. If the gradient is negative, then increased
output will reduce profits.
Solution:
𝑑𝜋
𝑑𝑄= −2𝑄 + 150
Substitute 𝑄 = 70 into this gradient function:
𝑑𝜋
𝑑𝑄 70 = −140 + 150 = 10 > 0
Since the gradient is positive, if output is
increased, profits will also increase.
Exercises: 1. Determine the gradient of the function
𝜋 = 5𝑄4 +1
𝑄− 7
At the points 𝑄 = 4, 𝑄 = 7. Give exact answers. 2. Determine the gradient of the function
𝑦 = 𝑒𝑥2 𝑥3 + 2 2
At the points 𝑄 = 2, 𝑄 = 5. Give exact answers and approximations.
3. A firm manufacturing cranes has current output of 10 and a profit function: 𝜋 = −207𝑄 log 0.03𝑄2 + 0.02𝑄 − 30𝑄 − 400
Will increasing output lead to increased profits? Justify your answer.
115
chapter five summary
Differentiation is finding the rate the 𝑦 − 𝑎𝑥𝑖𝑠 changes when there is a change in the 𝑥 − 𝑎𝑥𝑖𝑠. The gradient of a line tangent to a given point is the rate of change of 𝑦 for changes in 𝑥.
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐𝑎𝑛𝑔𝑒 = 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
The gradient function is a function that gives the gradient of the original function at all the points along the original function.
Differentiation by First Principles is the formula:
𝑓 ′ = lim𝑠→0
𝑓 𝑥 + 𝑠 − 𝑓 𝑥
𝑠
The power rule: For the original function 𝑦 = 𝑎𝑥𝑛 The derivative function is 𝑦′ = 𝑛𝑎𝑥𝑛−1
The derivative of a constant is zero
The chain rule: differentiate the function as if using the power rule, then multiply by the derivative of the inside of the brackets.
𝑑𝑦
𝑑𝑥=
𝑑𝑦
𝑑𝑢×
𝑑𝑢
𝑑𝑥
The product rule: For the function:
The derivative function is: 𝑓 ’(𝑥) = 𝑔’(𝑥). (𝑥) + 𝑔(𝑥). ’(𝑥)
𝑜𝑟 𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
The quotient rule is. For the function:
𝑦(𝑥) =𝑡(𝑥)
𝑏(𝑥)
The gradient function is:
𝑑𝑦
𝑑𝑥=
𝑡 ′ 𝑥 × 𝑏 𝑥 − 𝑏′(𝑥) × 𝑡(𝑥)
𝑏(𝑥) 2
𝑑𝑦
𝑑𝑥=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇𝑇𝑂𝑀 − 𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
The formal 𝑒 rule is:
For the function 𝑦 = 𝑒𝑓 𝑥
The derivative is 𝑦′ = 𝑓 ′ 𝑥 𝑒𝑓 𝑥
Formally, the ln rule is: For the function 𝑦 = ln 𝑓(𝑥)
The gradient function is 𝑦′ =𝑓 ′ 𝑥
𝑓 𝑥
The second derivative of a function is obtained by differentiating the function once, and then differentiating the derivative.
The gradient function is related to the original function: any value of the gradient function at a given 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is the gradient at that same 𝑥 − 𝑣𝑎𝑙𝑢𝑒 of the original function. A positive gradient on the original function gives a value above the 𝑥 − 𝑎𝑥𝑖𝑠 on the gradient function. A negative gradient on the original function gives a value below the 𝑥 − 𝑎𝑥𝑖𝑠 on the gradient function. When the original function is at a stationary point, the gradient function crosses the 𝑥 − 𝑎𝑥𝑖𝑠. An inflection point on the original function maps to a turning point on the gradient function (i.e. either a maximum or a minimum).
To find the gradient of a point 𝑥 = 𝑎 where 𝑎 is a number, differentiate the original function to get the gradient function, then put the number 𝑎 into the gradient function. The value from the gradient function is the gradient of the original function at 𝑥 = 𝑎.
chapter five questions
1. Order the points on the following graph from lowest gradient to highest gradient:
2. Use differentiation by first principles to determine
the gradient of the following functions: 𝑎) 𝑦 = 18𝑥 𝑏) 𝑦 = 𝑥2 𝑐) 𝑦 = (𝑥 + 3)(𝑥 + 4) 𝑑) 𝑦 = 𝑥3 + 4𝑥2
𝑒) 𝑦 =5
𝑥 𝑓) 𝑦 =
1
𝑥2
3. Using the power rule, differentiate the following functions: 𝑎) 𝑦 = 3𝑥2 𝑏) 𝑦 = 15𝑥2 + 𝑥 𝑐) 𝑦 = 2 + 𝑥2 + 𝑥3
𝑑) 𝑦 =17𝑥3
2+ 5𝑥2
𝑒) 𝑦 = 8𝑥−4 + 𝑥3 − 6 𝑓) 𝑦 = 12𝑥−2 + 𝑥−3 + 𝑥3 + 2
𝑔) 𝑦 =1
𝑥2
) 𝑦 =15
3𝑥2+ 𝑥2
𝑖) 𝑦 =1
𝑥+
1
𝑥2−
1
𝑥3
4. Using the chain rule, differentiate the functions: 𝑎) 𝑦 = 𝑥 + 1 2 𝑏) 𝑦 = 3𝑥 − 4 5 𝑐) 𝑦 = 2 𝑥 + 5 6 𝑑) 𝑦 = 12 1 − 4𝑥 3 𝑒) 𝑦 = 75 𝑥2 + 𝑥 4
𝑓) 𝑦 = −3 4𝑥 − 1 4
5
5. Using the product rule, differentiate the following functions (Hint: underline the two functions first): 𝑎) 𝑦 = 𝑥3 15𝑥 + 2 𝑏) 𝑦 = 𝑥3 − 1 𝑥9 + 1 𝑐) 𝑦 = 15𝑥4 1 − 𝑥3 𝑑) 𝑦 = 4𝑥−7 𝑥2 + 𝑥4
𝑒) 𝑦 = −1
3𝑥3 + 𝑥4 𝑥5 3
𝑦
𝑥
𝐴
𝐵
𝐶
𝐷 𝐸
𝐹
𝐻
𝐺
𝑓(𝑥) = 𝑔(𝑥). (𝑥)
116
𝑓) 𝑦 = 𝑥−6
−4+ 2𝑥 −1 − 𝑥−3
𝑔) 6 𝑥−6 − 6 𝑥6 − 6𝑥 6. Use the quotient rule to find the derivative of the
following functions:
𝑎) 𝑦 =𝑥 + 1
𝑥 − 1 𝑏) 𝑦 =
2𝑥 + 1
1 − 2𝑥
𝑐) 𝑦 =3𝑥 − 2
1 − 𝑥2 𝑑) 𝑦 =
𝑥2 + 1
𝑥2 − 1
𝑒) 𝑦 =
13𝑥2 + 1
5 − 𝑥 𝑓) 𝑦 =
𝑥2 − 𝑥3
𝑥2 − 2
𝑔) 𝑦 =13𝑥−3 − 𝑥3
1 + 𝑥−4+ 9
) 𝑦 = 𝑥2 + 1 3 + 𝑥
1 − 3𝑥2
7. Use the 𝑒 rule to find the derivative of the following functions: 𝑎) 𝑦 = 𝑒2𝑥 𝑏) 𝑦 = 𝑒5𝑥−1
𝑐) 𝑦 = 𝑒1−4𝑥 𝑑) 𝑦 = 𝑒𝑥2−1
𝑒) 𝑦 = 𝑒13𝑥−𝑥2 𝑓) 𝑦 = 2𝑒𝑥4−1+𝑥−3
𝑔) 𝑦 = 12𝑒4/𝑥 ) 𝑦 =2𝑒𝑥−3−
𝑥2
−7
8. Using the ln rule, find the derivative of the following functions: 𝑎) 𝑦 = ln 5𝑥 𝑏) 𝑦 = ln 𝑥2 𝑐) 𝑦 = 2 ln 5𝑥 − 1 𝑑) 𝑦 = −3 ln 𝑥2 − 𝑥
𝑒) 𝑦 = −4
−3ln 𝑥−3 + 𝑥
𝑓) 𝑦 = 2 ln 𝑥2 − 𝑥−3 +1
𝑥
𝑔) 𝑦 = 0.04 ln 8𝑥 − 2𝑥−2 9. Using any of the rules, determine the derivative of
the functions: 𝑎) 𝑦 = 2𝑥3 − 1 2 3𝑥 + 5 3 𝑏) 𝑦 = 1 − 𝑒−5𝑥−1 𝑥 − 1 4
𝑐) 𝑦 = ln 𝑒0.2𝑥3+𝑥
𝑑) 𝑦 =1 − 𝑒𝑥2
1 + 𝑒𝑥2
𝑒) 𝑦 = 15 ln 12 − 𝑥−3 4
𝑓) 𝑦 = 1
𝑥5 − 1 1 + ln 𝑥
𝑔) 𝑦 = 5𝑥2𝑒3𝑥−1
) 𝑦 =1
−3∙
𝑒𝑥 + 1
2𝑥−3 − 𝑥4
𝑖) 𝑦 = 2𝑥3 − 1 7 1 + ln 𝑒𝑥2+ 1
𝑗) 𝑦 =𝑒7𝑥−1 + 𝑥4
ln 𝑥5 + 𝑥−7+ 18𝑥4 − ln 𝑥7 − 1
10. Find the first and second derivatives of the following functions: 𝑎) 𝑦 = 𝑥4 + 15𝑥3 − 1
𝑏) 𝑦 =17
𝑥4+ ln 8𝑥
𝑐) 𝑦 = 0.03𝑥9 + 𝑥2 − 1 4
𝑑) 𝑦 = 𝑒2 1 − 𝑥4 + 𝑒𝑥2−1 1 − 𝑥
𝑒) 𝑦 = 12 − 𝑥
3𝑥2 + 1
𝑓) 𝑦 = ln 5𝑥2 + 𝑥 11. Determine the gradient at 𝑄 = 10 and at 𝑄 = 35
for the production function: 𝑄 = ln 𝑥0.25 + 4𝑥
12. For the following original function, sketch in the gradient function:
13. For the following gradient function, sketch in the
original function:
14. For the total cost function:
𝑇𝐶 = 𝑄3 − 15𝑄2 + 75𝑄 − 5 Determine: a) The gradient function (i.e. marginal cost
function). b) When the marginal cost function is at a
minimum. c) The gradient of the marginal cost function at
𝑄 = 15. d) Sketch the marginal cost function using
quadratic theory. e) Sketch the total cost function from the
marginal cost function. 15. A company producing fishing rods faces a cost
function of the form: 𝑇𝐶 = 50 ln 0.1𝑥2 + 2 3 + 15𝑥 − 18
Determine: a) The marginal cost function. b) The gradient of the marginal cost function
when 𝑄 = 30 and when 𝑄 = 60.
𝑥
𝑦
𝑓 ′ 𝑥
𝑥
𝑦
𝑦 = 𝑓(𝑥)
117
Chapter 6
Applications of differentiation
The usefulness of differentiation
6.1 Graphical Optimisation 118
6.2 Mathematical Optimisation 118
6.3 The Nature of an Optimal Point 120
6.4 Inflection Points 123
6.5 Combining All Theory 124
6.6 Applications – Profits 126
6.7 Applications – Break-Even 129
6.8 Applications: Marginal and Average Values 130
6.9 Differentiation and Elasticity 133
6.10 Elasticity and Total Revenue 135
Chapter Six Summary 137
Chapter Six Questions 138
118
6.1 graphical optimisation
Theory: optimisation is the process of finding an
optimal point, with that point usually being a
maximum or minimum.
A shop selling mobile phones aims to maximise
profit. As a consumer, you would like to minimise
how much you pay for a given phone. These are
two optimisation problems; in one case, the firm is
finding a maximum, and in the other case, the
consumer is finding a minimum.
The profit of a firm (usually denoted by 𝜋) is a
function of the quantity of goods sold to
consumers. The following is a typical profit curve,
with 𝜋 on the 𝑦 − 𝑎𝑥𝑖𝑠 and quantity, 𝑄, on the
𝑥 − 𝑎𝑥𝑖𝑠. How many units would you produce?
Any rational person would aim to sell 50 units. Any
other quantity, such as 25 or 90 units, does not
give maximum profit, so these are not optimal
points.
The maximum above is both a local and global
maximum.
Theory: a global optimum is a point that is the
optimal point over the entire number range (i.e.
𝑥 − 𝑎𝑥𝑖𝑠). A local optimum is an optimal point
only over a small range around that point.
Above, there are two maximum points and one
minimum point. Point 𝐴 is a local maximum (a
maximum over a small range around the point),
whereas point 𝐶 is a local and global maximum (a
maximum over the entire number range: 𝑥 −
𝑎𝑥𝑖𝑠). Point 𝐵 is a local minimum, but is not a
global minimum as the ends of the curve are
lower.
Be aware that an optimal point may not always be
the globally optimal point (most of the time it will
be, but you have to make sure).
Exercises: 1. For the graph below, determine which points are
local maxima/minima and global maxima/minima.
6.2 mathematical optimisation
Finding optimal points graphically is quite simple,
however most of the time only the original
function is given, which may be difficult to draw.
This is where differentiation becomes very useful.
Below is a graph where the gradients of optimal
point have been drawn in.
𝑦
𝑥
𝐴
𝐵
𝐶
𝐷
𝐸
𝐹
𝐺
𝑦
𝑥
𝐴
𝐵
𝐶
𝜋
𝑄
𝜋 𝑄
50 25 90
119
Theory: the gradient at any maximum or any
minimum is always zero.
This feature allows the original function to be
differentiated to obtain the gradient function; the
gradient function is set equal to zero to solve for
the optimal point(s).
Theory: the First Order Condition (FOC) is found by
setting the first derivative equal to zero:
𝑑𝑦
𝑑𝑥= 0
That is, differentiate the function and set it equal
to zero. Solve for the unknown(s).
Example 1: Find the optimal point(s) of the
quadratic
𝑦 = 2𝑥2 − 8𝑥 − 1
Plan: use the FOC to find the 𝑥 − 𝑣𝑎𝑙𝑢𝑒 𝑠 of the
optimal point(s)
𝑑𝑦
𝑑𝑥= 0
Solution: differentiate the original function:
𝑑𝑦
𝑑𝑥= 4𝑥 − 8
Set it equal to zero:
𝑑𝑦
𝑑𝑥= 4𝑥 − 8 = 0
Rearrange to solve for 𝑥:
4𝑥 = 8
𝑥 = 2
There is an optimal point at 𝑥 = 2.
A point also requires a 𝑦 − 𝑣𝑎𝑙𝑢𝑒 which comes
from the original function:
𝑦 2 = 2 2 2 − 8 2 − 1
𝑦 2 = −9
Thus 2, −9 is an optimal point.
Example 2: find the optimal point(s) for the
function
𝑦 = 𝑥3 + 3𝑥2 − 24𝑥 + 5
Plan: use the FOC to find the optimal point(s).
𝑑𝑦
𝑑𝑥= 0
Solution: differentiate the original function and set
it to zero (i.e. FOC):
𝑑𝑦
𝑑𝑥= 3𝑥2 + 6𝑥 − 24 = 0
Solving for 𝑥 requires solving this quadratic. To
solve the quadratic, the Quadratic Formula must
be used:
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−6 ± 62 − 4 3 (−24)
2(3)
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−6 ± 324
6
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = −4 OR 𝑥 − 𝑣𝑎𝑙𝑢𝑒 = 2
This gives two optimal points, but they also require
𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 which come from the original function:
𝑦 −4 = −4 3 + 3 −4 2 − 24 −4 + 5
𝑦 −4 = 85
𝑦 2 = 2 3 + 3 2 2 − 24 2 + 5
𝑦 2 = −23
Both −4,85 and 2, −23 are optimal points.
From the cubics section in Chapter 4, there should
be one local maximum and one local minimum.
Determining which value corresponds to the
maximum and which to the minimum is left to the
next section.
𝑦
𝑥
𝑓 𝑥
120
Example 3: find the optimal points of the function
𝑓 𝑥 = 𝑥2𝑒2𝑥+3 + 7
Plan: differentiate using the product rule as the
base rule:
𝑑𝑦
𝑑𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
Use the 𝑒 rule within the product rule:
𝑦′ = 𝑓 ′ 𝑥 𝑒𝑓 𝑥
Then apply the FOC:
𝑑𝑦
𝑑𝑥= 0
Solution: define the two functions:
Find the derivative of each of the two parts:
𝑑𝑒𝑟 1𝑠𝑡 = 2𝑥
𝑑𝑒𝑟 2𝑛𝑑 = 2𝑒2𝑥+3
Substitute into the product rule:
𝑑𝑓 𝑥
𝑑𝑥= 2𝑥𝑒2𝑥+3 + 2𝑒2𝑥+3 𝑥2
Simplify and factorise:
𝑑𝑓 𝑥
𝑑𝑥= 2𝑥𝑒2𝑥+3 1 + 𝑥
Apply the FOC:
2𝑥𝑒2𝑥+3 1 + 𝑥 = 0
For the solution, one of the products on the left
side must equal zero. There are three options:
1. 2𝑥 = 0, so 𝑥 = 0;
2. 𝑒2𝑥+3 = 0, but this will never occur as this
never crosses the 𝑥 − 𝑎𝑥𝑖𝑠; or
3. 1 + 𝑥 = 0, giving 𝑥 = −1.
There are two optimal points with 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠
𝑥 = 0 and 𝑥 = −1. For the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠:
𝑓 0 = 0 2𝑒2 0 +3 + 7
𝑓 0 = 7
𝑓 −1 = −1 2𝑒2 −1 +3 + 7
𝑓 −1 = 𝑒1 + 7 = 𝑒 + 7
Thus 0,0 and −1, 𝑒 + 7 are optimal points.
Exercises: 1. Find any stationary point(s) for the functions:
𝑎) 𝑦 = 𝑥2 − 13𝑥 + 15
𝑏) 𝑦 = 3𝑥3 + 2𝑥2 − 5𝑥 + 15
𝑐) 𝑦 = −7𝑥3
8+ 12𝑥 − 1
𝑑) 𝑦 = 3𝑥2𝑒4𝑥−1
𝑒) 𝑦 =𝑥2 + 1
𝑥 − 3
𝑓) 𝑦 = 𝑒3𝑥+1 + 2𝑒−2𝑥+1 − 6
𝑔) 𝑦 = −𝑥2 + 3𝑥 + 5 𝑒𝑥+5
) 𝑦 = 𝑒𝑥2 𝑥2 − 3
6.3 the nature of an optimal point
After finding an optimal point, it is important to
determine if that point is a minimum or maximum.
Profit is something which is usually maximised and
if an optimal point was chosen using the FOC
without checking it is a maximum, profit might in
fact be minimised.
The nature of an optimal point is found (i.e. if an
optimal point is a maximum or minimum) using
the second derivative.
Theory: the Second Order Condition (SOC) is used
to find the nature of an optimal point. The point in
question is:
A minimum if the second derivative evaluated
at the optimal point is greater than zero.
Minimum if 𝑑2𝑦
𝑑𝑥2> 0
A maximum if the second derivative evaluated
at the optimal point is less than zero.
Maximum if 𝑑2𝑦
𝑑𝑥2< 0
A stationary inflection point if the second
derivative evaluated at the optimal point is
equal to zero.
Stationary inflection point if 𝑑2𝑦
𝑑𝑥2= 0
The process is quite simple and is easily
demonstrated in an example.
𝑓 𝑥 = 𝑥2𝑒2𝑥+3 + 7
121
Example 1: find the optimal point and the nature
of this optimal point for the function
𝑦 = −𝑥2 + 2𝑥 + 15
Plan: use the FOC to find any stationary points
𝑑𝑦
𝑑𝑥= 0
Find the second derivative and evaluate at the
optimal point. Then apply the SOC to determine if
it is a maximum or minimum.
Solution: to find any stationary points:
𝑑𝑦
𝑑𝑥= −2𝑥 + 2 = 0
−2𝑥 = −2
𝑥 = 1
The 𝑦 − 𝑣𝑎𝑙𝑢𝑒 is obtained from the original
function:
𝑦 1 = − 1 2 + 2 1 + 15
𝑦 1 = 16
The second derivative is:
𝑑2𝑦
𝑑𝑥2= −2
Apply the SOC:
𝑑2𝑦
𝑑𝑥2= −2 < 0 ∴ maximum
Thus there is a maximum at 1,16 .
From Chapter 4, since the value of 𝑎 in the
quadratic is negative, a maximum is expected.
Example 2: find any optimal points and their
nature for the function
𝑦 = 𝑥3 − 12𝑥2 + 36𝑥 + 24
Plan: use the FOC to find any optimal points. Then
find the second derivative, and evaluate it at any
optimal points; apply the SOC to the results.
Solution: find the optimal points
𝑑𝑦
𝑑𝑥= 3𝑥2 − 24𝑥 + 36 = 0
Solve using the Quadratic Formula:
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−(−24) ± (−24)2 − 4 3 (36)
2(3)
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =24 ± 576 − 432
6
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =24 ± 12
6
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 = 2 𝑂𝑅 𝑥 − 𝑣𝑎𝑙𝑢𝑒 = 6
The two optimal points have 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of 𝑥 = 2
and 𝑥 = 6. The corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 are:
𝑦 2 = 2 3 − 12 2 2 + 36 2 + 24
𝑦 2 = 56
𝑦 6 = 6 3 − 12 6 2 + 36 6 + 24
𝑦 6 = 24
Thus the two points are (2,56) and (6,24).
Find the second derivative:
𝑑2𝑦
𝑑𝑥2= 6𝑥 − 24
Evaluate the second derivative at 𝑥 = 2:
𝑑2𝑦
𝑑𝑥2 2 = 6 2 − 24 = −12 < 0
∴ maximum at 𝑥 = 2
Evaluate the second derivative at 𝑥 = 6:
𝑑2𝑦
𝑑𝑥2 6 = 6 6 − 24 = 12 > 0
∴ minimum at 𝑥 = 6
Thus 2,56 is a maximum, and 6,24 is a
minimum.
To summarise the process:
1. Find the optimal 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 from the FOC
and their corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠.
2. Determined if the point(s) are maxima or
minima by substituting those 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠
from the FOC into the second derivative,
3. Applying the SOC.
Example 3: find and define any optimal points for
𝑦 = 𝑒𝑥2−3 + ln 4 − 𝑥2
Plan: find any optimal points using the FOC
𝑑𝑦
𝑑𝑥= 0
122
The 𝑒 rule and power rule need to be used. Then
find and evaluate the second derivative at the
optimal points, and apply the SOC.
Solution: find the optimal points using the FOC:
𝑑𝑦
𝑑𝑥= 2𝑥𝑒𝑥2−3 − 2𝑥 = 0
2𝑥𝑒𝑥2−3 − 2𝑥 = 0
2𝑥(𝑒𝑥2−3 − 1) = 0
So either 2𝑥 = 0 or 𝑒𝑥2−3 − 1 = 0
If 2𝑥 = 0, then 𝑥 = 0.
If 𝑒𝑥2−3 − 1 = 0 then
𝑒𝑥2−3 = 1
The only time when an exponential is equal to 1 is
when the exponent is equal to zero. So 𝑥2 − 3 = 0
is required:
𝑥2 = 3
𝑥2 = 3
𝑥 = − 3 and 𝑥 = + 3
There are three solutions for optimal points:
𝑥 = 0, 𝑥 = − 3 and 𝑥 = + 3
For the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠:
𝑦 0 = 𝑒 0 2−3 + ln 4 − 0 2
= 𝑒−3 + ln 4
𝑦 − 3 = 𝑒 − 3 2−3 + ln 4 − − 3
2
= ln 4 − 2
𝑦 3 = 𝑒 3 2−3 + ln 4 − 3
2
= ln 4 − 2
Giving points − 3, ln 4 − 2 , 3, ln 4 − 2 and
0, 𝑒−3 + ln 4 .
Use the second derivative to find the nature of the
three points:
𝑑2𝑦
𝑑𝑥2= 2𝑒𝑥2−3 + 2𝑥 𝑒𝑥2−3 2𝑥 − 2
= 2𝑒𝑥2−3 + 4𝑥2 𝑒𝑥2−3 − 2
The product rule was used to evaluate the second
derivative however simplification is not needed as
all that is required is to evaluate the second
derivative at each of the three optimal points:
1. For 𝑥 = 0:
𝑑2𝑦
𝑑𝑥2(0) = 2𝑒0−3 + 4(0)2 𝑒0−3 − 2
= 2𝑒−3 − 2 ≈ −1.9 < 0 ∴ maximum
The −1.9 can be obtained from a calculator.
2. For 𝑥 = − 3:
𝑑2𝑦
𝑑𝑥2(− 3) = 2𝑒(− 3)2−3 + 4(− 3)2 𝑒(− 3)2−3 − 2
= 2𝑒0 + 12𝑒0 − 2 = 12 > 0 ∴ minimum
3. For 𝑥 = 3:
𝑑2𝑦
𝑑𝑥2 + 3 = 2𝑒 + 3
2−3 + 4 + 3
2 𝑒 + 3 2−3 − 2
= 2𝑒0 + 12𝑒0 − 2 = 12 > 0 ∴ minimum
Thus − 3, ln 4 − 2 and 3, ln 4 − 2 are
minima, and 0, 𝑒−3 + ln 4 is a maxima. The two
minimums are flanking a maximum (the shape of a
W).
Note: a good way of remembering the SOC is that
it is not what is intuitively expected. If the second
derivative at the point in question is greater than
zero, it is a minimum (see, it’s not what is
intuitively expected). Similarly, if the second
derivative at the point in question is less than zero,
it is a maximum (again, not what is intuitively
expected).
Exercises: 1. Find any stationary points for the functions below,
then determine the nature of these stationary points.
𝑎) 𝑦 = 3𝑥2 − 11𝑥 + 10
𝑏) 𝑦 = 𝑥3 + 4𝑥2 − 5𝑥 + 1
𝑐) 𝑦 = −2𝑥3
9+ 2𝑥 − 19
𝑑) 𝑦 = 5𝑥2𝑒−7𝑥+4
𝑒) 𝑦 =4𝑥2 + 3
5 − 𝑥
𝑓) 𝑦 = 𝑒−3𝑥−6 + −5
6 𝑒7𝑥+1 − 5
𝑔) −3𝑥2 + 5𝑥 + 12 𝑒𝑥+4
) 𝑦 = 3𝑒𝑥2 2𝑥2 − 8
123
6.4 inflection points
Theory: Inflection points are points of a function
where a function changes from concave to convex
(or vice versa).
Concave means that the curve looks like part of a
cave when drawn on a set of axes; and convex is
when a “cave” cannot be drawn (for a technical
definition, consult a second year mathematics for
economists text book).
Below, the red lines are concave, as they could be
drawn into a smooth “cave” shape, whereas the
blue lines cannot, so they are convex.
Theory: an inflection point is where a function
changes from convex to concave (or concave to
convex).
But a much easier way of understanding this is to
draw an inflection point.
Notice that the function is initially concave (red)
and then changes to convex (blue). The point at
which this change occurs is the inflection point.
Theory: There are two main types of inflection
points:
1. Stationary inflection points have a gradient of
zero at the inflection point, but are not
maxima nor minima.
2. Non-stationary inflection points do not have a
gradient of zero at the inflection point.
The blue function has a non-stationary inflection
point, whereas the red function has a stationary
inflection point.
The mathematical way of finding inflection points
is a continuation of the SOC.
Theory: if the SOC is evaluated for a stationary
point and is found to be zero, that point is a
stationary inflection point.
A generic way of finding any inflection point (not
just a stationary inflection point) is a continuation
of the SOC:
SOC for any inflection point: 𝑑2𝑦
𝑑𝑥2= 0
That is, set the second derivative equal to zero,
then solve for 𝑥. This finds all inflection points.
Example 1: find any inflection point(s) (if they
exist) for the function
𝑦 = 𝑥3 − 3𝑥2 − 28𝑥 + 60
Plan: find the second derivative, set it to zero then
solve for 𝑥.
Solution: differentiate the function twice:
𝑦
𝑥
stationary inflection point
non-stationary inflection point
𝑦
𝑥
inflection point
𝑦
𝑥
124
𝑑𝑦
𝑑𝑥= 3𝑥2 − 6𝑥 − 28
𝑑2𝑦
𝑑𝑥2= 6𝑥 − 6
Set the second derivative equal to zero.
6𝑥 − 6 = 0
𝑥 = 1
There is an inflection point at 𝑥 = 1. To find the
corresponding 𝑦 − 𝑣𝑎𝑙𝑢𝑒 use the original function.
𝑦(1) = 1 3 − 3 1 2 − 28(1) + 60
= 1 − 3 − 28 + 60 = 30
There is an inflection point at (1,30).
Theory: if the FOC and SOC (inflection point) have
the same 𝑥 − 𝑣𝑎𝑙𝑢𝑒, then that inflection point is a
stationary inflection point.
Example 2: prove the following function has one
inflection point, and that it is a stationary
inflection point
𝑓 𝑥 = 𝑥 − 1 3 + 2
Plan: find any stationary points using the FOC
𝑑𝑓 𝑥
𝑑𝑥= 0
Then find any inflection points by setting the
second derivative equal to zero
𝑑2𝑓 𝑥
𝑑𝑥2= 0
If a single 𝑥 − 𝑣𝑎𝑙𝑢𝑒 is both an inflection and
stationary point, then it must be a stationary
inflection point.
Solution: find any stationary points using the FOC:
𝑑𝑓 𝑥
𝑑𝑥= 3 𝑥 − 1 2 = 0
For the left side to equal zero, (𝑥 − 1) must equal
zero, as then 3 0 2 = 0. So if 𝑥 − 1 = 0, 𝑥 = 1.
There is only one stationary point at 𝑥 = 1.
Expanding the brackets and applying the Quadratic
Formula will give the same solution.
To find any stationary points:
𝑑2𝑓 𝑥
𝑑𝑥2= 6 𝑥 − 1 = 0
Again, for the left side to be equal to zero (𝑥 − 1)
must equal zero. This means 𝑥 = 1.
Since 𝑥 = 1 is a stationary point (from the FOC)
and at that same point it is also an inflection point
(from the SOC), then it must be a stationary
inflection point.
Finally, to find the 𝑦 − 𝑣𝑎𝑙𝑢𝑒 at 𝑥 = 1, substitute
𝑥 = 1 in to the original function:
𝑓 1 = 1 − 1 3 + 2 = 2
There is a stationary inflection point at (1,2).
Exercises: 1. Determine any inflection points for the functions
below (Note: some functions do not have inflection points).
𝑎) 𝑦 = 𝑥3 + 5𝑥2 − 2𝑥 + 1
𝑏) 𝑦 = −3𝑥3 + 7𝑥 − 13
𝑐) 𝑦 =−13𝑥3
4+ 12𝑥2 − 54 + 𝑥
𝑑) 𝑦 = 2𝑥4 + 13𝑥3 − 12𝑥
𝑒) 𝑦 = 0.25𝑥4 − 𝑥3 − 10𝑥2 + 12𝑥 − 71
𝑓) 𝑦 =𝑥2 + 1
𝑥 − 3
𝑔) 𝑦 = −2𝑥2 + 4𝑥 + 10 𝑒𝑥+3
) 𝑦 = ln 𝑥2 − 9
6.5 combining all theory
Theory summary: The FOC finds the stationary
point(s) of a function.
𝐹𝑂𝐶: 𝑑𝑦
𝑑𝑥= 0
Evaluating the second derivative at any stationary
points, then applying the SOC finds the nature:
𝑑2𝑦
𝑑𝑥2> 0 is a minimum
𝑑2𝑦
𝑑𝑥2< 0 is a maximum
𝑑2𝑦
𝑑𝑥2= 0 is a stationary inflection point
To find any inflection points, set the second
derivative to zero and solve for 𝑥:
125
𝑑2𝑦
𝑑𝑥2= 0
To find 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠, substitute the respective
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 into the original function.
Example 1: find the nature of all stationary and
inflection points for the function
𝑓 𝑥 = 0.25𝑥4 − 18𝑥2 + 31
Plan: use the FOC to find any stationary points:
𝑑𝑦
𝑑𝑥= 0
Then evaluate the second derivative at these
points, and apply the SOC. Finally, set the second
derivative equal to zero to find any inflection
points.
Solution: use the FOC to find any stationary points:
𝑑𝑓 𝑥
𝑑𝑥= 𝑥3 − 36𝑥 = 0
𝑥3 − 36𝑥 = 0
𝑥(𝑥2 − 36) = 0
So either 𝑥 = 0 𝑂𝑅 𝑥2 − 36 = 0
Separate the second part into two answers:
𝑥2 − 36 = 0
𝑥2 = 36
𝑥 = ±6
The stationary points have 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 𝑥 = 0, 𝑥 =
−6 and 𝑥 = 6.
Find the nature of these points:
𝑑2𝑦
𝑑𝑥2= 3𝑥2 − 36
1. 𝑑2𝑦
𝑑𝑥2 0 = 3 0 2 − 36 = −36 < 0
∴ maximum
2. 𝑑2𝑦
𝑑𝑥2 −6 = 3 −6 2 − 36 = 72 > 0
∴ minimum
3. 𝑑2𝑦
𝑑𝑥2 6 = 3 6 2 − 36 = 72 > 0
∴ minimum
A maximum at 𝑥 = 0 is flanked by minima at
𝑥 = −6 and 𝑥 = 6.
To find the inflection points:
𝑑2𝑦
𝑑𝑥2= 3𝑥2 − 36 = 0
3𝑥2 = 36
𝑥2 = 12
𝑥 = ± 12
There are two inflection points; one at 𝑥 = − 12
and another at 𝑥 = + 12.
Having figured out the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of all the
important points, the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 can now be
found using the original function (see if you get the
same 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠):
minimum at (−6, −293)
inflection point at − 12, −149
maximum at 0,31
inflection point at ( 12, −149)
minimum at (6, −293)
Sketching this function is quite simple; plot these 5
points, and at each minimum draw a small happy
face, and then at each maximum draw a small sad
face:
Then simply extend the lines to form a curve.
Obviously, this sketch is not to scale, but as long as
the general shape is correct and the important
points labelled, that is all that matters. The
𝑦
𝑥
(0,31)
( 12, −149) (− 12, −149)
(6, −293) (−6, −293)
126
𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 could also be found, but they are
beyond the scope of this book.
Example 2: determine the nature of any stationary
points, as well as any inflection points for
𝑔 𝑥 = 3𝑥𝑒−𝑥−4
Plan: use the FOC and SOC to find the nature and
location of any stationary point(s), then set the
second derivative equal to zero to find any
inflection points.
Solution: the FOC using the product and 𝑒 rules:
𝑑𝑒𝑟 1𝑠𝑡 = 3
𝑑𝑒𝑟 2𝑛𝑑 = −1 𝑒−𝑥−4
Substituting into the product rule:
𝑔′ 𝑥 = 3𝑒−𝑥−4 + 3𝑥 −1 𝑒−𝑥−4
𝑔′ 𝑥 = 3𝑒−𝑥−4 1 − 𝑥 = 0
There are two possibilities:
1. 3𝑒−𝑥−4 = 0, which is not possible, as this
exponential never crosses the 𝑥 − 𝑎𝑥𝑖𝑠.
2. 1 − 𝑥 = 0 therefore 𝑥 = 1.
There is only one stationary point. To determine
the nature of this point, find the second derivative
(using the product rule on 𝑔′ 𝑥 above):
𝑔′′ 𝑥 = 3 −1 𝑒−𝑥−4 1 − 𝑥 + −1 3𝑒−𝑥−4
Simplify this to the extreme (easier to work with):
𝑔′′ 𝑥 = −3𝑒−𝑥−4 1 − 𝑥 − 3𝑒−𝑥−4
𝑔′′ 𝑥 = −3𝑒−𝑥−4 1 − 𝑥 + 1
𝑔′′ 𝑥 = −3𝑒−𝑥−4 2 − 𝑥
Evaluate the second derivative at the 𝑥 = 1:
𝑔′′ 1 = −3𝑒−1−4 2 − 1 ≈ −0.0202
𝑎𝑠 − 0.0202 < 0 ∴ maximum
To find any inflection points, set the second
derivative equal to zero:
𝑔′′ 𝑥 = 0
−3𝑒−𝑥−4 2 − 𝑥 = 0
For this to be true, either −3𝑒−𝑥−4 = 0 which
we know from our work with exponentials is never
the case, or 2 − 𝑥 = 0 which is true when 𝑥 = 2.
Thus there is an inflection point at 𝑥 = 2.
Finally, the 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 of the two points are
obtained from the original function:
𝑔 1 = 3 1 𝑒−1−4 =3
𝑒5
𝑔 2 = 3 2 𝑒−2−4 =6
𝑒6
Remember index rules; that is why the 𝑒’s become
denominators.
There is a maximum at (1,3
𝑒5) and an inflection
point at (2,6
𝑒6). The inflection point is not a
stationary inflection point as there is no
corresponding 𝑥 − 𝑣𝑎𝑙𝑢𝑒 in the FOC.
Exercises: 1. Determine the nature of any stationary points for
the following functions, then determine if there are any inflection points. 𝑎) 𝑦 = 3𝑥4 + 5𝑥3 − 2𝑥2 − 14
𝑏) 𝑦 = 4𝑥 − 3 4
𝑐) 𝑦 = ln 5𝑥 + 3 + 𝑥 − 2 𝑥 − 3
𝑑) 𝑦 = ln 8𝑥 + 7 + 𝑥 − 1.5 0.5𝑥 − 2.75
𝑒) 𝑦 = −4𝑥2 + 4𝑥 + 12 𝑒5+𝑥
2. Sketch the following functions after the nature of all stationary points and inflection points have been found. 𝑎) 𝑦 = 𝑥 𝑥 − 3 𝑥 + 7
𝑏) 𝑦 = 𝑥 − 5 𝑥 − 3 2 𝑥 + 5
𝑐) 𝑦 = 5𝑥4 − 3𝑥3 + 12𝑥2 − 8𝑥 + 18
6.6 applications – profit
Intro example: you have just finished your
business degree and all you want to do is lie on the
beach all day and not work. But you need to make
money, so you open a small shop on the sand and
you sell surfboards. You obviously want to
minimise your time in the shop and also maximise
your profits. You’ve done a business degree so you
sit down one day and determine the approximate
demand function for surfboards to be
𝑃𝑑 = 350 − 5𝑄𝑑
127
You purchase the surfboards from a supplier for
$50 each, and it costs you $400/month to rent the
sand from the local government. What price
should you set to maximise your profits? This is a
very common application of optimisation.
Theory: the general form of profit is:
𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 − 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡𝑠
𝜋 = 𝑇𝑅 − 𝑇𝐶
Total revenue can also be simplified:
𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑃𝑟𝑖𝑐𝑒 × 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑠𝑜𝑙𝑑
𝑇𝑅 = 𝑃 ∙ 𝑄
If you sold 10 surfboards at $200 each, total
revenue would be 10 × $200 = $2000. But there
is another layer: the quantity of goods sold
depends on the demand for the goods. Thus
𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 depends on the demand function.
To find profits, costs need to be defined.
Theory: in general, total costs are represented by:
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡𝑠 = 𝐹𝑖𝑥𝑒𝑑 𝐶𝑜𝑠𝑡𝑠 + 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐶𝑜𝑠𝑡𝑠
𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶
Variable costs are usually represented by:
𝑉𝐶 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑄
Where the constant is how much a single item
costs from the supplier.
Above, the $400/month to rent the area on the
sand is a fixed cost, because it does not matter
how many surfboards are sold, that $400 still
needs to be paid to the local government. The
variable costs are those costs which are affected
when a good is sold. When a surfboard is sold at
the beach, you as the owner would have bought
that from a supplier.
Theory summary:
𝜋 = 𝑇𝑅 − 𝑇𝐶
𝑇𝑅 = 𝑃 ∙ 𝑄
𝑑𝑒𝑚𝑎𝑛𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑡𝑒 𝑓𝑜𝑟𝑚 𝑃 𝑄
𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶
All functions are in terms of 𝑄 as profits are
directly determined by the quantity of goods sold.
To solve the above example:
𝑇𝑅 = 𝑃 ∙ 𝑄
Insert the demand function into the 𝑇𝑅 function to
substitute away the 𝑃 and have 𝑇𝑅 in terms of 𝑄:
𝑇𝑅 = 350 − 5𝑄 𝑄
Simplify:
𝑇𝑅 = 350𝑄 − 5𝑄2
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡𝑠 are:
𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶
𝑇𝐶 = 400 + 50𝑄
The 400 is the rent, and the 50𝑄 is the cost of
buying 𝑄 surfboards from the supplier (i.e. $50 for
each surfboard multiplied by the quantity of
surfboards sold, 𝑄).
Now that both 𝑇𝑅 and 𝑇𝐶 have been found, the
profit function can be determined.
𝜋 = 𝑇𝑅 − 𝑇𝐶
𝜋 = 350𝑄 − 5𝑄2 − 400 + 50𝑄
𝜋 = 350𝑄 − 5𝑄2 − 400 − 50𝑄
𝜋 = −5𝑄2 + 300𝑄 − 400
Note: the 𝑇𝐶 function was written in brackets, to
emphasise the negative sign goes into all terms
inside that bracket. This is where many students
screw up, so don’t be one of them.
The quantity of surfboards to be sold to maximise
profit is found by determining the maximum of this
profit function using the FOC and SOC:
𝑑𝜋
𝑑𝑄= −10𝑄 + 300 = 0
−10𝑄 = −300
𝑄 = 30
To make sure this is a maximum:
𝑑2𝜋
𝑑𝑄2= −10
128
𝑑2𝜋
𝑑𝑄2 30 = −10 < 0 ∴ maximum
Thus 30 surfboards need to be sold to maximise
profit. However, as a supplier, you can never
control how many surfboards you sell, but you can
control the price. The demand function gives the
price required to sell 30 surfboards per month,
which maximises profit.
𝑃 = 350 − 5𝑄
𝑃 30 = 350 − 5 30
𝑃 30 = 200
The surfboards should be priced at $200 each.
Lastly, how much profit will be made per month if
30 boards are sold? Using the profit function,
because as its name implies, it determines profit:
𝜋 = −5𝑄2 + 300𝑄 − 400
𝜋(30) = −5 30 2 + 300(30) − 400
𝜋(30) = −5 30 2 + 300(30) − 400
𝜋 30 = 4100
Example 1: Mr Watkins opens a cake store called
Sugar Overload, and he specialises in wedding
cakes. He figures out his monthly demand is
𝑄 = 80 −𝑃
5
The rent for the bakery he works in is
$500/month, and one cakes costs him
approximately $100 to make. Find the number of
cakes he needs to sell to maximise profit, the price
per cake, and the profit he will make at this price.
Plan: use the profit equation
𝜋 = 𝑇𝑅 − 𝑇𝐶
Replace 𝑇𝑅 with 𝑃 ∙ 𝑄 where 𝑃 is the rearranged
demand function (𝑃 in terms of 𝑄).
Apply the FOC to find any optimal points, then
apply the SOC to determine if it is a maximum.
Substitute the optimal quantity back into the profit
function to determine the maximum profit, then
into the rearranged demand function to get the
price.
Solution: the demand function needs to be
rearranged to isolate 𝑃:
𝑄 = 80 −𝑃
5
𝑄 − 80 = −𝑃
5
(−5)(𝑄 − 80) = 𝑃
𝑃 = −5𝑄 + 400
Substitute this into the 𝑇𝑅 equation:
𝑇𝑅 = 𝑃. 𝑄
𝑇𝑅 = (−5𝑄 + 400). 𝑄
𝑇𝑅 = −5𝑄2 + 400𝑄
Find the 𝑇𝐶 function:
𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶
𝑇𝐶 = 500 + 100𝑄
Set up the profit function:
𝜋 = 𝑇𝑅 − 𝑇𝐶
𝜋 = −5𝑄2 + 400𝑄 − 500 + 100𝑄
𝜋 = −5𝑄2 + 400𝑄 − 500 − 100𝑄
𝜋 = −5𝑄2 + 300𝑄 − 500
Apply the FOC to find any stationary points:
𝑑𝜋
𝑑𝑄= −10𝑄 + 300 = 0
−10𝑄 = −300
𝑄 = 30
Mr Watkins needs to make 30 cakes per month. To
make sure this quantity will give maximum profit,
the SOC must be used:
𝑑2𝜋
𝑑𝑄2= −10
𝑑2𝜋
𝑑𝑄2 30 = −10 < 0 ∴ maximum
What price should Mr Watkins charge for each of
these cakes? Using the (rearranged) demand
function:
𝑃 = −5𝑄 + 400
𝑃(30) = −5(30) + 400
129
𝑃 30 = −150 + 400 = 250
Looking back to what the question is asking, if Mr
Watkins sells 30 cakes at $250 each, the profit he
will make is:
𝜋 = −5𝑄2 + 300𝑄 − 500
𝜋(30) = −5 30 2 + 300(30) − 500
𝜋 30 = −5 900 + 9000 − 500 = 4000
Thus Mr Watkins should sell 30 cakes per month at
$250 each and he will make a maximum profit of
$4000 per month.
Exercises (remember to prove they are optimal points): 1. A company manufacturing mining truck trays has a
yearly demand of: 𝑃 = −0.05 𝑄 + 200 𝑄 − 200 + 20,000 If the costs labour and capital which average out to $15,500/𝑡𝑟𝑎𝑦 and fixed costs of $7,500/𝑤𝑒𝑒𝑘, determine: a) The yearly revenue function b) The annual cost function c) The annual profit function d) The maximum yearly profit attainable
2. A manufacturer of surfboards faces a monthly demand function of form:
𝑃𝑑 = −0.2𝑄𝑑2 + 110
If costs of making a board are $60 and the factory has fixed costs of $745/𝑚𝑜𝑛𝑡, determine: a) The revenue function b) The cost function c) The profit function d) The maximum profit attainable
3. A company imports motorbikes from China and sells them onto consumers at an inflated price. If they purchase each motorbike for $5,000, and when on-selling the motorbikes, they face a domestic demand function of: 𝑃𝑑 = − 0.1𝑄 + 15 2 0.1𝑄 + 13 (𝑄 − 22) If there are no other costs, determine: a) The profit function b) The profit maximising quantity.
6.7 applications – break-even
When a business “breaks even”, it means the
revenues just meet the costs. This implies that at
break-even, profit is zero.
Theory: break even is when 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 equals
𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒, which implies profit is zero.
𝜋𝐵𝐸 = 0
This is very similar to the previous section, except
that instead of maximising profit, the profit
function is simply set equal to zero.
Example 1: find the break-even quantities for a
firm with a profit function
𝜋 = −𝑄2 + 200𝑄 − 7500
Plan: set the profit function equal to zero and
solve.
Solution: set 𝜋 = 0:
0 = −𝑄2 + 200𝑄 − 7500
Use the Quadratic Formula:
𝑄𝐵𝐸 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑄𝐵𝐸 =−200 ± (200)2 − 4 −1 (−7500)
2(−1)
𝑄𝐵𝐸 =−200 ± (200)2 − 4 −1 (−7500)
2(−1)
𝑄𝐵𝐸 =−200 ± 10000
−2=
−200 ± 100
−2
𝑄𝐵𝐸 = 150 or 𝑄𝐵𝐸 = 50
Thus the two break-even quantities are 150 and
50. Mathematically, these are simply the roots of
the profit function:
Example 2: A firm produces wide-screen TV’s and
their estimated demand is
𝑄𝑑 =440 − 4𝑃𝑑
2
If they faces the increasing cost function
𝑇𝐶 = 0.5𝑄2 + 1800
find the break-even output levels for this firm.
𝜋
(50,0) (150,0)
𝐵𝐸1 𝐵𝐸2
𝜋𝑚𝑎𝑥
𝑄
(100,0)
130
Plan: find 𝑇𝑅 in terms of 𝑄, then replace the profit
equation 𝜋 = 𝑇𝑅 − 𝑇𝐶 to find the profit function.
Set this equal to zero and solve for break-even
output levels.
Solution: rearrange the function to isolate 𝑃:
𝑄 =440 − 4𝑃
2
2𝑄 = 440 − 4𝑃
2𝑄 − 440 = −4𝑃
𝑃 = −0.5𝑄 + 110
Replace the 𝑇𝑅 equation:
𝑇𝑅 = 𝑃. 𝑄
𝑇𝑅 = −0.5𝑄 + 110 𝑄
𝑇𝑅 = −0.5𝑄2 + 110𝑄
Having 𝑇𝑅 and 𝑇𝐶, the 𝜋 function can be formed:
𝜋 = 𝑇𝑅 − 𝑇𝐶
𝜋 = −0.5𝑄2 + 110𝑄 − 0.5𝑄2 + 1800
𝜋 = −0.5𝑄2 + 110𝑄 − 0.5𝑄2 − 1800
𝜋 = −𝑄2 + 110𝑄 − 1800
To find the break even quantities, set 𝜋 = 0:
𝜋𝐵𝐸 = 0
−𝑄2 + 110𝑄 − 1800 = 0
Use the quadratic formula to solve this quadratic:
𝑄𝐵𝐸 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑄𝐵𝐸 =−110 ± (110)2 − 4 −1 (−1800)
2(−1)
𝑄𝐵𝐸 =−110 ± 4900
−2=
−110 ± 70
−2
Thus the firm breaks even at 𝑄𝐵𝐸 = 90 or
𝑄𝐵𝐸 = 20.
Exercises (remember to prove they are optimal points): 1. A company selling financial services (e.g. financial
planning) has a monthly demand for the services (on a per minute basis) according to the function:
𝑃 = ln 7𝑄 + 130 If the costs are commissions ($100/𝑢𝑛𝑖𝑡) and fixed costs of $500/𝑚𝑜𝑛𝑡, determine: a) The revenue function b) The cost function c) The profit function d) The break-even profit e) The maximum profit attainable
2. A manufacturer of mobile phones faces a monthly demand function of form:
𝑃 = 5𝑒−0.5𝑄+6 If costs of making a single mobile phone are $10 and the factory has fixed costs of $1412/𝑚𝑜𝑛𝑡, determine: a) The revenue function b) The cost function c) The profit function d) The break-even profit e) The maximum profit attainable
3. A company imports cheap golf-clubs from China and sells them onto consumers at an inflated price. If they purchase the golf-clubs for $5 each, and when on-selling the clubs, they face a domestic demand function of:
𝑃 = 𝑄 + 800 𝑄 − 700 If there are no other costs, determine: a) The profit function b) The break-even quantities. c) The profit maximising quantity.
6.8 applications – marginal and
average values
Theory: the average of a group of variables (e.g.
cost) is the sum of all the individual variables (e.g.
sum all the costs) divided by the number of units
(e.g. the number of units sold). The “marginal”
value of a variable is the addition to the total from
an extra (marginal) unit. In mathematical terms:
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 =𝑠𝑢𝑚 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 =𝑑𝑦
𝑑𝑥
It is easier to remember that whenever you read
the word “marginal”, it means the derivative.
For example, the average output of labour is the
total output divided by the number of workers.
Similarly, marginal productivity of labour is the
additional output obtained from an extra
(marginal) person being employed (which is found
using the first derivative).
Intro example: a firm manufactures wooden
cabinets in a small factory. When the firm hires the
first worker, that worker produces three cabinets
per day. The firm then hires another worker, and
131
these two workers, together, produce eight
cabinets (as they are specialising: one cuts the
timber, the other sands it etc.). The firm hires
another person, and now the three employees
produce 12 cabinets. When one worker wants to
use the saw, another worker is using it. After hiring
yet another worker, all four people make 14
cabinets. Hiring any more workers will not at all
increase production, as some of them are always
waiting to use the equipment. In summary:
Worker 1:
o 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 =3
1= 3
o 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 3
Worker 2:
o 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 =8
2= 4
o 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 8 − 3 = 5
Worker 3:
o 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 =12
3= 4
o 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 12 − 8 = 4
Worker 4:
o 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 =14
4= 3.5
o 𝑚𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓 𝑙𝑎𝑏𝑜𝑢𝑟 = 14 − 12 = 2
Economic theory: as more labour (or other input)
is added to a production process, output initially
increases greatly. However, past some point,
additional labour adds less and less to total output.
Eventually, no additional output can be obtained
from additional labour (and in theory, output may
start to fall).
There are three distinct regions:
1. When output is increasing at an increasing rate
(i.e. convex).
2. When output is increasing at a decreasing rate
(i.e. concave)
3. When output begins to fall.
The point at which it changes from convex to
concave is the inflection point. This theory is
known, in general terms, as the Law of Diminishing
Marginal Returns.
The theory of marginal and average values can be
applied to mathematical functions.
Example 1: find the average cost and marginal cost
functions if the total cost function is
𝑇𝐶 =1
3𝑄3 +
5
2𝑄2 + 50
Plan: divide the 𝑇𝐶 function by 𝑄 to get average
cost, and differentiate 𝑇𝐶 with respect to 𝑄 to get
marginal cost.
Solution: for average costs:
𝐴𝐶 =𝑇𝐶
𝑄
𝐴𝐶 =
13 𝑄3 +
52 𝑄2 + 50
𝑄
𝐴𝐶 =1
3𝑄2 +
5
2𝑄 +
50
𝑄
For marginal costs:
𝑀𝐶 =𝑑𝑇𝐶
𝑑𝑄
𝑀𝐶 = 𝑄2 + 5𝑄
𝐼𝑛𝑝𝑢𝑡
𝑂𝑢𝑡𝑝𝑢𝑡
1. 2. 3.
1 2 3 4
3
8
12
14 𝑂𝑢𝑡𝑝𝑢𝑡
𝐿𝑎𝑏𝑜𝑢𝑟
132
Example 2: for the given function, find
1. the point where marginal costs are at a
minimum.
2. an expression for average cost.
𝑇𝐶 =𝑄3
100+
−3
10𝑄2 + 3𝑄 + 7
Plan:
1. differentiate 𝑇𝐶 to get 𝑀𝐶. Then find the
minimum of the marginal cost function using
FOC; make sure it is a minimum using the SOC.
2. Average cost is 𝑇𝐶
𝑄
Solution: differentiating 𝑇𝐶:
𝑑𝑇𝐶
𝑑𝑄= 𝑀𝐶 =
3
100𝑄2 −
6
10𝑄 + 3
To find where marginal cost is at a minimum, the
FOC must be applied to the marginal cost function:
𝑑𝑀𝐶
𝑑𝑄=
6
100𝑄 −
6
10= 0
Solve this for 𝑄:
6
100𝑄 =
6
10
𝑄 = 10
To prove this is a minimum:
𝑑2𝑀𝐶
𝑑𝑄2=
6
100> 0 ∴ mimimum
The 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 at this point is:
𝑇𝐶 10 = 10 3
100+
−3
10 10 2 + 3 10 + 7
𝑇𝐶 10 = 10 − 30 + 30 + 7 = 17
An expression of average costs is:
𝐴𝐶 =𝑇𝐶
𝑄=
𝑄3
100 +−310 𝑄2 + 3𝑄 + 7
𝑄
𝐴𝐶 =𝑄2
100−
3
10𝑄 + 3 +
7
𝑄
Example 3: find the level of output when
productivity of labour changes from increasing at
an increasing rate to increasing at a decreasing
rate, for the production function:
𝑄 = −𝐿3 + 4𝐿 + 10𝐿2 + 20
Plan: remember 𝐿 is the independent variable and
𝑄 the dependent variable. The inflection point
needs to be found (as it is at this point where the
function changes from increasing at an increasing
rate to increasing at a decreasing rate). We know
from Section 6.4 that this is done by setting the
second derivative equal to zero.
Solution: differentiate the function
𝑀𝑃𝐿 =𝑑𝑄
𝑑𝐿= −3𝐿2 + 4 + 20𝐿
Finding the inflection point of an original function
is the same as finding the maximum/minimum for
the first derivative (Chapter 5). So using the FOC
on the 𝑀𝑃𝐿 function:
𝑑𝑀𝑃𝐿
𝑑𝐿=
𝑑𝑄2
𝑑2𝐿= −6𝐿 + 20 = 0
Solving this gives:
6𝐿 = 20
𝐿 =20
6≈ 3.33
Thus between the third and forth unit of labour,
the productivity of labour changes from increasing
at an increasing rate to increasing at a decreasing
rate.
Example 4: find the point where average costs are
equal to marginal costs, for the cost function:
𝑇𝐶 = 0.02 𝑄3 − 36𝑄2 + 432𝑄
Plan: find 𝑀𝐶 and 𝐴𝐶, then set them equal to each
other.
Solution:
𝑀𝐶 =𝑑𝑇𝐶
𝑑𝑄= 0.06𝑄2 − 1.44𝑄 + 8.64
𝐴𝐶 =𝑇𝐶
𝑄
𝐴𝐶 =0.02𝑄3 − 0.72𝑄2 + 8.64𝑄
𝑄
𝐴𝐶 = 0.02𝑄2 − 0.72𝑄 + 8.64
133
Set 𝑀𝐶 = 𝐴𝐶:
0.06𝑄2 − 1.44𝑄 + 8.64 = 0.02𝑄2 − 0.72𝑄 + 8.64
Bring everything onto one side:
0.04𝑄2 − 0.72𝑄 = 0
Either factorise out 𝑄 or use the quadratic formula
to solve for 𝑄:
𝑄 0.04𝑄 − 0.72 = 0
So either:
1. 𝑄 = 0, or
2. 0.04𝑄 − 0.72 = 0
0.04𝑄 = 0.72
𝑄 = 18
Marginal cost is equal to average cost when 𝑄 = 0
and when 𝑄 = 18.
Exercises: 1. Sketch the function of marginal productivity of an
extra lawn mower (i.e. capital) for a lawn mowing firm with four employees. Justify your interpretation. Initially the firm has no lawn-mowers.
2. Find expressions for the marginal and average costs of the following functions.
𝑎) 𝐶 = 15𝑄2
𝑏) 𝐶 = 16𝑄3 + 12𝑄2 + 5𝑄
𝑐) 𝐶 = 0.05𝑄𝑒0.1𝑄+4
3. Find the marginal productivity of capital in the following production functions:
𝑎) 𝑄 = 14𝐾1.5
𝑏) 𝑄 = 0.01𝐾𝑒4𝐾−1
𝑐) 𝑄 = 𝐾 ln 𝐾 + 1
4. For the revenue and cost functions: 𝑅 𝑄 = 𝑄2 − 257𝑄 + 12,222 𝐶 𝑄 = 5,000 + 180𝑄 a) Determine when marginal revenue and
marginal costs are equal. b) Find the profit function, and find the quantity
where profit is maximised. c) Compare the answers from a) and b) above.
5. For the following functions, find where the average product of labour is equal to the marginal product or labour.
𝑎) 𝑄 = − 𝐿 + 5 2
𝑏) 𝑄 = 𝐿3 − 20𝐿2 + 125𝐿
𝑐) 𝑄 = 𝐿𝑒𝐿−5 − 𝐿2
6.9 differentiation and elasticity
Before reading this section, revise the concept of
elasticity in Chapter 2. Remember that the formula
for elasticity is either:
𝜀 =%∆𝑄
%∆𝑃 𝑂𝑅 𝜀 =
∆𝑄
∆𝑃×
𝑃
𝑄
The formula on the right can be rewritten as:
𝜀 =𝑑𝑄
𝑑𝑃×
𝑃
𝑄
This makes finding elasticities a great deal easier
and allows us to find elasticities of non-linear
functions.
Example 1: find and interpret the elasticity of
supply for the supply function 𝑄𝑠 = 𝑃2 + 𝑃 − 6
when output is 50 units.
Plan: use the elasticity formula:
𝜀𝑠 =𝑑𝑄
𝑑𝑃×
𝑃
𝑄
Differentiate the supply equation, then substitute
into the elasticity equation. Find 𝑃 by substituting
𝑄 = 50 into the supply equation.
Solution: differentiate the supply equation:
𝑑𝑄𝑠
𝑑𝑃= 2𝑃 + 1
Substitute this into the elasticity formula:
𝜀𝑠 = 2𝑃 + 1 𝑃
𝑄=
[2𝑃 + 1] 𝑃
𝑄
Find 𝑃 when 𝑄 = 50 from the supply equation:
𝑄𝑠 = 𝑃2 + 𝑃 − 6
50 = 𝑃2 + 𝑃 − 6
0 = 𝑃2 + 𝑃 − 56
Solve for 𝑃 using the quadratic formula:
𝑃 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑃 =−1 ± 12 − 4 1 −56
2 1
=−1 ± 225
2
𝑃 = −8 𝑜𝑟 𝑃 = 7
134
Price can never be negative so disregard 𝑃 = −8.
Price is therefore 𝑃 = 7.
Substitute 𝑃 = 7 and 𝑄 = 50, into the elasticity
formula:
𝜀𝑠 = [2𝑃 + 1] 𝑃
𝑄
𝜀𝑠 = 2 7 + 1 (7)
(50)
𝜀𝑠 = 2.1
Take the absolute value of 2.1 (it is still 2.1) and
apply the elasticity rules; the interpretation is that
this good is elastic in supply. That is, if price was to
increase by, say 10%, supply would increase by
21%.
Example 2: you have just got a new job at a big
finance firm, and your boss gives you the following
demand function of one of his client’s products (a
type of computer):
−200𝑃 + 400,000 = 𝑄𝑑 + 5 2
He wants to test your skills so asks you to find the
elasticity of demand at the current sales of 480
computers/month, and what it means.
Plan: use the elasticity formula:
𝜀𝑑 =𝑑𝑄
𝑑𝑃×
𝑃
𝑄
Differentiate the demand function, then substitute
it into the above equation. Find 𝑃 when 𝑄 = 480
from the demand function.
Solution: rearranging this demand function to
have 𝑄𝑑 by itself is very difficult (try for yourself
using the reverse of BIMDAS).
To make things simpler, isolate the other variable
𝑃 (which is a lot easier):
−200𝑃 = 𝑄𝑑 + 5 2 − 400,000
𝑃 = 𝑄𝑑 + 5 2 − 400,000
−200
𝑃 = −0.005 𝑄𝑑 + 5 2 + 2000
Differentiate 𝑃 with respect to 𝑄𝑑 (the opposite of
what is required, as the derivative of 𝑄𝑑 with
respect to 𝑃 is needed for the elasticity formula):
𝑑𝑃
𝑑𝑄𝑑= −0.005 2 𝑄𝑑 + 5
𝑑𝑃
𝑑𝑄𝑑= −0.01𝑄𝑑 − 0.05
Inverting the whole of both sides of this last
equation gives the derivative that is required:
𝑑𝑄𝑑
𝑑𝑃=
1
−0.01𝑄𝑑 − 0.05
Substitute everything into the elasticity formula:
𝜀𝑑 = 1
−0.01𝑄𝑑 − 0.05 ×
𝑃
𝑄
Having been given 𝑄𝑑 = 480, 𝑃 is obtained from
the rearranged demand equation:
𝑃 = −0.005 𝑄𝑑 + 5 2 + 2000
𝑃 = −0.005 480 + 5 2 + 2000
𝑃 = 823.875
Substitute 𝑃 = 823.875 and 𝑄𝑑 = 480 into the
elasticity formula above:
𝜀𝑑 = 1
−0.01 480 − 0.05 ×
823.875
480
𝜀𝑑 ≈ −0.354
This means demand is inelastic. That is, if price is
increased by a certain percentage, quantity
demanded would be reduced by only 0.354 of that
percentage.
Exercises: 1. Determine the elasticity of the following demand
functions at the shown output. Interpret the result.
𝑎) 𝑄𝑑 = −1
4𝑃𝑑 + 48 𝑎𝑡 𝑃 = 20
𝑏) 𝑄𝑑 =−2𝑃𝑑
7+ 71 𝑎𝑡 𝑃 = 23
𝑐) 𝑃𝑑 = −0.27𝑄𝑑 + 77 𝑎𝑡 𝑄 = 13
𝑑) 𝑃𝑑 = −𝑄2 + 4𝑄 + 54 𝑎𝑡 𝑄 = 5
𝑒) 𝑃𝑑 = −0.25𝑄2 + 4𝑄 + 121 𝑎𝑡 𝑄 = 28
2. Determine the elasticity of the following supply functions at the price indicated. Interpret the result.
𝑎) 𝑃𝑠 − 3
0.7= 𝑄𝑠 𝑎𝑡 𝑄 = 2
𝑏) 𝑃𝑠 = 𝑄2 + 5 𝑎𝑡 𝑄 = 13
𝑐) 𝑃𝑠 = 7 ln 𝑄 + 5 + 3 𝑎𝑡 𝑄 = 6
135
𝑑) 𝑃𝑠 = 𝑒3𝑄+4 + 19 𝑎𝑡 𝑄 = 14
𝑒) 𝑃𝑠 = 𝑄 + 5 𝑄2 + 4𝑄 + 4 𝑎𝑡 𝑄 = 15
6.10 elasticity and total revenue
Most retail businesses have a relatively constant
supply curve (especially when they import goods).
Such firms focus mainly on maximising total
revenue, which is determined by the quantity of
goods sold, which in turn comes from the demand
function.
The elasticity concept can be used to determine if
changes in price will lead to an increase (or
decrease) in total revenue.
Theory: given the definition of total revenue:
𝑇𝑅 = 𝑃 ∙ 𝑄
Take the log of the whole of both sides:
log 𝑇𝑅 = log 𝑃 ∙ 𝑄
log 𝑇𝑅 = log 𝑃 + log 𝑄
Taking the logs of an equation is a good
approximation to the percentage change of the
two sides, so:
%∆𝑇𝑅 = %∆𝑃 + %∆𝑄
Since from elasticity, %∆𝑃 and %∆𝑄 can be found
easily, then the %∆𝑇𝑅 can also be determined.
Example 1: given the elasticity of demand is
−0.69, and prices rise by 20%, by what
percentage will total revenue change?
Plan: find %∆𝑄 from the elasticity formula, then
applying the %∆𝑇𝑅 formula.
Solution:
𝜀𝑑 =%∆𝑄
%∆𝑃
−0.69 =%∆𝑄
20
%∆𝑄 = 20 −0.69 = −13.8%
Then applying the percentage change in total
revenue formula:
%∆𝑇𝑅 = %∆𝑃 + %∆𝑄
%∆𝑇𝑅 = 20 + −13.8 = 6.2%
For a 20% price increase, total revenue will
increase by 6.2%.
Even though the price has increased, demand is
inelastic so this means that for a given price
increase, quantity demanded will be reduced less
than the increase in price so overall revenue will
increase. But this is not always the case.
Example 2: determine the percentage change in
total revenue if the demand function for a retail
good is
𝑄𝑑 = −0.5𝑃 + 54
when initial quantity sold is 20, and the owner
wants to raise prices by 5%.
Plan: find the elasticity of demand at 𝑄 = 20 using
𝜀𝑑 =𝑑𝑄
𝑑𝑃×
𝑃
𝑄
Then use the original definition of elasticity to
determine %∆𝑄
𝜀𝑑 =%∆𝑄
%∆𝑃
which can then be used in the %∆𝑇𝑅 formula
%∆𝑇𝑅 = %∆𝑃 + %∆𝑄
Solution: find the elasticity:
𝜀𝑑 = −0.5 ×𝑃
20
To find 𝑃, use the demand equation:
20 = −0.5𝑃 + 54
𝑃 =20 − 54
−0.5= 68
This gives an elasticity of:
𝜀𝑑 = −0.5 ×68
20= −1.7
This means that demand is elastic at this point
(why?). Find the %∆𝑄:
−1.7 =%∆𝑄
5
%∆𝑄 = −1.7 ∙ 5 = −8.5%
Finally, apply the %∆𝑇𝑅 formula:
136
%∆𝑇𝑅 = 5 + −8.5 = −3.5%
So for a 5% price increase, there would be a fall of
3.5% in total revenue.
These last two examples show that for a price
increase, total revenue can either increase or
decrease. It all depends on the elasticity of
demand. In Example 2, the percentage change
could have been found using the original demand
equation, however this is difficult.
Theory: for an inelastic demand function:
A price increase will increase total revenue
(the diagram below shows that as price rises
from 𝑃1 to 𝑃2, the area 𝑃 ∙ 𝑄, which is also 𝑇𝑅,
increases. Initially, it is the red and green area,
then after the price rise, it is the green and
blue area. Since the blue area is larger than the
red area, 𝑇𝑅 must increase).
A price decrease will decrease total revenue
(work through the same logic as above, but
backwards).
For an elastic demand function:
A price increase will decrease total revenue (in
the diagram below, as price increases from 𝑃1
to 𝑃2, the 𝑇𝑅 rectangle changes from the
green and red, to the green and blue. Since the
red rectangle is larger than the blue one, 𝑇𝑅
must decrease).
A price decrease will increase total revenue
(using the same diagram, work the other way
to make clear that a decrease in price will
increase 𝑇𝑅).
When elasticity is equal to 1 (in absolute terms), a
change in price will leave total revenue
unchanged.
Example 3: given the demand function
𝑄 = −0.4𝑃2 + 77
and current sales of 𝑄 = 37, how will a 15%
decrease in prices affect total revenue?
Plan: find the elasticity of demand at 𝑄 = 37 using
𝜀𝑑 =𝑑𝑄
𝑑𝑃×
𝑃
𝑄
then use the original definition of elasticity
𝜀𝑑 =%∆𝑄
%∆𝑃
to determine %∆𝑄, which can then be used in the
%∆𝑇𝑅 formula: %∆𝑇𝑅 = %∆𝑃 + %∆𝑄.
Solution: find the derivative of the demand
function:
𝑑𝑄
𝑑𝑃= −0.8𝑄
Find the value of 𝑃 when 𝑄 = 37 by rearranging
the demand equation:
37 − 77 = −0.4𝑃2
−40
−0.4= 𝑃2 = 100
𝑃 = ±10
Price must be positive, so 𝑃 = 10.
Substitute this in to find the elasticity:
𝜀𝑑 = −0.8𝑄 ×𝑃
𝑄
𝜀𝑑 = −0.8 ∙ 10 ×10
37
𝜀𝑑 = −80
37≈ −2.162
𝑃1
𝑃2
𝑄2
𝐷𝐷
𝑄1
𝑃
𝑄
𝑃1
𝑃2
𝑄2 𝑄1
𝐷𝐷 𝑃
𝑄
137
To find %∆𝑄 when price decreases by 15%, use
the original elasticity formula:
−80
37=
%∆𝑄
−15
%∆𝑄 = −15 −80
37
%∆𝑄 =1200
37≈ 32.43%
Finally, apply the %∆𝑇𝑅 formula:
%∆𝑇𝑅 ≈ −15% + 32.43% = 17.43%
Thus, total revenue will increase by approximately
17.43% for a 15% price decrease.
Exercises:
1. Given the following elasticities, determine how 𝑇𝑅 will change for a 17% price decrease. 𝑎) 𝜀𝑑 = −0.4
𝑏) 𝜀𝑑 = −1.9
𝑐) 𝜀𝑑 = −0.9
𝑑) 𝜀𝑑 = −1.01
𝑒) 𝜀𝑑 = −1
2. For the weekly demand function
𝑄𝑑 =1000
3−
2𝑃𝑑
0.6
and sales of 55 units per week, determine how a 5% price decrease will affect total revenue.
3. For the demand function 𝑃𝑑 = 𝑄2 + 10𝑄 − 3000 and sales of 33 units per week, determine how a 24% price increase will affect total revenue.
4. For the monthly demand function 𝑃𝑑 = −5 ln 𝑄2 + 5 + 20 And sales of 5 units per week, determine how a 15% price increase will affect total revenue.
chapter six summary
Optimisation is the process of finding an optimal point, with that point usually being a maximum or minimum. A global optimum is a point that is the optimal point over the entire number range (i.e. 𝑥 − 𝑎𝑥𝑖𝑠). A local optimum is an optimal point only over a small range around that point.
The gradient at any maximum or any minimum is always zero. The First Order Condition (FOC):
𝑑𝑦
𝑑𝑥= 0
Then solving for the unknowns. The nature of an optimal point is found using the Second Order Condition (SOC):
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑖𝑓 𝑑2𝑦
𝑑𝑥2> 0
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑖𝑓 𝑑2𝑦
𝑑𝑥2< 0
𝑆𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡 𝑖𝑓 𝑑2𝑦
𝑑𝑥2= 0
Any inflection point is using the SOC:
𝑆𝑂𝐶 𝑓𝑜𝑟 any 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡: 𝑑2𝑦
𝑑𝑥2= 0
Inflection points are points of a function where a function changes from concave to convex (or vice versa). The two types of inflection points: 1. Stationary inflection points have a gradient of zero
at the inflection point. 2. Non-stationary inflection points do not have a
gradient of zero at the inflection point.
The general form of profit is: 𝜋 = 𝑇𝑅 − 𝑇𝐶 Total revenue can also be simplified:
𝑇𝑅 = 𝑃 ∙ 𝑄 In general, total costs are represented by: 𝑇𝐶 = 𝐹𝐶 + 𝑉𝐶 Variable costs are usually represented by: 𝑉𝐶 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 × 𝑄 All functions are in terms of 𝑄 as profits are directly determined by the quantity of goods sold. Break even is when 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 equals 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒, which implies profit is zero. 𝜋𝐵𝐸 = 0
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 =𝑠𝑢𝑚 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 =𝑑𝑦
𝑑𝑥
As more labour is added to a production process, output initially increases greatly. However, past some point, additional labour adds less and less to total output. Eventually, no additional output can be obtained from additional labour (and in theory, output may start to fall). The point at which it changes from convex to concave is the inflection point. This theory is known, in general terms, as the Law of Diminishing Marginal Returns.
%∆𝑇𝑅 = %∆𝑃 + %∆𝑄 For an inelastic demand function:
A price increase will increase total revenue
A price decrease will reduce total revenue. For an elastic demand function:
A price increase will reduce total .
A price decrease will increase total . When elasticity is equal to 1 (in absolute terms), a change in price will leave total revenue unchanged.
138
chapter six questions
1. Determine which of the following points are local/global maxima/minima:
2. Determine the FOC for the following functions:
𝑎) 𝑦 = 𝑥2 + 3𝑥 − 1 𝑏) 𝑦 = 15𝑥3 + 1 − 𝑥 4
𝑐) 𝑦 = 12𝑥2𝑒1−𝑥2
𝑑) 𝑦 = 2 𝑥 − 1 2 𝑥3 − 1 2
𝑒) 𝑦 =𝑥 − 1
𝑥 + 1
3. Determine the nature of any stationary points for the following functions: 𝑎) 𝑦 = 𝑥 − 1 𝑥 + 3 𝑏) 𝑦 = 𝑥3 + 3𝑥2 − 45𝑥 + 18 𝑐) 𝑦 = −2𝑥𝑒𝑥+3 𝑑) 𝑦 = 𝑒4−𝑥 𝑥 − 3 2
𝑒) 𝑦 = 1
14.5𝑥 𝑒𝑥2+3 + 19
𝑓) 𝑓 𝑥 =1 − 𝑥
𝑥2 + 1
4. Find all stationary and inflection points for the following functions, and determine their nature where applicable: 𝑎) 𝑦 = 𝑥3 − 𝑥2 + 15𝑥 − 79 𝑏) 𝑦 = 𝑥 − 1 𝑥 − 5 𝑥 + 4 𝑐) 𝑦 = 𝑥 + 3 2 1 − 𝑥 𝑑) 𝑓 𝑥 = −𝑥𝑒4−𝑥 𝑒) 𝑓 𝑥 = 𝑥2𝑒1−𝑥
𝑓) 𝑦 = ln 𝑒4𝑥2−3𝑥+2
5. A company has determined their profit function to be:
𝜋 = −100𝑄2 + 10,000𝑄 − 47,500 Find: a) The break-even quantities. b) The quantity that maximises profit. c) The maximum profit.
6. A factory manufactures music players and sells them at $350. The factory faces a cost function of the form:
𝑇𝐶 = 0.2𝑄2 + 40 Determine: a) The total revenue function. b) The profit function. c) The break-even quantities. d) The quantity for profit maximisation. e) The maximum profit. Sketch the profit function, and plot all the important points.
7. A company faces a demand function of the form
𝑃𝑑 =1
0.1𝑄+ 0.25
And a total cost function of the form: 𝑇𝐶 = 0.2𝑒0.01𝑄+4
Determine: a) The total revenue function. b) The profit function. c) The output for maximum profit. d) The price corresponding to maximum profit. e) The maximum profit attainable.
8. Determine the average and marginal values for the following functions: 𝑎) 𝜋 = −𝑄2 + 20𝑄 𝑏) 𝑇𝐶 = 0.1 𝑞 + 20 𝑞 + 5 𝑐) 𝑇𝑅 = 𝑄0.5 𝑄 + 1 𝑑) 𝑇𝐶 = 0.01𝑥2 ln 𝑥 + 1 𝑒) 𝜋 = −0.1 𝑒𝑄−5 + 𝑒5−𝑄 + 10 𝑓) 𝜋 = 𝑞𝑒−𝑞+1
9. For the profit function 𝜋 = −3𝑄3 + 75𝑄2
Determine the values of 𝑄 where average profit is equal to marginal profit.
10. Determine when the average product of labour is equal to the marginal product of labour, for the function:
𝑄 = 5𝐿0.5 + 𝐿2 + 4𝐿 11. A firm manufacturing microwave dishes determines
its profit functions to be 𝜋 = −1.5𝑄3 + 50𝑄2
Determine: a) An expression for average and marginal profit. b) The break-even quantities. c) The quantity corresponding to maximum
profit. d) The maximum profit attainable. e) The output level where average profit and
marginal profit are equal. 12. A production function has the form:
𝑄 = 15𝐿3 + 3𝐿2 +1
4𝐿
Determine the levels of labour where the marginal productivity of labour is double the average productivity of labour.
13. Given the demand function
𝑃𝑑 = 𝑒−0.01𝑄2−0.05𝑄+3 + 180 Determine and interpret the elasticity at 𝑄 = 8.
14. Determine and interpret the elasticity for the following supply function, at output of 20 units:
𝑃𝑠 = 100 ln 0.1𝑄 + 4 + 0.9𝑄 15. The owner of a firm producing car rims wants to
increase price by 5% from a level of 𝑄 = 20. Given the demand function is
𝑃𝑑 =100
𝑄 + 1
Determine if total revenue will increase or decrease with the 5% increase in price.
16. For the following demand function 𝑃𝑑 = − 𝑄 + 50 2 𝑄 − 40 𝑒−𝑥
If output is at 4, and the owner of the factory wants to increase price by 25%. How will total revenue be affected? Give a numerical answer, and relate it
𝑦
𝑥 𝐴
𝐵
𝐶 𝐷
𝐸
𝐹
139
back to the elasticity obtained. Hint: to differentiate this function, the product rule has to be used within the product rule.
17. A firm produces top quality surfboards and has a demand function of the form:
𝑃𝑑 = −0.5𝑄 + 2500 Each surfboard costs $250 to make, and the factory in which they are manufactured has a rent of $1500. Determine: a) The total cost function b) The total revenue function. c) The profit function. d) The break-even quantities. e) The quantity for maximum profit. f) The maximum profit. g) The price per surfboard to attain this profit. h) The marginal profit function. i) The average profit function. j) Where the marginal and average profit
functions are equal. If output is initially 1000, and the owner is thinking of raising prices by 10%, determine: k) The elasticity at 𝑄 = 20, and its interpretation. l) The extent to which total revenue would
change, if the 10% price rise went ahead.
18. A firm produces sandstone sculptures, and has a demand function of the form:
𝑃𝑑 = − − 0.4𝑞 + 4276 Each sandstone sculpture costs $80 to make, and the factory in which they are manufactured has a rent of $200. Determine: a) The total cost function b) The total revenue function. c) The profit function. d) The break-even quantities. e) The quantity for maximum profit. f) The maximum profit. g) The price each sculpture is sold at, to attain
this profit. h) The marginal profit function. i) The average profit function. j) Where the marginal and average profit
functions are equal. If output is initially 4000, and the owner is thinking of raising prices by 25%, determine: k) The elasticity at 𝑄 = 24, and its interpretation. l) The extent to which total revenue would
change, if the 25% price rise went ahead.
140
Chapter 7
Multiple Variable Differentiation
Differentiation with more than one variable 7.1 Additional Variables 141
7.2 Simple Partial Differentiation 142
7.3 Complex Partial Differentiation 144
7.4 Second Order Partial Derivatives 146
7.5 Application of Partial Differentiation 147
7.6 Total Differentiation 148
7.7 Optimisation with Many Variables 151
7.7 Economic Applications 155
Chapter Seven Summary 158
Chapter Seven Questions 158
141
7.1 additional variables
Theory: multivariable or multivariate simply
means more than one variable. Multivariable
functions means that a dependent variable is
determined by two or more other variables.
All the previous chapters have only had a single
independent variable 𝑥 determining the
dependent variable 𝑦 through a certain function:
𝑦 = 𝑓 𝑥
The function could be quadratic, cubic, etc.
Theory: the simplest multivariable functions have
the form:
𝑧 = 𝑓(𝑥, 𝑦)
Where 𝑧 is a dependent variable, and it depends
on two independent variables 𝑥 and 𝑦. That is, 𝑧 is
a function of 𝑥 and 𝑦, and can also be written as:
𝑧(𝑥, 𝑦)
This is now a three-dimensional function (i.e. it is
drawn on three axes):
Mathematically, an example of a multivariable
function is:
𝑧 = 𝑥2 + 𝑦2
This is a multivariable equation (𝑧 is a function of 𝑥
and 𝑦). Because there are now three variables,
these graphs are three dimensional in nature. The
above multivariable function is the shape of a cup.
Similar to a quadratic, the function above has a
minimum which can be visually determined.
Multivariable functions are very useful as they
represent the real world much better. Most goods
require many different inputs. For example, a
block of chocolate needs cocoa powder, sugar,
water, milk, heat to melt it, paper to wrap it, ink to
print on the paper etc. Each one of these inputs
has its own market; paper prices change up and
down, cocoa prices might be high from low rainfall
in South America etc.
A very common economic function is called the
Cobb-Douglas Production Function which is an
approximation for production of a good from the
inputs of capital equipment (e.g. a factory) and
labour.
Theory: the Cobb-Douglas Production Function has
the general form:
𝑄 = 𝐴𝐿𝛼𝐾𝛽
where the values of 𝐴, 𝛼 and 𝛽 are all positive
constants.
A numerical example could be:
𝑄 = 5𝐿0.43𝐾0.6
Which specifies that the production of a certain
good (e.g. 𝑄𝑡𝑜𝑛𝑛𝑒𝑠 𝑜𝑓 𝑤𝑒𝑎𝑡 ) is a function of capital
(𝐾: tractors, storage sheds etc.) and labour (𝐿:
people driving the tractors, fixing the sheds etc.).
Obviously, there are other factors, such as rainfall,
soil condition etc., but they are ignored for now.
𝑥
𝑦
𝑧
𝑥
𝑥
𝑦
𝑦 𝑧
𝑧
142
Exercises: 1. List all the different variables that a baker needs to
take into account when baking a cake. 2. Think of a few goods that that do not require many
different types of inputs. What do they have in common?
3. Which of these are not Cobb-Douglas Production Functions?
𝑎) 𝑄 = 5𝐿0.2𝐾0.4
𝑏) 𝑄 = 5𝐿𝐾0.4
𝑐) 𝑄 =5𝐿0.7
𝐾−0.4
𝑑) 𝑄 =5𝐿−0.7
𝐾−0.4
𝑒) 𝑄 = −3𝐿0.4𝐾0.6
7.2 simple partial differentiation
An intuitive explanation of partial differentiation is
best to get your head around this concept.
Intro example 1: two factories, an aircraft
manufacturer (such as Boeing) and a metal
manufacturing company (such as Metalex), are
located close to one another; Metalex supplies
Boeing with the metal sheeting to cover the
aircraft. Both companies require electricity to run
their large production factories (Boeing to build
the planes, and Metalex to create the highest
quality metal) but the electricity system cannot
handle any more current (electrical flow).
Say Boeing wanted to introduce electrical heating
to their factory. That would mean more electricity
would need to flow to Boeing, costing them more.
Assuming prices of electricity are constant, this is
the direct effect of increasing the use of electricity
(that is, the extra cost of using more electricity for
heating). What is not being taken into account is
that when Boeing increases their use of electricity,
this will reduce the amount of electricity available
for Metalex, making the metal more expensive.
Metalex must then charge Boeing more for the
metal, making it more expensive to produce
aircraft. This is the indirect effect of Boeing using
more electricity.
When talking about partial differentiation, any
possible indirect effects are ignored, and assumed
not to exist. This is a big assumption, and later in
this chapter it will be removed.
Theory: partial differentiation is finding the rate at
which the dependent variable changes when an
independent variable changes assuming all other
variables are held constant.
That is, it is assumed that the other “independent”
variables do not affect the independent variable
that is changing (even though in reality, they
might).
Partial differentiation is written in one of two
ways:
1. 𝜕𝑧
𝜕𝑥 𝑂𝑅
𝜕𝑧
𝜕𝑦
This is read as “the derivative of 𝑧 with respect to
𝑥 holding everything else constant”. The “curvy
dees” are “deltas” and are not the same as normal
𝑑’s used in differentiation.
2. 𝑧𝑥 𝑂𝑅 𝑧𝑦
In this method of defining a partial derivative, the
subscript is the variable being differentiated, and it
is assumed that all other variables are constant.
Intro example 2: partially differentiate 𝑧 with
respect to 𝑥 for the function
𝑧 = 𝑥𝑦 + 5𝑥 − 7𝑦
The 𝑦 has to be treated as if it were a constant,
even though it obviously is not:
ELECTRICITY
BOEING METAL MANUFACTURER
direct effect (cost of extra electricity)
indirect effect (increase in metal costs)
143
𝜕𝑧
𝜕𝑥= 1𝑦 + 5
The 1𝑦 comes from differentiating 𝑥𝑦 with respect
to 𝑥 treating 𝑦 as if it were a constant; if 3𝑥 had to
be differentiated, the answer would be 3 1 = 3.
Similarly above, the 𝑦 is treated as if it were a 3,
but in fact it is still 𝑦. An absurd example might
help.
Intro example 3: find the partial derivative of 𝑧
with respect to 𝑥 for the function:
𝑧 = 𝑥 𝑦 𝑤 𝑝 𝑞 𝑟 𝑒
This equation has lots of variables, all multiplying
one another, but all that is relevant is the 𝑥, as
everything else is held constant.
𝜕𝑧
𝜕𝑥= 𝑦 𝑤 𝑝 𝑞 𝑟 𝑒
You may be wondering why the answer is not zero,
as everything else is a constant? The reason is that
these “constants” are multiplying 𝑥. If they were
added to 𝑥, then the derivative would be zero.
Theory: to partially differentiate a multivariable
function with respect to a single variable,
differentiate the function with respect to that
variable while treating all other variables as
constants.
Example 1: find the partial derivative with respect
to 𝑥 for the function
𝑧 = 𝑥2𝑦2𝑝𝑘 + 𝑝𝑓𝑑 + 3𝑥
Plan: differentiate using the normal rules, but
assume everything other than 𝑥 is a constant.
Solution:
𝜕𝑧
𝜕𝑥= 2𝑥𝑦2𝑝𝑘 + 3
The first part of the original equation has 𝑥2 and
all the other variables 𝑦2𝑝𝑘 are assumed to be
constant. Thus 𝑥2 is differentiated and then
multiplied by the remaining “constants”. The
second part of the original equation is 𝑝𝑓𝑑 and
since there are no 𝑥’s and differentiating a
constant (remember, everything other than an 𝑥 is
assumed to be a constant), gives zero. The last part
of the original equation is 3𝑥, so it is differentiated
to give the constant 3.
Example 2: find the partial derivative with respect
to 𝑥 in the following equation
𝑧 = 𝑥2 9 4 5 + 4 8 7 + 3𝑥
Plan: simplify the numbers then differentiate using
the normal rules, but assume everything other
than 𝑥 is a constant.
Solution: simplify the numbers:
𝑧 = 180𝑥2 + 224 + 3𝑥
Differentiate using the normal rules:
𝜕𝑧
𝜕𝑥= 2 180 𝑥 + 3 = 360𝑥 + 3
Look back at the last two examples. The second
example is identical to the first, except the
variables (other than 𝑥) are replaced with actual
numbers. If Example 2 was done without
simplifying first, the answer would be:
𝜕𝑧
𝜕𝑥= 2𝑥 9 4 5 + 3
Which is very similar to the answer in Example 1.
It is very important you understand that the other
variables are held constant when partially
differentiating.
Example 3: partially differentiate with respect to 𝑥
and then with respect to 𝑦 for the function
𝑧 = 𝑥2𝑦 − 3𝑦3𝑥2 + 5𝑥 − 7𝑦
Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding
𝑦 constant. Then for 𝜕𝑧/𝜕𝑦 differentiate the
function holding 𝑥 constant.
Solution: finding 𝜕𝑧/𝜕𝑥:
𝜕𝑧
𝜕𝑥= 2𝑥𝑦 − 3𝑦32𝑥 + 5
In the original function:
144
The first red section: differentiate 𝑥2 then
multiply by the rest of the red section (i.e. 𝑦)
giving: 2𝑥𝑦
The first blue section: differentiate 𝑥2 then
multiply by everything else in that section (i.e.
−3𝑦3) giving: −3𝑦32𝑥
The second red section: differentiate the 𝑥,
then multiply by everything else in that section
(i.e. 5) giving: 5.
The second blue section: there are no 𝑥’s so
the derivative of a constant is zero.
Put these together and simplify:
𝜕𝑧
𝜕𝑥= 2𝑥𝑦 − 6𝑦3𝑥 + 5
Now for 𝜕𝑧/𝜕𝑦, differentiate the original equation
holding 𝑥 as a constant:
𝜕𝑧
𝜕𝑦= 𝑥2 − 9𝑦2𝑥2 − 7
Figure out how this last answer was obtained, one
section at a time.
In the above example, you may have noticed that
when looking at each section, 𝑥 was isolated and
differentiated, then everything else was brought
back in. This is a good technique to use, but only
use it where there are no division signs.
Example 4: find both partial derivatives of
𝑧 = 2𝑥3𝑦2 + 2𝑥𝑦 − 3𝑥 + 2𝑦 + 15
Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding
𝑦 constant. Then for 𝜕𝑧/𝜕𝑦 differentiate the
function holding 𝑥 constant.
Solution: for 𝜕𝑧/𝜕𝑥:
𝜕𝑧
𝜕𝑥= 3𝑥2 ∙ 2𝑦2 + 2𝑦 − 3
𝜕𝑧
𝜕𝑥= 6𝑥2𝑦2 + 2𝑦 − 3
And for 𝜕𝑧/𝜕𝑦:
𝜕𝑧
𝜕𝑦= 2𝑦 ∙ 2𝑥3 + 2𝑥 + 2
𝜕𝑧
𝜕𝑦= 4𝑦𝑥3 + 2𝑥 + 2
It is vital you understand this simple partial
differentiation before moving on.
Exercises: 1. Find both first order partial derivatives for the
functions: 𝑎) 𝑧 = 𝑥2𝑦 + 3𝑦2𝑥 + 7𝑥𝑦
𝑏) 𝑧 = 3𝑥2 + 3𝑦2 + 3 𝑥𝑦 7
𝑐) 𝑧 = 5𝑥𝑦 + 𝑥2𝑦 + 11
𝑑) 𝑧 = 2𝑥𝑦 − 2𝑥2𝑦3 + 𝑥−1
𝑒) 𝑧 = 2𝑥𝑦−2 +3
𝑥𝑦− 2
2. Differentiate the following with respect to both 𝑥 and 𝑦 (each individually):
𝑎) 𝑧 = 3𝑥7𝑦−3 + 5𝑥 + 5𝑥𝑦 − 3
𝑏) 𝑧 = −2𝑥−2𝑦 + 7𝑦−2𝑥
𝑐) 𝑧 = 15𝑥2𝑦3 − 15𝑥−2𝑦−3
7.3 complex partial differentiation
This section applies the more complex rules
(product, chain etc.) to partial differentiation.
Example 1: find both partial derivatives of
𝑧 = 𝑥2 + 𝑦3 3𝑥4 − 𝑦 + 3 2
Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding
𝑦 constant. Then for 𝜕𝑧/𝜕𝑦 differentiate the
function holding 𝑥 constant.
Use the product rule as the base rule, with the
chain rule within it.
Solution: for 𝜕𝑧/𝜕𝑥:
𝜕𝑧
𝜕𝑥= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
𝑑𝑒𝑟 1𝑠𝑡 = 2𝑥
𝑑𝑒𝑟 2𝑛𝑑 = 2 3𝑥4 − 𝑦 + 3 1 . 3 4𝑥3
= 24𝑥3 3𝑥4 − 𝑦 + 3
The 𝑑𝑒𝑟 2𝑛𝑑 was obtained by using the chain rule,
and treating 𝑦 as a constant. The 𝑦 is still there as
due to the chain rule, the bracketed terms are
rewritten with the power reduced by one.
For the solution, substitute into the base (product)
rule:
145
𝜕𝑧
𝜕𝑥=
2𝑥 3𝑥4 − 𝑦 + 3 2 + 24𝑥3 3𝑥4 − 𝑦 + 3 𝑥2 + 𝑦3
And for 𝜕𝑧/𝜕𝑦:
𝜕𝑧
𝜕𝑦= 𝑑𝑒𝑟 1𝑠𝑡 × 2𝑛𝑑 + 𝑑𝑒𝑟 2𝑛𝑑 × 1𝑠𝑡
𝑑𝑒𝑟 1𝑠𝑡 = 3𝑦2
𝑑𝑒𝑟 2𝑛𝑑 = 2 3𝑥4 − 𝑦 + 3 1(−1)
= −2 3𝑥4 − 𝑦 + 3
Substitute this back into the base function:
𝜕𝑧
𝜕𝑦=
3𝑦2 3𝑥4 − 𝑦 + 3 2 + −2 3𝑥4 − 𝑦 + 3 𝑥2 + 𝑦3
Both the solutions above can be simplified by
factorisation (see Chapter 1 and 5).
Example 2: find both partial derivatives of
𝑧 = 𝑥2 + 𝑦 3
𝑥 − 𝑦
Plan: for 𝜕𝑧/𝜕𝑥 differentiate the function holding
𝑦 constant. Then for 𝜕𝑧/𝜕𝑦 differentiate the
function holding 𝑥 constant. Use the quotient rule
as the base rule, and the chain rule within the
quotient rule.
Solution: for 𝜕𝑧/𝜕𝑥:
𝜕𝑧
𝜕𝑥=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇 − 𝑑𝑒𝑟 𝐵𝑂𝑇 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
𝑑𝑒𝑟 𝑇𝑂𝑃 = 3 𝑥2 + 𝑦 2 2𝑥
= 6𝑥 𝑥2 + 𝑦 2
𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = 1
The chain rule was used to obtain the derivative of
the top. Remember the chain rule: bring the power
out front of the brackets, take one off the power,
then multiply by the derivative of the inside of the
brackets.
Substitute everything into the Quotient Rule:
𝜕𝑧
𝜕𝑥=
6𝑥 𝑥2 + 𝑦 2 𝑥 − 𝑦 − (1) 𝑥2 + 𝑦 3
𝑥 − 𝑦 2
Factorise 𝑥2 + 𝑦 2 out of the top:
𝜕𝑧
𝜕𝑥=
𝑥2 + 𝑦 2 6𝑥 𝑥 − 𝑦 − 𝑥2 + 𝑦
𝑥 − 𝑦 2
Further simplify the square brackets:
𝜕𝑧
𝜕𝑥=
𝑥2 + 𝑦 2 5𝑥2 − 6𝑥𝑦 − 𝑦
𝑥 − 𝑦 2
For 𝜕𝑧/𝜕𝑦:
𝜕𝑧
𝜕𝑦=
𝑑𝑒𝑟 𝑇𝑂𝑃 × 𝐵𝑂𝑇 − 𝑑𝑒𝑟 𝐵𝑂𝑇 × 𝑇𝑂𝑃
[𝐵𝑂𝑇𝑇𝑂𝑀]2
𝑑𝑒𝑟 𝑇𝑂𝑃 = 3 𝑥2 + 𝑦 2 1
= 3 𝑥2 + 𝑦 2
𝑑𝑒𝑟 𝐵𝑂𝑇𝑇𝑂𝑀 = −1
Substitute into the base (quotient) rule:
𝜕𝑧
𝜕𝑦=
3 𝑥2 + 𝑦 2 𝑥 − 𝑦 − (−1) 𝑥2 + 𝑦 3
𝑥 − 𝑦 2
Factorise and simplify the top (do it yourself!):
𝜕𝑧
𝜕𝑦=
𝑥2 + 𝑦 2 3𝑥 − 2𝑦 + 𝑥2
𝑥 − 𝑦 2
Make sure you fully understand the last two
examples before trying the following exercises.
Exercises: 1. Differentiate with respect to both 𝑥 and 𝑦
(individually) using the Chain Rule: 𝑎) 𝑧 = 𝑥 + 𝑦 3
𝑏) 𝑧 = 3𝑥2 + 𝑦 4
𝑐) 𝑧 = 7𝑥2 + 7𝑦2 5
𝑑) 𝑧 = −2𝑥−2 + 𝑦−2 4
𝑒) 𝑧 = 3𝑥𝑦 + 2𝑥2𝑦3 −3
2. Differentiate with respect to both 𝑥 and 𝑦 (individually) using the Product Rule: 𝑎) 𝑧 = 𝑥2 + 5𝑦 3𝑥 + 7𝑦2
𝑏) 𝑧 = 𝑥2 + 𝑦 𝑦2 + 𝑥
𝑐) 𝑧 = 3𝑥𝑦 − 𝑥2 𝑥𝑦3 + 1
𝑑) 𝑧 = 3𝑥2 +1
𝑦 𝑥𝑦 − 𝑦4
𝑒) 𝑧 = 3
𝑥+
3
𝑥𝑦
4
𝑥2−
4
𝑥𝑦
3. Differentiate with respect to both 𝑥 and 𝑦 (individually) using the Quotient Rule:
𝑎) 𝑧 =𝑥 + 𝑦
𝑥 − 𝑦
𝑏) 𝑧 =3𝑥2 − 2𝑦3
𝑥2𝑦
𝑐) 𝑧 =2𝑥𝑦 − 2𝑥2𝑦
15𝑥𝑦 + 1
𝑑) 𝑧 =4𝑥𝑦4
1 − 𝑥𝑦
𝑒) 𝑧 =1 + 𝑥𝑦2
1 − 𝑥𝑦2
146
4. Differentiate with respect to both 𝑥 and 𝑦 (individually) using the 𝑒 and ln rules:
𝑎) 𝑧 = 𝑒𝑥𝑦−𝑥2
𝑏) 𝑧 = 𝑒𝑥2𝑦−𝑦2𝑥
𝑐) 𝑧 = ln 𝑥𝑦 − 𝑥2𝑦
𝑑) 𝑧 = 𝑒 ln 𝑥𝑦−1
𝑒) 𝑧 = ln 𝑒𝑥3𝑦3+3𝑥2𝑦 − 1
5. Differentiate with respect to both 𝑥 and 𝑦 (individually) using any rules:
𝑎) 𝑧 = 𝑒𝑥2𝑦−1 + 𝑥𝑦 3𝑥𝑦 − 1 3
𝑏) 𝑧 = 5𝑥2𝑦 − 𝑥 ln 15𝑥𝑦 + 𝑒𝑥𝑦
𝑐) 𝑧 = ln 15𝑥𝑦 + 𝑒1
𝑥𝑦 + 𝑒𝑥𝑦2
2
𝑑) 𝑧 =𝑒
𝑥𝑦 − 𝑥2𝑦
𝑥𝑦 − ln 𝑥𝑦
𝑒) 𝑧 = 3𝑥𝑦 − 𝑥2 4
1 − 𝑥𝑦
7.4 second order partial derivatives
This is a very simple concept. In theory it is the
same as a normal second derivative.
Theory: the second order partial derivative is
simply partially differentiating the original function
twice.
The second order partial derivative is denoted in
one of two ways:
𝜕2𝑧
𝜕𝑥2 ,
𝜕2𝑧
𝜕𝑦2 𝑂𝑅 𝑧𝑥𝑥 , 𝑧𝑦𝑦
Example 1: find all second order partial derivatives
for the function
𝑧 = 𝑥3𝑦7 − 3𝑥𝑦 + 2𝑥 − 𝑦2
Plan: differentiate the function with respect to 𝑥
twice, then differentiate the function with respect
to 𝑦 twice.
Solution: differentiate with respect to 𝑥:
𝜕𝑧
𝜕𝑥= 3𝑥2𝑦7 − 3𝑦 + 2
Differentiate with respect to 𝑥 again:
𝜕2𝑧
𝜕𝑥2= 6𝑥𝑦7
For the second order partial derivative of 𝑦:
𝜕𝑧
𝜕𝑦= 7𝑦6𝑥3 − 3𝑥 − 2𝑦
Differentiate this equation with respect to 𝑦 again:
𝜕2𝑧
𝜕𝑦2= 42𝑦5𝑥3 − 2
This is very similar to what was learnt in Chapter 5,
however there are two more second order partial
derivatives. These are called mixed partial
derivatives.
Theory: mixed partial derivatives are found by
differentiating with respect to one variable first,
then differentiating with respect to the other
variable. They are denoted in a similar way to the
straight second order partial derivatives:
𝜕2𝑧
𝜕𝑥𝜕𝑦 ,
𝜕2𝑧
𝜕𝑦𝜕𝑥 𝑂𝑅 𝑧𝑥𝑦 , 𝑧𝑦𝑥
Example 1 cont: finding both mixed partial
derivatives:
1. Differentiate with respect to 𝑥 first:
𝜕𝑧
𝜕𝑥= 3𝑥2𝑦7 − 3𝑦 + 2
Then differentiate with respect to 𝑦:
𝜕2𝑧
𝜕𝑥𝜕𝑦= 3 7 𝑥2𝑦6 − 3
= 21𝑥2𝑦6 − 3
2. Differentiate with respect to 𝑦 first:
𝜕𝑧
𝜕𝑦= 7𝑦6𝑥3 − 3𝑥 − 2𝑦
Then differentiate with respect to 𝑦:
𝜕2𝑧
𝜕𝑦𝜕𝑥= 7(3)𝑦6𝑥2 − 3
= 21𝑥2𝑦6 − 3
Notice that the two mixed partial derivatives are
the same. This is always the case, however it is a
good idea to always write out the theory.
Theory: mixed partial derivatives are the same, no
matter which way they are found.
𝑧𝑥𝑦 = 𝑧𝑦𝑥
147
The following example has minimal narration so
use scrap paper to see if you can get the same
answers.
Example 2: find all second order partial derivates
for the function
𝑧 = ln 𝑥2 +1
𝑦3+ 5
4
Plan: find the first order partial derivatives, then
differentiate again to find the straight second
order partial derivatives. Then find the mixed
second order partial derivatives.
Solution: for the first order partial derivatives, the
ln rule is used, with the chain rule within it:
𝑧𝑥 =
4 2𝑥 𝑥2 +1𝑦3 + 5
3
𝑥2 +1𝑦3 + 5
4
𝑧𝑥 =8𝑥
𝑥2 +1𝑦3 + 5
𝑧𝑦 =
−3𝑦−4 4 𝑥2 +1𝑦3 + 5
3
𝑥2 +1𝑦3 + 5
4
𝑧𝑦 =−12𝑦−4
𝑥2 +1𝑦3 + 5
For the straight second order partial derivatives
(using the quotient and chain rules):
𝑧𝑥𝑥 =8 𝑥2 +
1𝑦3 + 5 − 2𝑥 8𝑥
𝑥2 +1𝑦3 + 5
2
𝑧𝑥𝑥 =−8𝑥2 +
8𝑦3 + 40
𝑥2 +1𝑦3 + 5
2
𝑧𝑦𝑦 =
48𝑦−5 𝑥2 +1𝑦3 + 5 − −3𝑦−4 −12𝑦−4
𝑥2 +1𝑦3 + 5
2
𝑧𝑦𝑦 =
48𝑥2
𝑦5 +12𝑦8 +
48𝑦5
𝑥2 +1𝑦3 + 5
2
For the mixed second order partial derivative:
𝑧𝑥𝑦 = 𝑧𝑦𝑥 =24𝑥𝑦−4
𝑥2 +1𝑦3 + 5
2
Exercises: 1. Determine all four second order partial derivatives
for the following functions (and make sure 𝑧𝑥𝑦 = 𝑧𝑦𝑥 ):
𝑎) 𝑧 = 5𝑥𝑦 + 3𝑦 3
𝑏) 𝑧 = 12 𝑥2𝑦 + 𝑥𝑦2 𝑥 + 𝑦 2
𝑐) 𝑧 =𝑥2 + 𝑥𝑦
𝑦 − 𝑥
𝑑) 𝑧 = 𝑒𝑥2𝑦+4𝑥
𝑒) 𝑧 = ln 7𝑥2𝑦 + 4𝑦
2. Determine all four second order straight partial derivatives, and make sure that the mixed second order partial derivatives are the same. 𝑎) 𝑧 = 𝑒5𝑥𝑦 + 4𝑥𝑦 3
𝑏) 𝑧 = ln 𝑒𝑥2𝑦2+5𝑥𝑦
𝑐) 𝑧 = 1
𝑥+
1
𝑥𝑦
2
𝑥−3𝑦−2 + 1
7.5 application of partial differentiation
The Cobb-Douglas Production Function is a very
common application of partial derivatives.
Theory: the partial derivative of the Cobb-Douglas
Production Function gives the marginal product of
labour (𝜕𝑄/𝜕𝐿) and the marginal product of
capital (𝜕𝑄/𝜕𝐾). The higher the value of the
marginal product, the greater the increase in
output from an extra unit of that input.
Economic Theory: returns to scale are defined as
the extent output changes when inputs are
changed by a certain ratio. For the Cobb-Douglas
Production Function
𝑄 = 𝐴𝐿𝛼𝐾𝛽
148
If:
1. 𝛼 + 𝛽 < 1 then there are decreasing returns
to scale; if inputs are doubled, output will less
than double.
2. 𝛼 + 𝛽 = 1 then there are constant returns to
scale; if inputs double, output will double.
3. 𝛼 + 𝛽 > 1 then there are increasing returns to
scale; if inputs double, output will more than
double.
Example 1: find the marginal productivity of labour
for the function
𝑄 = 4𝐿0.3𝐾0.7
when 𝐿 = 30 and 𝐾 = 46. Also determine the
returns to scale.
Plan: partially differentiate 𝑄 with respect to 𝐿.
Add the indices to determine the returns to scale.
Solution:
𝜕𝑄
𝜕𝐿= 4 0.3 𝐿−0.7𝐾0.7
= 1.2𝐿−0.7𝐾0.7
Substitute 𝐿 = 30 and 𝐾 = 46:
𝜕𝑄
𝜕𝐿= 1.2 30 −0.7 46 0.7 ≈ 1.619 3𝑑. 𝑝
Output will increase by 1.619 units for an extra
unit of labour.
Since the addition of the indices is 0.3 + 0.7 = 1,
there are constant returns to scale.
Example 2: find the marginal productivities of
labour and capital for the following production
function
𝑄 = 12𝐿0.5𝐾0.75
at a labour level of 46 and a capital level of 71.
Plan: differentiate with respect to 𝐿 then
substitute in 𝐿 = 46, 𝐾 = 71 to find the marginal
productivity of labour. Similarly for 𝐾.
Solution: for marginal productivity of labour:
𝜕𝑄
𝜕𝐿= 6𝐿−0.5𝐾0.75
𝜕𝑄
𝜕𝐿= 6 46−0.5 710.75 ≈ 21.64 2𝑑. 𝑝.
And for marginal productivity of capital:
𝜕𝑄
𝜕𝐾= 9𝐿0.5𝐾−0.25
𝜕𝑄
𝜕𝐾= 9 460.5 71−0.25 ≈ 21.03 2𝑑. 𝑝.
Thus for an extra unit of labour, output would
increase by approximately 21.64 units, and for an
extra unit of capital, output would increase by
approximately 21.06 units.
In this last example, if the cost of a unit of labour
and a unit of capital were the same, it would be
better to employ more labour, as they will add
(slightly) more to output than capital.
Exercises: 1. Determine the marginal productivity of labour and
capital in the following Cobb-Douglas Production Functions and also the returns to scale. 𝑎) 𝑄 = 4𝐿0.5𝐾0.5
𝑏) 𝑄 = 13𝐿0.6𝐾0.4
𝑐) 𝑄 = 12.5𝐿0.7𝐾0.4
𝑑) 𝑄 = 75𝐿0.8𝐾0.5
𝑒) 𝑄 = 0.8𝐿0.6𝐾0.3
𝑓) 𝑄 = 100𝐿0.2𝐾0.9
𝑔) 𝑄 = 5.73𝐿0.79𝐾0.41
) 𝑄 = 2.118𝐿0.812𝐾0.417
If you had the option of increasing either labour or capital by one unit in each of the above questions, which one would you increase? Why?
7.6 total differentiation
At the beginning of the chapter, the electricity
example was used to demonstrate the theory of
partial differentiation. Boeing had installed electric
heaters in their plant, which reduced the amount
of electricity available for the metal manufacturer
Metalex. Boeing assumed that their decision
would not affect costs other than the use of
electricity (i.e. direct effect). However, Metalex
will not produce as much metal, so will charge
Boeing more per sheet of metal. This is the indirect
149
effect of using more electricity. Adding the direct
and indirect effects gives the total effect; this is
what total differentiation finds.
As a manager, better decisions are made when
variables are not assumed to be constant. Knowing
the total effect is better than pretending variables
are constant, when in reality, they are not.
Theory: the following is a formula that allows us to
move from partial differentiation to total
differentiation. The total derivative equation is:
𝑑𝑧 =𝜕𝑧
𝜕𝑥𝑑𝑥 +
𝜕𝑧
𝜕𝑦𝑑𝑦
It is much simpler than it seems.
In the Boeing example, the variables are: total
costs (𝑧), cost of electricity (𝑥) and the cost of
metal (𝑦).
Note: this equation finds the gradient function of a
multivariable equation. For small changes in either
or both of the input variables, this equation is a
good approximation. When the changes are large,
then the approximation will be less accurate. For
the reason, consult a second year text book
dealing with “linear approximations and errors”.
Example 1: for the following function, determine
the extent total production will change if
electricity use increases by 5 units from 150, and
metal prices decrease by 3 units from 400.
𝑃 = 5𝐸2𝑀 + 5𝐸2 − 2𝑀2 + 3𝑀 − 6𝐸 + 550
Plan: partially differentiate the function 𝑃, then
substitute the partial derivatives, the changes in
electricity and metal, and the original levels of
electricity and metal into the total derivative
function.
Solution: the partial derivatives are:
𝜕𝑃
𝜕𝐸= 10𝐸𝑀 + 10𝐸 − 6
𝜕𝑃
𝜕𝑀= 5𝐸2 − 4𝑀 + 3
The small change in 𝐸 is 𝑑𝐸 = 5 and the small
change in 𝑀 is 𝑑𝑀 = −3. It is negative because it
is a decrease.
Substitute the partial derivatives into the total
derivative equation:
𝑑𝑃 =𝜕𝑃
𝜕𝐸𝑑𝐸 +
𝜕𝑃
𝜕𝑀𝑑𝑀
𝑑𝑃 = 10𝐸𝑀 + 10𝐸 − 6 𝑑𝐸 + 5𝐸2 − 4𝑀 + 3 𝑑𝑀
Further substitution of 𝑑𝐸 = 5 and 𝑑𝑀 = −3:
𝑑𝑃 = 10𝐸𝑀 + 10𝐸 − 6 (5) + 5𝐸2 − 4𝑀 + 3 (−3)
Finally, substitute the original values of 𝐸 = 150
and 𝑀 = 400:
𝑑𝑃 = 10 150 400 + 10 150 − 6 5
+ 5 150 2 − 4 400 + 3 −3
𝑑𝑃 = 3,007,470 − 332,709 = 2,674,761
Production will increase by approximately two and
a half million units.
This is still an approximation. To get the actual
change in production, find the production at the
original input level of 𝐸 = 150, 𝑀 = 400 and then
the production at the new input level of
𝐸 = 155, 𝑀 = 397 (these numbers come from the
original plus the change in the inputs; 𝐸 = 150 +
5, 𝑀 = 400 − 3). Finding the difference between
these two production levels gives the precise
change in production (actually do it and you should
get 2.702million units (3d.p.)).
The total derivative approximation method is used
because often, the 𝑧 function (the cost 𝑃 in this
example) cannot be isolated easily, but the partial
derivatives can easily be found with other
𝑑𝑧 =𝜕𝑧
𝜕𝑥𝑑𝑥 +
𝜕𝑧
𝜕𝑦𝑑𝑦
very small change in 𝑥
very small change in 𝑦
Partial derivative with respect to 𝑥
very small
change in 𝑧
Partial derivative with respect to 𝑦
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techniques. You will learn this in second year, if
you continue with economics.
Theory: to get the actual change in a multivariable
function, find the value of the function at the New
variable level then subtract the value of the
function at the Original variable level:
∆𝑧 = 𝑧 𝑥𝑁 , 𝑦𝑁 − 𝑧 𝑥𝑂 , 𝑦𝑂
Example 2: find the approximate change in output
for the production function
𝑄 = 5𝐿0.6𝐾0.5
when the original levels of the inputs are 𝐿 = 500,
𝐾 = 700 and when 𝐿 increases by 2 units, and 𝐾
decreases by 3 units.
Also, find the exact change in 𝑧.
Plan: find the partial derivatives of 𝑧, then
substitute everything into the total derivative
function to solve for 𝑑𝑧.
To find the actual change in 𝑧, find the difference
between the values of 𝑧𝑁 𝑥 = 502, 𝑦 = 697 and
𝑧𝑂 𝑥 = 500, 𝑦 = 700 .
Solution: the partial derivatives are:
𝑄𝐿 = 3𝐿−0.4𝐾0.5
𝑄𝐾 = 2.5𝐿0.6𝐾−0.5
Substitute these into the total derivative equation:
𝑑𝑄 =𝜕𝑄
𝜕𝐿𝑑𝐿 +
𝜕𝑄
𝜕𝐾𝑑𝐾
𝑑𝑄 = 3𝐿−0.4𝐾0.5 𝑑𝐿 + 2.5𝐿0.6𝐾−0.5 𝑑𝐾
Then substitute 𝐿 = 500, 𝐾 = 700, 𝑑𝐿 = 2 and
𝑑𝐾 = −3:
𝑑𝑄 = 3 500 −0.4 700 0.5 2
+ 2.5 500 0.6 700 −0.5 −3
𝑑𝑄 = 1.416 3𝑑. 𝑝.
For the actual change:
𝑄 𝐿 = 500, 𝐾 = 700 = 5 500 0.6 700 0.5
≈ 5506.8219
𝑄 𝐿 = 502, 𝐾 = 697 = 5 502 0.6 697 0.5
≈ 5508.1863
∆𝑄 = 𝑄 𝐿 = 502, 𝐾 = 697 − 𝑄 𝐿 = 500, 𝐾 = 700
∆𝑄 = 5508.1863 − 5506.8219 = 1.364 (3𝑑. 𝑝. )
Example 3: using the following Cobb-Douglas
Production Function
𝑄 = 40𝐿0.7𝐾0.6
with the current use of labour at 400 units, and
the current use of capital at 300 units, find an
approximation for the change in production when
labour is increased by 1% and capital is reduced by
0.5%.
Plan: use the percentage changes in 𝐿 and 𝐾 to
find the actual changes (in units). Partially
differentiate the function, then substitute
everything into the total derivative equation.
Solution: The actual changes in 𝐿 and 𝐾 are:
∆𝐿 = 1
100 400 = 4
∆𝐾 = −0.5
100 300 = −1.5
the partial derivatives are:
𝜕𝑄
𝜕𝐿= 40 0.7 𝐿−0.3𝐾0.6 = 28𝐿−0.3𝐾0.6
𝜕𝑄
𝜕𝐾= 40 0.6 𝐿0.7𝐾−0.4 = 24𝐿0.7𝐾−0.4
Substitute these along with 𝐿 = 400, 𝐾 = 300,
𝑑𝐿 = 4 and 𝑑𝐾 = −1.5 into the total derivative
equation:
𝑑𝑄 =𝜕𝑄
𝜕𝐿𝑑𝐿 +
𝜕𝑄
𝜕𝐾𝑑𝐾
𝑑𝑄 = 28𝐿−0.3𝐾0.6 4 + 24𝐿0.7𝐾−0.4 −1.5
𝑑𝑄 = 28(400)−0.3(300)0.6 4
+ 24(400)0.7(300)−0.4 −1.5
𝑑𝑄 = 324.96 (2𝑑. 𝑝. )
This is the approximate change in quantity from
changes in both inputs. To determine the exact
change, find 𝑄𝑂(𝐿 = 400, 𝐾 = 300) then subtract
it from 𝑄𝑁 𝐿 = 404, 𝐾 = 298.5 . Do this
151
calculation, and see if you get
∆𝑄 = 322.16 2𝑑. 𝑝. .
Exercises: 1. For the function
𝑧 = 0.3𝑥𝑦2 + 0.5𝑥2𝑦 Determine an expression for the total derivative (leaving any unknowns as variables).
2. The following production function approximates the output of a farm: 𝑄 = 4.75𝐾0.7𝐿0.3 Determine: a) An expression for the total derivative. b) An expression for the total derivative when
𝐿 = 50, 𝐾 = 70. c) The approximate change in output when 𝐿
increases by 1unit and 𝐾 increases by 2 units. d) Find the actual change in output. (Hint: find
the output at the two different 𝐿, 𝐾 combinations)
e) The returns to scale. 3. The following is a production function which
approximates the production of a cabinet maker: 𝑄 = 53.5𝐿0.71𝐾0.32 Determine: a) An expression for the total derivative. b) An expression for the total derivative when
𝐿 = 300, 𝐾 = 140 c) The approximate change in output when 𝐿
decreases by 2 units and 𝐾 increases by 1 unit. d) Find the actual change in output. (Hint: find
the output at the two different 𝐿, 𝐾 combinations)
e) The returns to scale.
7.7 optimisation with many variables
Partial derivatives play an important role in finding
the optimal points of functions. Similar to
optimisation in Chapter 5, optimisation for
multivariable function has first and second order
conditions, but with some modifications.
Theory: the First Order Condition (FOC) is to set all
first order partial derivatives to zero.
𝜕𝑧
𝜕𝑥= 0 and
𝜕𝑧
𝜕𝑦= 0
Then solve for 𝑥 and 𝑦 using simultaneous
equations, to find any stationary points.
Example 1: find the stationary point(s) for
𝑧 = 0.5𝑦2 + 𝑥𝑦 + 𝑥2 − 𝑦 − 3𝑥 + 5
Plan: use the FOC by setting both partial
derivatives equal to zero, then solving for 𝑥 and 𝑦.
Solution: using the FOC:
𝜕𝑧
𝜕𝑥= 𝑦 + 2𝑥 − 3 = 0
𝜕𝑧
𝜕𝑦= 𝑦 + 𝑥 − 1 = 0
Use simultaneous equations to solve for 𝑥 and 𝑦:
′ : 𝑦 = −2𝑥 + 3
Substitute ′ → :
−2𝑥 + 3 + 𝑥 − 1 = 0
−2𝑥 + 3 + 𝑥 − 1 = 0
𝑥 = 2
Substitute back into ′ : 𝑦 = −2 2 + 3 = −1
There is one stationary point at 𝑥 = 2, 𝑦 = −1. A
“point” in a three dimensional graph also requires
the value of the dependent variable (i.e. the 𝑧).
This is found from the original function:
𝑧 2, −1 = 0.5𝑦2 + 𝑥𝑦 + 𝑥2 − 𝑦 − 3𝑥 + 5
= 0.5 −1 2 + 2 −1 + 2 2 − −1
− 3 2 + 5
= 2.5
Thus there is a stationary point at 𝑥 = 2, 𝑦 = −1,
𝑧 = 2.5 or in coordinate form 𝑥, 𝑦, 𝑧 =
2, −1, 2.5 .
The question now becomes: is this stationary point
a maximum or a minimum (or something called a
saddle point)?
This is unknown yet, but firstly, what is a saddle
point?
Theory: a saddle point is a stationary point that is
neither an overall maximum nor an overall
minimum. It looks like a saddle:
152
Look closely: in one direction there is a minimum,
but it the other direction, there is a maximum. This
is a stationary point, but it is not an optimal point.
Theory: to find out whether a stationary point is a
maximum, a minimum or a saddle point, the
Second Order Condition (SOC) is used:
Find the value of all straight second order partial
derivatives at the point(s) obtained from the FOC,
then apply the following:
1. The point is a possible maximum if:
𝜕2𝑧
𝜕𝑥2< 0 𝐴𝑁𝐷
𝜕2𝑧
𝜕𝑦2< 0
That is, if both second order straight partial
derivatives are less than zero, it is a possible
maximum.
2. The point is a possible minimum if:
𝜕2𝑧
𝜕𝑥2> 0 𝐴𝑁𝐷
𝜕2𝑧
𝜕𝑦2> 0
That is, if both second order straight partial
derivatives are greater than zero, it is a possible
minimum.
3. The point is a saddle point if the signs of the two
straight first order partial derivatives do not
match, or if one is equal to zero.
Example 2: for the function from Example 1
𝑧 = 0.5𝑦2 + 𝑥𝑦 + 𝑥2 − 𝑦 − 3𝑥 + 5
determine the possible nature of the stationary
point at 2, −1, 2.5 .
Plan: evaluate the straight second order partial
derivatives at 𝑥 = 2, 𝑦 = −1.
Solution: straight second order partial derivatives:
𝜕2𝑧
𝜕𝑥2= 2 > 0
𝜕2𝑧
𝜕𝑦2= 1 > 0
Both straight second order partial derivatives are
greater than zero, so this is a possible minimum.
The process used up to now has been similar to
that in Chapter 5, but the following is a new rule.
Theory: once the straight second order partial
derivatives have been evaluated, and a possible
minimum or possible maximum has been
established, to make sure it is a minimum or
maximum, the delta test must be used:
∆= 𝜕2𝑧
𝜕𝑥2 𝜕2𝑧
𝜕𝑦2 − 𝜕2𝑧
𝜕𝑥𝜕𝑦
2
The delta test is: the two straight second partial
derivatives multiplied together with the mixed
partial second derivative squared, subtracted.
𝐼𝑓 ∆ > 0, the point is what was suspected.
𝐼𝑓 ∆ < 0, ignore everything, the point is a saddle.
Written simpler:
∆= 𝑓𝑥𝑥 𝑓𝑦𝑦 − 𝑓𝑥𝑦 2
Example 2 cont: confirm the nature of the point at
2, 7, 34.5 .
Solution: the mixed partial second derivative:
𝜕2𝑧
𝜕𝑥𝜕𝑦= 1
Apply the delta test:
∆= 2 1 − 1 2 = 1 > 0
Thus the point is what it was thought to be. Since
it was thought to be a minimum in this example,
the delta test confirms this.
𝑥
𝑦
𝑧
153
Theory: the delta test does NOT tell us whether a
point is a maximum or a minimum, the SOC
determines this. The delta test just proves or
disproves what was suspected from the SOC.
Example 3: determine the nature of all stationary
points for the function
𝑧 =1
3𝑥3 −
1
2𝑦2 −
3
2𝑥2 − 𝑥𝑦 − 20𝑥 − 5𝑦 − 10
Plan: using the FOC, find the partial derivatives
and set them to zero. Solve for the variables.
Next, find the straight second order partial
derivatives and evaluate them at the stationary
point(s). From their signs, determine the possible
nature of the stationary point.
Find the mixed second order partial derivative, and
apply the delta test to confirm the nature.
Solution: find the FOC:
𝜕𝑧
𝜕𝑥= 𝑥2 − 3𝑥 − 𝑦 − 20 = 0
𝜕𝑧
𝜕𝑦= −𝑦 − 𝑥 − 5 = 0
Use simultaneous equations:
′ : 𝑦 = −𝑥 − 5
Substitute′ → :
𝑥2 − 3𝑥 − (−𝑥 − 5) − 20 = 0
𝑥2 − 3𝑥 + 𝑥 + 5 − 20 = 0
𝑥2 − 2𝑥 − 15 = 0
Solve for 𝑥 using the Quadratic Formula:
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−(−2) ± (−2)2 − 4 1 (−15)
2(1)
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =2 ± 64
2
𝑥 = 5 𝑂𝑅 𝑥 = −3
These are the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 of two stationary points.
The 𝑦 − 𝑣𝑎𝑙𝑢𝑒s are obtained from ′.
For 𝑥 = 5: 𝑦 = − 5 − 5 = −10
For 𝑥 = −3: 𝑦 = − −3 − 5 = −2
Thus the two stationary points are:
Point 1: 𝑥 = 5, 𝑦 = −10
Point 2: 𝑥 = −3, 𝑦 = −2
Lastly, get the 𝑧 − 𝑣𝑎𝑙𝑢𝑒𝑠 for these two points
(see if you get these same values):
𝑧 5, −10 = −555
6
𝑧 −3, −2 = 29.5
The SOC needs to be used to determine the nature
of these two points:
𝜕2𝑧
𝜕𝑥2= 2𝑥 − 3
𝜕2𝑧
𝜕𝑦2= −1
Look at each of the two points individually:
Point 1: 5, −10, −555
6
𝜕2𝑧
𝜕𝑥2 𝑥 = 5, 𝑦 = −10 = 2 5 − 3 = 7 > 0
𝜕2𝑧
𝜕𝑦2 𝑥 = 5, 𝑦 = −10 = −1 < 0
The two signs do not match, so this is a saddle
point.
Point 2: −3, −2,29.5
𝜕2𝑧
𝜕𝑥2 𝑥 = −3, 𝑦 = −2 = 2 −3 − 3 = −9 < 0
𝜕2𝑧
𝜕𝑦2 𝑥 = −3, 𝑦 = −2 = −1 < 0
Both signs are less than zero, so this point is a
possible maximum. To confirm this, the delta test
is used:
∆= 𝜕2𝑧
𝜕𝑥2 𝜕2𝑧
𝜕𝑦2 − 𝜕2𝑧
𝜕𝑥𝜕𝑦
2
The mixed second order partial derivative is:
𝜕2𝑧
𝜕𝑥𝜕𝑦= −1
Put everything into the delta formula:
∆= −9 −1 − −1 2 = 9 − 1 = 8 > 0
Since ∆ > 0, this confirms what was suspected
from the SOC; since the SOC gave a possible
154
maximum, the delta test confirms this.
𝑃𝑜𝑖𝑛𝑡 1: 5, −10, −555
6 (𝑠𝑎𝑑𝑑𝑙𝑒 𝑝𝑜𝑖𝑛𝑡)
𝑃𝑜𝑖𝑛𝑡 2: −3, −2,29.5 (𝑚𝑎𝑥𝑖𝑚𝑢𝑚)
Example 4: determine the nature of any stationary
points for the function
𝑧 = 𝑥2𝑒𝑥 + 𝑦2 − 𝑒0.5𝑥𝑦
(𝑒 is the number, not a variable).
Plan: use the FOC to find any stationary points:
find both partial derivatives, set them equal to
zero and solve using simultaneous equations.
The 𝑒 rule will have to be used.
To find the nature of these: find the straight
second order partial derivatives and evaluate them
at the points found using the FOC. Determine if
any of the stationary points are possible
maxima/minima. Use the delta test to make sure
of their nature.
Solution: the partial derivatives are:
𝜕𝑧
𝜕𝑥= 2𝑥𝑒𝑥 + 𝑥2𝑒𝑥 − 0.5𝑒0.5𝑥𝑦 = 0
𝜕𝑧
𝜕𝑦= 2𝑦 − 𝑒0.5𝑥 = 0
Solve these simultaneously:
′ : 𝑦 =𝑒0.5𝑥
2
Substitute ′ → :
2𝑥𝑒𝑥 + 𝑥2𝑒𝑥 − 0.5𝑒0.5𝑥 𝑒0.5𝑥
2 = 0
Simplify:
2𝑥𝑒𝑥 + 𝑥2𝑒𝑥 − 0.25𝑒𝑥 = 0
Factorise:
𝑒𝑥 2𝑥 + 𝑥2 − 0.25 = 0
For the left side to be equal to zero, either:
1. 𝑒𝑥 = 0 which is impossible, or
2. 𝑥2 + 2𝑥 − 0.25 = 0
Solve using quadratics:
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−2 ± 22 − 4 1 (−0.25)
2(1)
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 =−2 ± 5
2
𝑥1 = −1 − 5/2 𝑜𝑟 𝑥2 = −1 + 5/2
This gives two stationary points at the two
𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 above with approximations
𝑥1 ≈ −2.118 and 𝑥2 ≈ 0.118. But each of these
points also needs a 𝑦 − 𝑣𝑎𝑙𝑢𝑒 which is found from
either of the simultaneous equations (′ is easiest
as 𝑦 is already isolated):
𝑦 𝑥1 = −1 − 5
2 =
𝑒0.5 −1−
52
2≈ 0.1734
𝑦 𝑥2 = −1 + 5
2 =
𝑒0.5 −1+
52
2≈ 0.5304
So the two stationary points are:
Point 1: 𝑥 = −1 − 5/2, 𝑦 =𝑒
0.5 −1− 52
2
Point 2: 𝑥 = −1 + 5
2, 𝑦 =
𝑒0.5 −1+
52
2
To obtain the 𝑧 − 𝑣𝑎𝑙𝑢𝑒𝑠, substitute each of these
pairs of 𝑥 and 𝑦 − 𝑣𝑎𝑙𝑢𝑒𝑠 into the original
equation and simplify. Do it and you should get:
𝑧1
−1 − 5
2,𝑒
0.5 −1− 52
2
= 𝑒−1− 52 2 + 5
𝑧2
−1 + 5
2,𝑒
0.5 −1+ 52
2
= 𝑒−1+ 52 2 − 5
To find the nature of these points, use the SOC:
𝜕2𝑧
𝜕𝑥2= 2𝑒𝑥 + 2𝑥𝑒𝑥 + 2𝑥𝑒𝑥 + 𝑥2𝑒𝑥 − 0.25𝑒0.5𝑥𝑦
𝜕2𝑧
𝜕𝑦2= 2
155
Evaluate each of the stationary points at each of
the two partial derivatives (approximate values are
used as it is very difficult to work with the exact
values; the reason will be clear later):
Point 1: 𝑥1 ≈ −2.118, 𝑦1 ≈ 0.1734
𝜕2𝑧
𝜕𝑥2≈ 2𝑒−2.118 + 2 −2.118 𝑒−2.118
+ 2 −2.118 𝑒−2.118
+ −2.118 2𝑒−2.118
− 0.25𝑒0.5 −2.118 0.1734
≈ −0.254 < 0
𝜕2𝑧
𝜕𝑦2= 2 > 0
The signs do not match up so this is a saddle point.
Point 2: (𝑥2 ≈ 0.118, 𝑦2 ≈ 0.5304)
𝜕2𝑧
𝜕𝑥2≈ 2𝑒0.118 + 2 0.118 𝑒0.118 + 2 0.118 𝑒0.118
+ 0.118 2𝑒0.118 − 0.25𝑒0.5 0.118 0.5304
≈ 2.657 > 0
𝜕2𝑧
𝜕𝑦2= 2 > 0
Since the signs match up and both are greater than
zero, this point is a possible minimum.
To prove that this point is a minimum, the delta
test needs to be evaluated at this point. The mixed
partial derivative is:
𝜕2𝑧
𝜕𝑥𝜕𝑦= −0.5𝑒0.5𝑥
Evaluated at Point 2 (𝑥2 ≈ 0.118, 𝑦2 ≈ 0.5304):
𝜕2𝑧
𝜕𝑥𝜕𝑦≈ −0.5𝑒0.5(0.118) ≈ −0.5304
Apply the delta test:
∆= 𝜕2𝑧
𝜕𝑥2 𝜕2𝑧
𝜕𝑦2 − 𝜕2𝑧
𝜕𝑥𝜕𝑦
2
∆≈ 2.657 2 − −0.5304 2 ≈ 5.033 > 0
This proves what was suspected from the SOC.
Since the SOC gave a possible minimum, the delta
test proves this is true. Thus:
Saddle point: −1 − 5
2,𝑒
0.5 −1− 52
2, 𝑒−1−
5
2 2 + 5
Minimum: −1 + 5
2,𝑒
0.5 −1+ 52
2, 𝑒−1+
5
2 2 − 5
Note: the only reason approximations were used
(to at least three decimal places!) when evaluating
the SOC and the delta test was that the sign of the
value was needed, not the precise value at that
point. If the approximation gives a SOC result quite
a bit above/below zero, then that
positive/negative sign can be taken with certainty.
However, if the SOC was close to zero (around
0.01), then more decimal places or the exact
answer would have to be used to be 100% sure
that the SOC is really positive/negative.
This was a very difficult example, but look over it
again and write out what is being found and why in
each of the steps before trying the following
exercises.
Exercises: 1. Determine if the following functions have any
stationary points, and determine their nature. 𝑎) 𝑧 = 4𝑥𝑦 − 3𝑥2 − 5𝑦2 + 7 𝑏) 𝑧 = 4𝑥 + 5𝑦 − 3𝑥2 − 2𝑦2 𝑐) 𝑧 = 𝑥3 + 𝑦3 + 18 𝑑) 𝑓 𝑥, 𝑦 = 2.5𝑥 + 3.5𝑦 − 2.2𝑥2 − 1.8𝑦2 𝑒) 𝑧 = −4𝑥4 − 5𝑦4 + 32𝑥 + 40𝑦 𝑓) 𝑓 𝑥, 𝑦 = 3𝑥𝑒𝑥+𝑦 + 2𝑦 + 𝑥
2. Given the profit function: 𝜋 = 𝐾0.4𝐿0.5 − 0.3𝐾 − 0.4𝐿 Determine any stationary points, and if there is a maximum.
7.8 economic applications
Theory: the total revenue from two different
goods is simply the sum of the revenues of the
individual goods.
Example 1: a direct computer
manufacturer/retailer sells two different types of
156
computers; laptops (𝐿) and desktops (𝐷). They sell
these computers at the prices:
𝑃𝐿 = $1197.00
𝑃𝐷 = $864.00
The company pays $560 in rent and face a variable
cost function:
𝑉𝐶 = 0.3𝐿2 + 0.45𝐷2 + 0.3𝐿𝐷
Determine:
a) The profit function
b) The optimal levels of 𝐿 and 𝐷 (prove this).
c) The maximum profit possible.
Plan: find the 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 function using
𝑇𝐶 = 𝑉𝐶 + 𝐹𝐶
Then find the 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 function (which just
comes from 𝑃 × 𝑄 for each of the two computers).
Determine the profit function using:
𝜋 = 𝑇𝑅 − 𝑇𝐶
Find any stationary points using the FOC then
prove their nature using the SOC and the delta
test. Substitute the (𝐿, 𝐷) combination back into
the profit function to determine the maximum
profit.
Solution: the 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 function is:
𝑇𝐶 = 0.3𝐿2 + 0.45𝐷2 + 0.3𝐿𝐷 + 560
The 𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 function is the sum of 𝑃 × 𝑄
for each of the two goods:
𝑇𝑅 = 1197𝐿 + 863.10𝐷
Thus the profit function is:
𝜋 = 1197𝐿 + 864𝐷 − 0.3𝐿2 − 0.45𝐷2 − 0.3𝐿𝐷 − 560
To find stationary points, use the FOC:
𝜋𝐿 = 1197 − 0.6𝐿 − 0.3𝐷 = 0
𝜋𝐷 = 864 − 0.9𝐷 − 0.3𝐿 = 0
′ : 0.3𝐷 = 1197 − 0.6𝐿
′′ : 𝐷 = 3990 − 2𝐿
Substitute ′′ → :
864 − 0.9 3990 − 2𝐿 − 0.3𝐿 = 0
864 − 3591 + 1.8𝐿 − 0.3𝐿 = 0
𝐿 = 1818
Substitute this into ′′ to give:
𝐷 = 3990 − 2 1818 = 354
To determine the nature of this stationary point,
the SOC must be used:
𝜋𝐿𝐿 = −0.6 < 0
𝜋𝐾𝐾 = −0.9 < 0
Both straight second order partial derivatives are
negative so this is a possible maximum. To prove it,
the delta test must be used:
𝜋𝐿𝐾 = −0.3
∆= 𝜋𝐿𝐿𝜋𝐾𝐾 − 𝜋𝐿𝐾 2
∆= −0.6 −0.9 − −0.3 2
∆= 0.54 − 0.09 = 0.45 > 0
Since the delta test is positive, it confirms what
was suspected from the SOC (i.e. the point is a
maximum).
To determine the actual profit attainable,
substitute 𝐿 = 1458 and 𝐷 = 1074 into the profit
function:
𝜋 1818,354 = 1197 1818 + 864 354
− 0.3 1818 2 − 0.45 354 2
− 0.3 1818 354 − 560
𝜋 1818,354 = 1,240,441
Thus to attain a maximum profit of approximately
$1.24million, output should be set at 1818 laptops
and 354 desktops.
Example 2: a company faces a demand function of
the form
𝑃𝑑 =1000
𝑄𝑑+ 20
and a production function of the form:
𝑄 = −𝐿2 − 2𝐾2 + 𝐿𝐾 + 5𝐾 + 10𝐿
If each labour unit 𝐿 costs $20 and each capital
unit 𝐾 costs $30, and fixed costs are $600,
determine the profit maximising level of capital
and labour, and also find the maximum profit.
157
Plan: find the 𝑇𝐶 function using the fixed costs and
variable costs. Then find the 𝑇𝑅 function, using the
demand function and the production function.
Use 𝜋 = 𝑇𝑅 − 𝑇𝐶 to determine the maximum
profit levels of 𝐿 and 𝐾, then substitute back to get
the maximum profit.
Solution: to find total revenue:
𝑇𝑅 = 𝑃 ∙ 𝑄 = 1000
𝑄+ 20 ∙ 𝑄
𝑇𝑅 = 1000 + 20𝑄
To find 𝑇𝑅 in terms of labour and capital,
substitute in for 𝑄:
𝑇𝑅 = 1000 + 20 −𝐿2 − 2𝐾2 + 𝐿𝐾 + 5𝐾 + 10𝐿
𝑇𝑅 = 1000 − 20𝐿2 − 40𝐾2 + 20𝐿𝐾 + 100𝐾 + 200𝐿
Total costs come from the cost per unit costs and
the fixed costs:
𝑇𝐶 = 20𝐿 + 30𝐾 + 600
The profit function is:
𝜋 = 𝑇𝑅 − 𝑇𝐶
𝜋 = 1000 − 20𝐿2 − 40𝐾2 + 20𝐿𝐾 + 100𝐾 + 200𝐿
− 20𝐿 + 30𝐾 + 600
𝜋 = 400 − 20𝐿2 − 40𝐾2 + 20𝐿𝐾 + 70𝐾 + 180𝐿
To find any stationary point, use the FOC:
𝜋𝐿 = −40𝐿 + 20𝐾 + 180 = 0
𝜋𝐾 = −80𝐾 + 20𝐿 + 70 = 0
Rearrange 𝜋𝐿 to isolate 𝐾:
𝐾 =40𝐿 − 180
20= 2𝐿 − 9
Substitute into 𝜋𝐾:
−80 2𝐿 − 9 + 20𝐿 + 70 = 0
−160𝐿 + 720 + 20𝐿 + 70 = 0
𝐿 =79
14
Substitute this into the isolated 𝐾 equation:
𝐾 = 2𝐿 − 9 = 2 79
14 − 9 =
16
7
Thus there is a stationary point at
𝐾 =16
7 , 𝐿 =
79
14
Use the SOC to determine the nature of this
stationary point:
𝜋𝐿𝐿 = −40 < 0
𝜋𝐾𝐾 = −80 < 0
Since both are less than zero, this stationary point
is a possible maximum.
To make sure it is a maximum, use the delta test.
The mixed second order partial derivative:
𝜋𝐿𝐾 = 20
∆= 𝜋𝐿𝐿𝜋𝐾𝐾 − 𝜋𝐿𝐾 2
∆= −40 −80 − 20 2 = 2800 > 0
This proves that what was suspected from the SOC
is actually correct. The SOC gave a possible
maximum, and the delta test confirmed this.
To find the profit at 𝐿 =79
14 and 𝐾 =
16
7 substitute
into the profit equation:
𝜋 = 400 − 20𝐿2 − 40𝐾2 + 20𝐿𝐾 + 70𝐾 + 180𝐿
𝜋 = 400 − 20 79
14
2
− 40 16
7
2
+ 20 79
14
16
7
+ 70 16
7 + 180
79
14
𝜋 = 9876
7≈ 987.86 2𝑑. 𝑝.
Exercises: 1. A farmer in a perfectly competitive industry grows
wheat and barley. If the rent on the farm is $4,000 per year, and the prices for the two crops are:
𝑃𝑊 = $225/𝑡𝑜𝑛 𝑃𝐵 = $190/𝑡𝑜𝑛
With variable costs being approximated by: 𝑉𝐶 = 0.5𝑊2 + 0.4𝐵2 + 0.6𝑊𝐵
Determine: a) The profit function b) The optimal levels of production c) The maximum profit attainable.
2. A shipyard uses two inputs, 𝐿 and 𝐾. If the demand for their services are approximated by
𝑃𝐷 =2500
0.5𝑄𝐷
+ 50
And the firm has a service production function of the form
𝑄 = −0.5𝐿2 − 2𝐾2 + 2𝐿𝐾 + 8𝐾 + 12𝐿 If each 𝐿 costs $25 and each 𝐾 costs $40, determine: a) The profit function in terms of 𝐿 and 𝐾. b) The optimal levels of 𝐿 and 𝐾. c) The maximum profit attainable.
158
3. A firm imports two types of televisions from Malaysia [plasma (𝑀) and LED (𝐿)] at a cost of:
𝑃𝑀;𝑖𝑚𝑝𝑜𝑟𝑡 = $300
𝑃𝐿;𝑖𝑚𝑝𝑜𝑟𝑡 = $500
If the domestic demand functions for each of the two types is approximated by:
𝑃𝑀 = 𝑀 −2𝐿2
𝑀+ 𝐿 + 1300
𝑃𝐿 = 𝐿 −5𝑀2
𝐿+ 2𝑀 + 1000
If the cost to rent the shop is $500,000, determine: a) The two total revenue functions b) The total cost function c) The profit function d) The optimal sales of the two types of
televisions. e) The two domestic prices of the televisions. f) The maximum profit attainable.
chapter seven summary
Multivariable functions means that a dependent variable is determined by two or more other variables. 𝑧 = 𝑓(𝑥, 𝑦) 𝑜𝑟 𝑧(𝑥, 𝑦)
The Cobb-Douglas Production Function has the general form:
𝑄 = 𝐴𝐿𝛼𝐾𝛽 where the values of 𝐴, 𝛼 and 𝛽 are all positive constants.
Partial differentiation is finding the rate at which the dependent variable changes when an independent variable changes assuming all other variables are held constant. 𝜕𝑧
𝜕𝑥 ,
𝜕𝑧
𝜕𝑦 𝑂𝑅 𝑧𝑥 , 𝑧𝑦
To partially differentiate a multivariable function with respect to a single variable, differentiate the function with respect to that variable while treating all other variables as constants.
The second order partial derivative is simply partially differentiating the original function twice.
𝜕2𝑧
𝜕𝑥2 ,
𝜕2𝑧
𝜕𝑦2 𝑂𝑅 𝑧𝑥𝑥 , 𝑧𝑦𝑦
Mixed partial derivatives are found by differentiating with respect to one variable first, then differentiating with respect to the other variable.
𝜕2𝑧
𝜕𝑥𝜕𝑦 ,
𝜕2𝑧
𝜕𝑦𝜕𝑥 𝑂𝑅 𝑧𝑥𝑦 , 𝑧𝑦𝑥
Mixed partial derivatives are the same, no matter which way they are found.
The higher the value of the marginal product, the greater the increase in output from an extra unit of that input.
Returns to scale are defined as the extent output changes when inputs are changed by a certain amount. For the Cobb-Douglas Production Function
𝑄 = 𝐴𝐿𝛼𝐾𝛽 1. 𝛼 + 𝛽 < 1 then there are decreasing returns to
scale. 2. 𝛼 + 𝛽 = 1 then there are constant returns to scale.
3. 𝛼 + 𝛽 > 1 then there are increasing returns to scale.
The total derivative equation is:
𝑑𝑧 =𝜕𝑧
𝜕𝑥𝑑𝑥 +
𝜕𝑧
𝜕𝑦𝑑𝑦
To find the actual change in a function: ∆𝑧 = 𝑧 𝑥𝑁 , 𝑦𝑁 − 𝑧 𝑥𝑂 , 𝑦𝑂
To find an optimal point: 1. The First Order Condition (FOC) is:
𝜕𝑧
𝜕𝑥= 0 𝐴𝑁𝐷
𝜕𝑧
𝜕𝑦= 0
Then solve for 𝑥 and 𝑦 using simultaneous equations, to find any stationary points. A saddle point is a stationary point that is neither an overall maximum nor an overall minimum. It looks like a saddle. 2. To find out whether a stationary point is a maximum, a minimum or a saddle point, the Second Order Condition (SOC) is used:
a) The point is a possible maximum if: 𝜕2𝑧
𝜕𝑥2< 0 𝐴𝑁𝐷
𝜕2𝑧
𝜕𝑦2< 0
b) The point is a possible minimum if: 𝜕2𝑧
𝜕𝑥2> 0 𝐴𝑁𝐷
𝜕2𝑧
𝜕𝑦2> 0
c) The point is a saddle point if the signs of the two straight first order partial derivatives do not match, or if one is equal to zero.
Use the delta test to make sure the point is what is suspected:
∆= 𝜕2𝑧
𝜕𝑥2
𝜕2𝑧
𝜕𝑦2 −
𝜕2𝑧
𝜕𝑥𝜕𝑦
2
𝐼𝑓 ∆ > 0, the point is what was suspected. 𝐼𝑓 ∆ < 0, ignore everything, the point is a saddle.
The total revenue from two different goods is simply the sum of the revenues of the individual goods.
chapter seven questions
1. Find both first order partial derivatives for: 𝑎) 𝑧 = 2𝑥𝑦 + 3𝑥 − 𝑦 + 4𝑥2𝑦3
𝑏) 𝑓 𝑥, 𝑦 = 12𝑥4 + 𝑥𝑒𝑦 − 12
𝑐) 𝑝 = 15𝑟0.3𝑚0.7 + 15𝑚𝑟
𝑑) 𝑧 = 18𝑥0.4𝑦0.8𝑥2 𝑦4 − 3𝑥𝑦 + 𝑥2
𝑒) 𝑧 = 𝑥 − 𝑦 𝑥 + 𝑦 4
𝑓) 𝑓 𝑥, 𝑦 = 3𝑥 − 4𝑦2 5 𝑥𝑦
𝑔) 𝑧 =𝑥2 − 𝑦3
1 − 𝑥 − 𝑦
) 𝑧 =12𝑥𝑦2 − 𝑦
1 − 𝑥𝑦
𝑖) 𝑧 = ln 1 − 𝑥2𝑦2 + 𝑥
159
𝑗) 𝑧 = 𝑒4𝑥−𝑦 𝑥3 + 4𝑥𝑦 − 3
𝑘) 𝑧 = 𝑥2𝑦2 − 1 𝑒4𝑥𝑦
𝑙) 𝑧 =9𝑥𝑦
𝑥𝑦 − 1+ ln 15 − 𝑥 + 15 ln 4 − 2
𝑚) 𝑧 =1 − ln 1 + 𝑥𝑦
1 − 𝑒𝑥𝑦
2. Find all four second order partial derivatives for: 𝑎) 𝑧 = 𝑥2 + 𝑦2 − 3𝑥𝑦 − 𝑥 + 4𝑦 + 12
𝑏) 𝑧 = 2𝑥𝑦2 + 𝑦𝑥2 − 3
𝑐) 𝑧 = 𝑥3𝑦4 + 2𝑥𝑦 − 4𝑥 − 6𝑦 + 8
𝑑) 𝑓 𝑥, 𝑦 = 2𝑥𝑦2 4 + 18𝑥𝑦2 − 13𝑥𝑦
𝑒) 𝑧 = 𝑒2𝑥2+𝑦
𝑓) 𝑧 = 𝑒𝑥2+𝑦2− 2𝑥𝑦2
𝑔) 𝑓 𝑥, 𝑦 = ln 2𝑥2 + 𝑦3
) 𝑧 = ln 13 − 𝑥𝑦 + 𝑥2 + 15 𝑥 − 𝑦 2
3. For the following functions, find any stationary points and determine their nature: 𝑎) 𝑧 = 2.2𝑥𝑦 − 3.3𝑥2 − 5.5𝑦2 + 1
𝑏) 𝑧 = 2𝑥 + 2𝑦 − 3𝑥2 − 3𝑦2 + 2𝑥𝑦
𝑐) 𝑧 = 8𝑥3 + 3𝑦3 − 𝑥𝑦 + 500
𝑑) 𝑓 𝑥, 𝑦 = 1.2𝑥 + 2.4𝑦 − 2.4𝑥2 − 1.8𝑦2 + 1.2𝑥𝑦
𝑒) 𝑧 = −2𝑥4 − 3𝑦4 + 10𝑥 + 20𝑦
𝑓) 𝑓 𝑥, 𝑦 = 5𝑦𝑒𝑥−𝑦 + 3𝑦 − 8𝑥
4. A pharmaceutical company uses capital equipment and labour as inputs into the manufacture of paracetamol. If the company has a demand function for the final product as:
𝑃𝑑 =1
𝑄+ 20
The firm also has a production function of the form: 𝑄 = 𝐾0.5𝐿0.4
And the cost function is approximated by: 𝑇𝐶 = 5𝐾 + 10𝐿 + 20
Determine in terms of 𝐾 and 𝐿: a) The total revenue function b) The profit function c) When profit is at a maximum. d) The price at the maximum (if the maximum
exists). e) The quantity sold at the maximum.
5. Given the profit function: 𝜋 = 𝐾0.4𝐿0.5 − 0.3𝐾 − 0.4𝐿
Determine any stationary points and their nature. Is maximum profit a possibility?
6. Use the total derivative function to find the approximate change in 𝑧 for the function
𝜋 = 𝐾0.7𝐿0.4 + 0.02𝐾𝐿 − 0.3𝐿 − 0.4𝐾 when 𝐿 is increased by 0.2 from 154 and 𝐾 is increased by 0.1 from 310. Also determine the exact change in 𝜋, and compare the two answers.
7. Approximate the change in the objective function 𝑅 𝑅 𝑔, 𝑘 = 0.1𝑔2𝑘 + 0.3𝑘2𝑔 − 4𝑔𝑘 + 15
when 𝑔 is decreased by 0.5 and 𝑘 is increased by 0.9, given the original levels of 𝑔 and 𝑘 being 78 and 104 respectively. Also determine the exact change in 𝑅.
8. For the production function
𝑄 = 8𝐾0.4𝐿0.5 Find a) an expression for the approximate change in
output if 𝐿 is decreased by 0.4% and 𝐾 is increased by 0.2% (i.e. only in terms of 𝐿 and 𝐾).
b) the exact change in output, given 𝐿 was originally 200 and 𝐾 was 180.
9. A profit function has the form 𝜋 = 4𝐾 + 7𝐿 − 12𝐾2 − 14𝐿2 − 0.01
Determine the level of utilisation of the two inputs at maximum profit, and also the maximum profit.
10. A firm imports two goods from China, televisions (𝑇) and radios (𝑅), and sells them on the domestic market. The costs to the firm are related to the amount they import, and the approximate cost function is
𝑇𝐶 = 3𝑇2 + 4𝑅2 + 400 + 𝑅𝑇 The two goods sell on the domestic market at the following prices:
𝑃𝑇 = 300 𝑃𝑅 = 150
Determine: a) The total revenue function. b) The profit function. c) The levels of 𝑇 and 𝑅 for profit maximisation. d) The maximum profit attainable.
11. A factory producing portable music players has a demand function of the form
𝑃𝐷 =1100
𝑄𝐷
+ 45
with a production function of the form: 𝑄 = −2𝐿2 − 3𝐾2 + 2.5𝐿𝐾 + 8𝐿 + 12𝐾
If labour costs $50 and capital costs $35, and if fixed costs are $550, determine: a) The total revenue function in terms of 𝐿 and 𝐾. b) The total cost function in terms of 𝐿 and 𝐾. c) The profit function. d) The levels of 𝐿 and 𝐾 for maximum profit. e) The maximum profit attainable.
12. A firm imports two similar products from Brazil at fixed costs: refrigerators (𝑅) cost $650 and air conditioners (𝐴) cost $350. Given the domestic demand for refrigerators is:
𝑃𝑅 = 𝑅 −4𝐴2
𝑅+ 𝐴 + 1400
And the domestic demand for air conditioners is:
𝑃𝐴 = 𝐴 −6𝑅2
𝐴+ 3𝑅 + 1200
With fixed costs amounting to $10,000, determine: a) The two revenue functions. b) The total cost function. c) The profit function. d) The number of 𝑅 and 𝐴 that need to be sold to
maximise profit. e) The domestic prices of the two goods. f) The maximum attainable profit. g) Do these prices make sense?
160
Chapter 8
Financial Mathematics Mathematics and money
8.1 Index Numbers and Averages 161
8.2 Series and Sums 163
8.3 Simple Interest 167
8.4 Compound Interest 168
8.5 Annual Interest Rates 170
8.6 Net Present Value 172
8.7 Internal Rate of Return 174
Chapter Eight Summary 177
Chapter Eight Questions 177
161
8.1 index numbers and averages
In this context, index numbers are not related to
indices learnt in Chapter 1.
Theory: in a finance sense, index numbers
represent values in terms of a base value.
For example, the Consumer Price Index provides a
value of the price level of a basket of goods in a
certain year in terms of that same basket of goods
in a given base year (e.g. the year 2000).
Theory: to find the indexed value of a given
commodity, use the formula:
𝐼𝑛𝑑𝑒𝑥𝑡 =𝑉𝑎𝑙𝑢𝑒𝑡
𝑉𝑎𝑙𝑢𝑒0× 100
This is very similar to finding a percentage, except
that it is always to a base year.
The base year is given an index value of 100.
Example 1: an index is to be constructed with the
year 2000 as the base year (with a value of 100). If
gold prices in 2000 were $731.53 and in 2009
were $1091.20, determine the value of the index
in 2009.
Plan: use the index formula
𝐼𝑡 =𝑃𝑡
𝑃0× 100
Solution: substitute all the known values:
𝐼2009 =1091.20
731.53× 100
𝐼2009 = 149.17
This is not a difficult idea. However, in finance,
usually you will only have access to an index and
not the raw data. You will have to know how to
manipulate an index to extract useful information.
Theory: to obtain the percentage difference
between two time periods in an index, use the
formula:
%∆𝑡→𝑡+1=𝐼𝑡+1 − 𝐼𝑡
𝐼𝑡× 100
You might be thinking why not just take the two
indices from one another as they are already in
percentage form. This is correct but the difference
between two index entries gives the percentage
difference between the two years in terms of the
base year, which in most cases, is useless.
Example 2: given the following data, determine
the percentage change in crude oil prices between
2002 and 2003
Plan: use the percentage change formula
%∆2002→2003 =𝐼2003 − 𝐼2002
𝐼2002× 100
Solution: substitute the index values:
%∆2002→2003 =119.2 − 112.3
112.3× 100
%∆2002→2003 = 6.14%
If the difference between the two years was taken,
the value would be 6.9%, but this is in terms of
the price in 2000, which is irrelevant.
Example 3: given the following Consumer Price
Index values, determine the percentage change in
prices between:
a) 2005 and 2006
b) 2008 and 2009
Plan: use the percentage change formula
%∆𝑡→𝑡+1=𝐼𝑡+1 − 𝐼𝑡
𝐼𝑡× 100
𝑌𝑒𝑎𝑟 𝐶𝑃𝐼 2005 149.1 2006 154.4 2007 157.9 2008 164.8 2009 168.1
𝑌𝑒𝑎𝑟 𝐶𝑟𝑢𝑑𝑒 𝑂𝑖𝑙 𝐼𝑛𝑑𝑒𝑥 2000 100 2001 105.6 2002 112.3 2003 119.2
162
Solution:
a) Substitute 𝐼𝑡+1 = 154.4 and 𝐼𝑡 = 149.1
%∆2005→2006 =154.4 − 149.1
149.1× 100
= 3.55% 2𝑑. 𝑝.
b) Substitute 𝐼𝑡+1 = 168.1 and 𝐼𝑡 = 164.8
%∆2008→2009=168.1 − 164.8
164.8× 100
= 2.00% 2𝑑. 𝑝.
Example 4: determine the percentage change in
the growth of gold prices from 2005 to 2009, given
the following index entries for the two years:
𝐼2005 = 123.2
𝐼2009 = 141.3
Plan: use the percentage change formula:
%∆𝑡→𝑡+𝑛 =𝐼𝑡+𝑛 − 𝐼𝑡
𝐼𝑡× 100
Solution: substitute 𝐼𝑡+𝑛 = 141.3 and 𝐼𝑡 = 123.2,
and solve:
%∆=141.3 − 123.2
123.2× 100 = 14.69% 2𝑑. 𝑝.
In the example above, there was growth of
14.69% over the four years from 2005 to 2009.
The average growth rate in gold prices over these
four years can be found in one of two ways.
Theory: the arithmetic average is the traditional
method of finding an average:
𝐴𝐴 =𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠
The geometric average is a concept similar to
compound interest (see later in Chapter). It
assumes that the percentage change is
compounded, thus to reverse the compounding:
𝐺𝐴 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒
100+ 1
𝑛
− 1
With 𝑛 being the number of years. The geometric
average is more realistic as it assumes growth is
relative to the most recent year, rather than the
first year.
Example 5: Find the arithmetic and geometric
average percentage change per year in the CPI
given the percentage change over four years was
21%.
Plan: apply the two formulae for averages.
Solution: for the arithmetic average:
𝐴𝑟𝑖𝑡𝑚𝑒𝑡𝑖𝑐 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =21
4= 5.25%
For the geometric average:
𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 21
100+ 1
4
− 1
= 1.048809 − 1 = 0.048809
Thus a geometric average of about 4.88% p.a.
The concept of the arithmetic average in this last
example is the simplest kind of average, so will not
be explained in depth. The concept of the
geometric average in this last example, however,
will be. In the first year, the CPI grew by 4.88%
from some base amount, say 100. So after one
year, it is 104.88. During the second year, the CPI
grows another 4.88%, however, it is now growing
from a base of 104.88, and this gives, at the end of
the second year, a value of 110. During the third
year, the CPI grows again at 4.88% relative to the
previous year (i.e. 1.10), so at the end of the third
year, it has a value of 115.4. At the end of the
fourth year, the value of the CPI is 121, or 21%
higher over the four years. The geometric average
is similar to compound interest, but in reverse.
Compare the geometric average to the arithmetic
average, where the arithmetic average is finding
only the average percentage relative to the first
year.
163
Example 6: the consumer price index table is
shown below with some entries missing (which is
common in data sets).
Determine
a) the percentage change in the CPI from 1994 to
1999.
b) The arithmetic average percentage change per
annum from 1994 to 1999.
c) The geometric average percentage change per
annum from 1994 to 1999.
Plan: use the percentage change formula:
%∆𝑡→𝑡+1=𝐼𝑡+1 − 𝐼𝑡
𝐼𝑡× 100
Then use the arithmetic average formula:
𝐴𝐴 =𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠
then the geometric average formula:
𝐺𝐴 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒
100+ 1
𝑛
− 1
Solution: for the percentage change:
%∆=122.9 − 111.6
111.6× 100 = 10.13% 2𝑑. 𝑝.
For the arithmetic average over the five years:
𝐴𝐴 =10.13%
5= 2.03%
For the geometric average over the five years:
𝐺𝐴 = 10.13
100+ 1
5
− 1
= 0.0195 4𝑑. 𝑝.
Thus a percentage growth per year of
approximately 1.95%.
Exercises: 1. Given the following data table:
𝑌𝑒𝑎𝑟 𝐶𝑃𝐼 1960 13.8 1961 14.1 1962 − 1963 − 1964 14.5 1965 15.1 1966 15.5 1967 − 1968 16.7 1969 17.1
Determine: a) The percentage change from 1960 to 1961. b) The percentage change from 1961 to 1964. c) The percentage change from 1960 to 1969. d) The percentage change from 1966 to 1968. e) The arithmetic average from 1961 to 1964. f) The geometric average from 1961 to 1964. g) The arithmetic average from 1960 to 1969. h) The geometric average from 1960 to 1969.
8.2 series and sums
Series are extremely common in business
(especially accounting and investments) and
simply put, they are the repetition of a certain
mathematical function.
Theory: a series is list of numbers with a constant
pattern. Each individual entry of a series is
important but so is the sum of the series.
Each of the following is a series; try find the
pattern.
2, 4, 6, 8, 10, 12
3, 10, 17, 24, 31, 38, 45
2, 4, 8, 16, 32, 64
100, 50, 25, 12.5, 6.75
Theory:
An arithmetic series has a constant difference
between terms.
A geometric series has a constant ratio
between terms.
Looking at each of the following arithmetic
progressions, you will see a common feature:
𝟓, 11, 17, 23, 29, 35, 41 +6
𝑌𝑒𝑎𝑟 𝐶𝑃𝐼 1993 109.5 1994 111.6 1995 − 1996 119.8 1997 − 1998 121.1 1999 122.9
164
𝟏, 1.5, 2, 2.5, 3, 3.5, 4 +0.5
𝟓𝟎𝟎, 490, 480, 470 (−10)
Each of these sequences has an initial number
(bolded) and also a constant addition/subtraction
between terms (in brackets).
Theory: the 𝑛𝑡 term of a sequence is given by the
formula:
𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑
Where 𝑎 is the initial number, 𝑛 the term you want
to find, and 𝑑 the difference between terms in a
sequence.
Example 1: find the 6𝑡 term of the following
arithmetic progression:
8, 17, 26, 35 …
Plan: determine the first term and the difference
between each term, then apply the formula
𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑
Solution: the initial number is 𝑎 = 8, the
difference between all the numbers is:
𝑑 = 9, and the 6𝑡 term is 𝑛 = 6. Substitute into
the equation:
𝑇6 = 8 + 6 − 1 ∙ 9
= 8 + 5 ∙ 9 = 53
Add two more 9’s to 35 to see if it is correct.
Example 2: find the 27𝑡 term of the following
progression:
3, 10.5, 18, 25.5, 33 …
Plan: determine the first term and the difference
between terms, then apply the formula:
𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑
Solution: the first term is 𝑎 = 3, the difference is:
𝑑 = 7.5, and 𝑛 = 27. Substitute this into the
formula:
𝑇27 = 3 + 27 − 1 ∙ 7.5 = 198
Example 3: find the 13𝑡 term of the sequence:
100, 97, 94, 91, 88 …
Solution: the first term is 𝑎 = 100, the difference
is:
So 𝑟 = −3, and 𝑛 = 13. Substituting this into the
equation:
𝑇13 = 100 + 13 − 1 ∙ −3 = 64
Summing a sequence is important in many
business disciplines. For example, the revenue of a
company might grow at a certain amount each
month, so to find out how much revenue the
company will earn over the whole year requires
summing a series.
Theory: the sum of an arithmetic series is found
using the formula:
𝑆𝑛 =𝑛
2 2𝑎 + 𝑛 − 1 𝑑
Example 4: find the sum of the sequence:
4, 9, 14, 19, 24, 29
Plan: determine the values of 𝑎 and 𝑑, then
substitute into the formula:
𝑆𝑛 =𝑛
2 2𝑎 + 𝑛 − 1 𝑑
Solution: the first term is 𝑎 = 4, the difference is
+5 (make sure of this) and there are 𝑛 = 6 terms.
Substitute into the formula:
𝑆6 = 6
2 2 ∙ 4 + 5 ∙ 5 = 99
Use your calculator to actually sum up the
sequence and makes sure 99 is correct.
Example 5: find the sum of the first 15 terms of
the sequence
100, 97, 94, 91, 88 …
−3 −3 −3 −3
3, 10.5, 18, 25.5, 33 …
7.5 7.5 7.5 7.5
8, 17, 26, 35 …
9 9 9
165
200,179,158,137, …
Plan: determine the values of 𝑎, 𝑑 and 𝑛 then
substitute into the summation formula.
Solution: 𝑎 = 200, 𝑑 = −21 and 𝑛 = 15.
Substitute into the summation formula:
𝑆15 =15
2 400 + 14 −21 = 795
Example 6: Bentley, a luxury car manufacturer,
increases production by ten cars each year,
relative to the previous year. If Bentley
manufactured 316 cars this year, what is the total
number of cars Bentley will make in two decades
(including this year)?
Solution: 𝑎 = 316, 𝑑 = 10, 𝑡 = 20, applying the
sum formula:
𝑠20 = 20
2 2 ∙ 316 + 19 ∙ 10 = 8220
Moving on to geometric progressions; the
following three progressions are all geometric:
𝟑, 6,12,24,48,96,192,384 (𝑟 = 2)
𝟓, 35,245,1715,12005 (𝑟 = 7)
𝟏𝟎𝟐𝟒, 256,64,16,4,1,0.25 𝑟 = 0.25
Remember that a geometric progression has a
constant ratio between terms. Above, all three
progressions have an initial term (in bold) and a
constant ratio between each term (in brackets).
Theory: the 𝑛𝑡 term of a geometric progression is
found using the formula:
𝑇𝑛 = 𝑎𝑟𝑛−1
Where 𝑎 is the first term, 𝑟 the ratio between
terms and 𝑛 the term being found. To find the
ratio, divide a term by the one before it:
𝑟 =𝑇𝑛
𝑇𝑛−1
Example 6: find the 8𝑡 term of the geometric
progression:
2, 6, 18, 54, 162
Plan: determine the values of 𝑎, 𝑟 and 𝑛 then
substitute into
𝑇𝑛 = 𝑎𝑟𝑛−1
Solution: 𝑎 = 2. To find the ratio, divide one term
by the one before it:
162
54=
54
18=
18
6=
6
2= 3
So 𝑟 = 3, and 𝑛 = 8. Substitute this into the
formula:
𝑇8 = 2 ∙ 37 = 4374
Sometimes, the ratio will need to be determined,
but consecutive terms will be unavailable.
Theory: for two terms with 𝑝 terms missing
between them, the ratio is determined by:
𝑟 = 𝑙𝑎𝑡𝑒𝑟 𝑡𝑒𝑟𝑚
𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑡𝑒𝑟𝑚
𝑝+1
Example 7: determine the ratio of the GP:
2, ∎, ∎, 54, …
Plan: use the ratio formula above.
Solution: the later term is 54, and the earlier term
is 2. The two terms are separated by 2 terms, so:
𝑟 = 54
2
3
= 273
= 3
Example 8: find the 12𝑡 term of:
256, ∎, 64, …
Plan: find and substitute the constant ratio 𝑟, the
first term 𝑎 and 𝑛 into the formula:
𝑇𝑛 = 𝑎𝑟𝑛−1
Solution: the constant ratio is:
𝑟 = 64
256
2
= 1
4
2
= 0.5
𝑎 is 256 and 𝑛 is 12:
𝑇12 = 256 0.5 12−1
= 0.125
166
Theory: the sum of a geometric progression is
found by applying the formula:
𝑆𝑛 =𝑎 𝑟𝑛 − 1
𝑟 − 1
Example 9: find the sum of the geometric
progression
4, 8, 16, 32, 64, 128, 256, 512
Plan: find 𝑟, 𝑎 and 𝑛, then apply the sum formula:
𝑆𝑛 =𝑎 𝑟𝑛 − 1
𝑟 − 1
Solution: the constant ratio is:
8
4=
16
8= 2 = 𝑟
And 𝑎 = 4, 𝑛 = 8, so substitute this into the
formula:
𝑆𝑛 =4 2 8 − 1
2 − 1= 1020
Example 10: if sales of computers this year are
400, and sales are expected to increase by 20%
each year for the next 9 years, how many
computers will be sold in the tenth year, and how
many computers will be sold in total over the next
ten years (including the base year).
Plan: find 𝑎, 𝑟 and 𝑛, then apply the geometric
progression term formula, then the sum formula.
Solution: the ratio 𝑟 is 1.2 (as it is a 20% increase
over the previous year), the first term is 𝑎 = 400,
and 𝑛 = 10 (as the base year is included).
Apply the term formula:
𝑇10 = 400 1.2 9 = 2064 𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑢𝑝
Apply the sum formula:
𝑆10 =400 1.2 10 − 1
1.2 − 1
= 10,383 𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑑𝑜𝑤𝑛
Example 11: a financial services firm can invest in
one of two schemes which run for 15 years;
Scheme 1: has an initial return of $1000 which
grows at 5% per year.
Scheme 2: has an initial return of $1100 which
grows at $45 per year.
Determine:
a) Which of these two schemes has a better
overall return over the fifteen years.
b) Which of the schemes has a higher return in
the tenth year.
Plan:
a) determine the sum of both series over the
fifteen years using the two sum formulas.
b) determine the tenth term of both schemes
using the two term formulae.
Solution: since Scheme 1 increases by 5% each
year, the Geometric Progression formula is used.
For scheme 1, 𝑎 = 1000, and 𝑟 = 1.05. Similarly,
since Scheme 2 increases by a constant $45 each
year, the Arithmetic Progression formula is used.
For scheme 2, 𝑎 = 1100 and 𝑑 = 45.
a) Applying the sum formulae for 𝑛 = 15:
Scheme 1:
𝑆15 =1000 1.05 15 − 1
1.05 − 1
= $21,578.56
Scheme 2:
𝑆15 =15
2 2 1100 + 15 − 1 45
= $21,225
Thus Scheme 1 is better overall.
b) Applying the term formulae for 𝑛 = 10:
Scheme 1:
𝑇10 = 1000 1.05 10−1
= $1,551.33
Scheme 2:
𝑇10 = 1100 + 10 − 1 45
= 1,505.00
Thus Scheme 1 is better in the tenth year.
167
Exercises: 1. Given the sequence:
3.1,11.9,20.7,29.5, … Determine: a) The 15𝑡 term.
b) The 25𝑡 term. c) The sum of the first 19 terms.
2. Given the sequence: 10,15,22.5,33.75, …
Determine:
a) The 12𝑡 term. b) The sum of the first 10 terms. c) The sum of the first 20 terms.
3. A firm can invest money into one of two projects each which runs for 12 years and each which has a return of: Project 1: 3500 in the first year, and this increases by 9% every subsequent year. Project 2: 4000 in the first year, and this increase by 300 every subsequent year. Determine a) which project has a larger overall income.
b) Which project has a larger return in the 5𝑡 year.
8.3 simple interest
When working with interest rate calculations,
some basic theory is required.
Theory:
The “principal” is the initial amount of money
invested.
The interest rate is the rate at which this
money increases in value per time period.
In simple interest, the amount of money which the
interest rate applies to does not change, so the
extra money earned from interest each year is
constant. This is similar to arithmetic progressions
in that a constant amount is earned each year.
Theory: the simple interest formula is:
𝐼𝑡 = 𝑃0 ∙ 𝑟 ∙ 𝑡
In words, the interest earned after 𝑡 years from
today (𝐼𝑡) is equal to the principal (𝑃0) times the
per annum interest rate (𝑟) times the number of
years it has been invested for (𝑡).
Example 1: find the interest earned, and also the
total amount of money in the bank when $1000 is
saved for 4 years at 7% p.a.
Plan: use the simple interest formula.
Solution: the principal 𝑃0 = 1000, 𝑡 = 4 and
𝑟 = 0.07. The rate must be in decimal form.
𝐼4 = 1000 0.07 4 = 280
This is the interest earned, but the total amount is
the interest earned plus the principal:
𝑃4 = 𝑃0 + 𝐼4 = 1000 + 280 = 1280
Theory: the total value of an investment is the
principal plus the interest earned over the given
time.
𝑃𝑡 = 𝑃0 + 𝐼𝑡
Example 2: a $4000 savings account is opened
with a bank which offers an interest rate of 9.3%
p.a., with a minimum period of 5 years. Using
simple interest, determine:
a) the interest earned over five years.
b) the total amount available after five years.
Plan: use the simple interest formula to calculate
the interest over the five years. Then add the
principal amount to determine the total amount in
the account.
Solution: to determine the interest earned:
𝐼𝑡 = 𝑃0 ∙ 𝑟 ∙ 𝑡
𝐼5 = 4000 0.093 5 = $1860
Add this amount to the principal to determine the
total amount available:
𝑃5 = 𝑃0 + 𝐼5
𝑃5 = 4000 + 1860 = $5860
Exercises: 1. A bank offers an interest rate of 4.6%, and you
invest $100. Using simple interest, determine the amount of interest you will have earned over 2.5 years, and also the total amount you will have available.
168
8.4 compound interest
Simple interest is not realistic in most situations.
Intro example 1: you win $1000 from a lottery and
decide to save it until you finish university. The
interest rate at which you save it is 20% p.a. How
much will you have at the end of the three years?
Solution: break the three year period into three
shorter periods, each a year in length:
Year 1: at the end of the year, the amount in the
bank will be:
1000 1.20 = 1200
The 1.20 comes from the original principal (the 1)
plus the interest rate (0.20).
Year 2: at the beginning of this year, the initial
amount available is the final amount of the
previous year. This amount is 1200, which is the
new principal. The value of this amount at the end
of the second year will be:
1200 1.20 = 1440
The 1200 can be rewritten in the way it was
originally found:
1000 1.20 1.20 = 1440
The left side can be simplified to:
1000 1.20 2 = 1440
Year 3: the initial amount in the third year is the
amount at the end of the year before:
1440 1.20 = 1728
This can be re-written as:
1000 1.20 1.20 1.20 = 1728
1000 1.20 3 = 1728
After three years, the value of the 1000
investment is 1728.
This example was to show what is meant by
compounding the principal each year. Notice that
to find the value after three years, the rate was to
the power of 3.
Theory: the value of an investment after 𝑡 years is
given by:
𝑉𝑡 = 𝑉0 1 + 𝑟 𝑡
Where 𝑉𝑡 is the value of the investment after 𝑡
years, 𝑉0 is the initial investment amount, and 𝑟 is
the interest rate. The interest rate must be written
as a decimal.
Example 1: find the value of $2000 invested at an
interest rate of 5% for 10 years.
Plan: determine the values of 𝑉0, 𝑟 and 𝑡, then
substitute into the compound formula. The
interest rate, 𝑟, must be written as a decimal.
Solution: substituting 𝑉0 = 2000, 𝑟 = 0.05, 𝑡 = 10
into the compounding formula:
𝑉10 = 2000 1 + 0.05 10 = 2000 1.05 10
𝑉10 = 3257.79
It should become evident that compound interest
is similar in theory to a geometric progression and
also a geometric average.
Example 2: a finance firm invests $1million for five
years in a Dubai construction project, which has an
expected return of 8% p.a. compounded annually.
Determine the value of the investment after 5
years, and also the interest earned over this time.
Plan: use the compound interest rate formula to
find the total value of the investment. Then to find
the interest earned, take the principal away from
the total value of the investment.
Solution: substitute 𝑉0 = 1, 𝑟 = 0.08 and 𝑡 = 5:
𝑉𝑡 = 𝑉0 1 + 𝑟 𝑡
𝑉5 = 𝑉0 1 + 0.08 5
𝑉5 = $1.469million
The interest earned is the total value take the
principal:
𝐼5 = 𝑉5 − 𝑉0
𝐼5 = 1.469 − 1 = $0.469million
169
Theory: the interest earned from compound
interest is the total value of the investment
subtract the principal amount:
𝐼𝑡 = 𝑉𝑡 − 𝑉0
Compounding every year is not what many banks
do with savings accounts. Rather, they set an
annual interest rate but compound monthly.
Theory: for an annual interest rate 𝑟 compounding
𝑚 times per year gives the formula:
𝑉𝑡 = 𝑉0 1 +𝑟
𝑚
𝑚𝑡
This is simpler than it looks.
Example 3: find the value of a $5000 investment if
it is invested at 6% p.a. for three years,
compounded monthly.
Plan: find the values of 𝑉0 , 𝑡 and 𝑟, as with normal
compounding, but then determine the number of
times, 𝑚, per time period, 𝑡, compounding is
occurring.
Solution: the initial amount is $5000 (𝑉0 = 5000),
the yearly interest rate is 6% (or 𝑟 = 0.06), the
time the investment lasts is 𝑡 = 3, and it is
compounded 𝑚 = 12 times per year. Substitute
into the formula:
𝑉3 = 5000 1 +0.06
12
12 3
= 5000 1.005 36
𝑉3 = 5983.40
Example 4: find the number of years (to the
nearest month) when an investment of $1000 will
reach $2000 at an interest rate of 9% p.a.
compounded every quarter.
Plan: use the compound formula (𝑚 times per
year) and substitute all known variables.
Rearrange to isolate 𝑡 to find the solution.
Solution: having 𝑉𝑡 = 2000, 𝑉0 = 1000, 𝑚 = 4
and 𝑟 = 0.09, substitute into the compound
formula:
𝑉𝑡 = 𝑉0 1 +𝑟
𝑚
𝑚𝑡
2000 = 1000 1 +0.09
12
4𝑡
Simplify:
2 = 1.0075 4𝑡
Take the log of both sides:
log 2 = 4𝑡 log 1.0075
Rearrange and solve:
4𝑡 =log 2
log 1.0075
𝑡 =
log 2log 1.0075
4≈ 23.19𝑦𝑒𝑎𝑟𝑠
Rounding up to the nearest quarter as that is what
the question is asking. Therefore the answer is 23
years and 1 quarter.
Theory: whenever finding when an investment will
exceed a certain amount, always round UP to the
nearest given time period. If it was to be rounded
down, the investment would not exceed the given
amount.
Example 5: a company can save $4000 in one of
two banks for a five year period. The two banks
offer the following returns:
Bank 1: 15% p.a. compounded weekly.
Bank 2: 15.5% p.a. compounded semi-annually.
Determine which of the projects is a better
investment.
Plan: use the compound formula (𝑚 times per
year) to determine which of the investments will
give the largest return after five years.
170
Solution:
Bank 1: 𝑟 = 0.15, 𝑚 = 52, 𝑉0 = 4000 and 𝑡 = 5:
𝑉𝑡 = 4000 1 +0.15
52
5 52
𝑉𝑡 ≈ $8,458.86
Bank 2: 𝑟 = 0.155, 𝑚 = 2, 𝑉0 = 4000 and 𝑡 = 5:
𝑉𝑡 = 4000 1 +0.155
2
2 5
𝑉𝑡 = $8,437.87
Notice that just because Bank 2 offers a higher
advertised interest rate, Bank 1 has a better actual
return due to the number of compounding periods
per year.
Theory: the more times an investment gets
compounded, the closer it is to being continuously
compounded, in which case the number 𝑒 is used:
𝑉𝑡 = 𝑉0𝑒𝑟𝑡
Which is identical to exponential growth from
Chapter 4.
Example 6: an investment offers an interest rate of
10%, compounded every minute of every day. The
investment has an initial value of $4000, and is
invested for one year. Find the value of the
investment using the compound interest formula,
then use the exponential formula. Compare the
answers.
Plan: find the number of minutes in a year, then
set this as 𝑚. Substitute this and all other known
variables into the compound interest formula, to
find 𝑉𝑡 .
Then, substitute all the known variables into the
exponential formula, to find 𝑉𝑡 .
Solution: number of minutes in a
year=minutes×hours×days = 60 ∙ 24 ∙ 365 =
525,600, 𝑉0 = 4000, 𝑟 = 0.10, then:
𝑉𝑡 = 4000 1 +0.10
525600
525600 ∙1
𝑉𝑡 ≈ 4420.68363
Now use the continuous formula above:
𝑉𝑡 = 4000𝑒0.10∙1 = 4420.68367
The two are very close, and the continuous
formula can be used as an accurate approximation
to very short time period compounding.
Example 7: find the value of a $4,000 savings
account which is compounded every second, if the
savings account is left for 5 years at 4.65% p.a.
Plan: use the continuous compounding formula.
Solution: 𝑉0 = 4000, 𝑟 = 0.0465, 𝑡 = 5 is
substituted into the continuous compound
formula:
𝑉𝑡 = 𝑉0𝑒𝑟𝑡
𝑉5 = 4000𝑒0.0465 ∙5
𝑉5 = 5047.00 2𝑑. 𝑝.
Exercises: 1. A bank offers an interest rate of 4.75% p.a., and
you decide to save $4,300 for four years. Determine: a) The interest earned after four years. b) The total value of the savings account.
2. A conservative company decides to save a cash-flow of $12,391 in a term deposit account which has a return of 8.2% p.a., compounded weekly. If the term deposit is for 10 years, determine: a) The value of the term deposit account after ten
years. b) The interest earned over the ten years.
3. You have purchased new car worth $12,000. If the interest rate is 7.25% p.a. and is compounded monthly, determine the amount you would owe after a three year loan, assuming you made no repayments.
4. A disreputable agent loans you $10,500 at 5.5% p.a. compounded continuously. Determine the amount you would have to pay back after two years.
8.5 annual interest rates
Example 5 from the previous section is a good
indicator that the advertised interest rate is not
always the realised interest rate.
171
Theory: the annual interest rate is lower than the
realised interest rate if the principal is
compounded more than once per year.
Intro example 1: for an advertised interest rate of
5%, $100 is invested for a year, compounded
semi-annually (every half year). Find the value of
the investment after one year.
Solution: 𝑉0 = 100, 𝑡 = 1, 𝑚 = 2, 𝑟 = 0.05:
𝑉1 = 100 1 +0.05
2
2∙1
= 100 1.025 2
𝑉1 = 105.06
From this, we can also see that the realised annual
interest rate is 5.06% rather than the advertised
5%.
If this was simple interest, the value of the total
investment would be:
𝑉1 = 𝑉0 + 𝐼1 = 100 + 5 = 105
It is obvious that with compound interest, the
realised interest rate (5.06%) is larger than the
advertised 5%. This makes a big difference with
large sums of money, such as home loans.
Theory: the realised interest rate is found using:
𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 +𝑟
𝑚
𝑚
− 1
This is very similar to the compound interest
formula except that there is no 𝑡 nor are there any
amounts (𝑉’s). The −1 gets rid of the original
amount, and just leaves the realised interest rate.
Example 1: find the realised annual interest rate
for an investment where the advertised interest
rate is 4.75%, and is compounded monthly.
Plan: apply the realised interest rate formula.
Solution:
𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 +0.0475
12
12
− 1 ≈ 0.0485(4 𝑑. 𝑝. )
Thus the realised interest rate is 4.85%, which is
higher than the advertised rate.
Example 2: a company has a cash inflow from its
customers tomorrow morning, the precise amount
is yet unknown. This company has two savings
accounts at two different banks. The offered
interest rates are:
Bank 1: 7% compounded quarterly
Bank 2: 6.9% compounded weekly
Which is a better investment?
Plan: use the realised interest rate formula to
determine the realised interest rates of the two
banks.
𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 +𝑟
𝑚
𝑚
− 1
Solution:
Bank 1: given 𝑟 = 0.07 and 𝑚 = 4:
𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 +0.07
4
4
− 1
𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 0.07186 5𝑑. 𝑝.
Thus a realised rate of approximately 7.186%.
Bank 2: given 𝑟 = 6.9% and 𝑚 = 52:
𝑟𝑟𝑒𝑎 𝑙𝑖𝑠𝑒𝑑 = 1 +0.069
52
52
− 1
𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 0.07139 5𝑑. 𝑝.
Thus a realised rate of approximately 7.139%.
Given the realised rates, Bank 1 is a better option.
Exercises: 1. A bank advertises an interest rate of 5.45% p.a.
compounded monthly. Determine the realised annual interest rate.
2. An investment promises a return of 6.24% p.a. compounded once every 4 months. Determine the realised interest rate.
3. Two banks are in competition, one with an advertised interest rate of 3.45% p.a. compounded monthly, and the other with an interest rate of 5.1% p.a. compounded once every six months. If you are borrowing money, which bank is a better choice?
172
8.6 net present value
Theory: Net Present Value is a way of finding the
value of income in the future, in terms of today’s
dollars. It is similar to the concept of
compounding, except backwards.
That is, $1000 today has a higher value than
$1000 some time in the future. The reason is that
if that $1000 dollars was invested at the given
interest rate, the value of the $1000 would grow
(through earning interest).
Intro example 1: for an investment of $1000 at an
interest rate of 10% per annum, compounded at
the end of each year, how much is that investment
worth in a year’s time?
Solution:
𝑉1 = 1000 1 + 0.1 1 = 1100
This means that getting $1000 today is the same
as getting $1100 dollars in one years time.
Intro example 2: if you need to repay a loan worth
$2000 in one year’s time, how much do you need
to invest today at 7% p.a. compounded only once,
at the end of the year?
Solution: 𝑉0 is unknown but 𝑉1 is known (= 2000),
𝑟 = 0.07, 𝑡 = 1:
𝑉𝑡 = 𝑉0 1 + 𝑟 𝑡
Rearranging to isolate 𝑉0 gives:
𝑉𝑡
1 + 𝑟 𝑡= 𝑉0
Substitute in to get:
𝑉0 =2000
1 + 0.07 1= 1869.16
So $1869.16 would need to be invested today to
have $2000 in one year’s time. This means that
$1869.16 today has the same value as $2000 in
one year.
Theory: the formula for the net present value of a
single future cash flow is:
𝑁𝑃𝑉 =𝑉𝑡
1 + 𝑟 𝑡
The 𝑁𝑃𝑉 is the same as 𝑉0 from the rearranged
annual compound formula.
Example 1: find the net present value of $5000 in
three years’ time at an interest rate of 4.25% p.a.
Plan: apply the NPV formula.
Solution: 𝑉3 = 5000, 𝑟 = 0.0425, 𝑡 = 3:
𝑁𝑃𝑉 =5000
1 + 0.0425 3=
5000
1.04253
𝑁𝑃𝑉 = 4413.08
This means that $5000 in three years’ time has the
same value as $4413.08 today.
Theory: if there are multiple cash-flows at
different years, the NPV of all the cash flows is the
sum of the NPVs of the individual cash flows.
𝑁𝑃𝑉 = 𝑁𝑃𝑉𝑛
𝑛
The overall NPV is simply the sum of all 𝑛
individual net present values.
Example 2: find the NPV of the two cash-flows:
Cash-inflow 1: $4000 in 3 years
Cash-inflow 2: $2400 in 7 years
The interest rate is assumed constant at 5.25%.
Plan: find the NPV of each of the two cash-inflows
individually, then sum them to find the overall
NPV.
Solution: discount each of the cash-inflows:
Cash-inflow 1:
𝑁𝑃𝑉1 =4000
1 + 0.0525 3≈ 3430.79
Cash-inflow 2:
𝑁𝑃𝑉2 =2400
1 + 0.0525 7≈ 1677.48
The NPV of both cash-inflows is:
173
𝑁𝑃𝑉 = 𝑁𝑃𝑉1 + 𝑁𝑃𝑉2 = 3430.79 + 1677.48
𝑁𝑃𝑉 = $5108.26
Theory: any project with a positive net present
value is accepted, as it is profitable.
Example 3: find the Net Present Value of the
project which has the following cash flows:
Assume that the interest rate is 6.25%.
Plan: find the NPV of each of the cash flows, then
sum them to find the overall NPV.
Solution:
Year 0: the NPV of a cash-flow at year 0 is the
value of the cash-flow. Since it is an outflow, the
NPV is:
𝑁𝑃𝑉0 = −4000
Year 1:
𝑁𝑃𝑉1 =4000
1 + 0.0625 1
𝑁𝑃𝑉1 = $3764.71
Year 2:
𝑁𝑃𝑉2 =500
1 + 0.0625 2
𝑁𝑃𝑉2 = $442.91
Thus the overall NPV is the sum of all the
individual NPV:
𝑁𝑃𝑉 = 𝑁𝑃𝑉0 + 𝑁𝑃𝑉1 + 𝑁𝑃𝑉2
= −4000 + 3764.71 + 442.91
= $207.62
Since the overall NPV is positive, this project is
worthwhile.
Example 4: determine the Net Present Value for
the following investment opportunity:
Assuming the interest rate is a constant 10%.
Plan: find the NPV of the individual cash-flows,
then sum them.
Solution:
Year 0:
𝑁𝑃𝑉0 = −983
Year 1:
𝑁𝑃𝑉1 =267.3
1 + 0.1 1= 243
Year 2:
𝑁𝑃𝑉2 =532.4
1 + 0.1 2= 440
Year 3:
𝑁𝑃𝑉3 =399.3
1 + 0.1 3= 300
Thus the overall NPV is:
𝑁𝑃𝑉 = 𝑁𝑃𝑉0 + 𝑁𝑃𝑉1 + 𝑁𝑃𝑉2 + 𝑁𝑃𝑉3
= −983 + 243 + 440 + 300 = 0
Thus the NPV of the project is zero, so there is no
point going ahead with the project, as the same
return could be obtained by saving the initial cash
outflow in a bank at an interest rate of 10%.
Exercises: 1. Given the following cash-flow data, determine the
NPV of the project and if it is worthwhile. Assume the interest rate is 4.5%.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $6000 0 1 0 $4000 2 0 $1000 3 0 $1500
2. Your supervisor gives you the following data on a project introducing a new advertising campaign. They want you to find out the NPV of this project, and if it is worthwhile to go ahead with it. Assume that the bank is offering an interest rate of 6.1% p.a.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤
0 $983.00 0 1 0 $267.30 2 0 $532.40 3 0 $399.30
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤
0 4000 0 1 0 4000 2 0 500
174
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $10,000 0 1 0 $3000 2 0 $4000 3 0 $5000
4 $2000 $2500 5 0 $3000
3. Given the following data about cash-flows from customers and suppliers, and an interest rate of 7.25%, determine the NPV of all the cash-flows.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $1500 $2000 1 $2500 $1000 2 0 $3000 3 0 $800
4 $4000 $1000 4. Describe, in your own words, the meaning of net
present value.
8.7 internal rate of return
In the last example of the previous section, there
was a NPV of zero. This is the definition of internal
rate of return.
Theory: the internal rate of return is the interest
rate that would give a net present value of zero.
The internal rate of return is the rate of return of
the project, and is unrelated to the market interest
rate.
Intro example 1: determine the internal rate of
return of the following project:
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤
0 $3000 0 1 0 $3500
Solution: the market rate has not been given, so
there is no way the NPV can be found. However,
the internal return of the project is unrelated to
the market rate. To find the IRR of the project, use
the NPV formula and set the overall NPV equal to
zero. Also, instead of using a number for the
interest rate, use the letter 𝑖:
𝑁𝑃𝑉 = −3000 +3500
1 + 𝑖 1
Set this equal to zero:
0 = −3000 +3500
1 + 𝑖 1
Then solve for 𝑖:
3000 =3500
1 + 𝑖
3000 1 + 𝑖 = 3500
3000 + 3000𝑖 = 3500
𝑖 =1
6= 0.1667
Thus the rate of return of the project is 16.67%,
and this is unrelated to the market rate.
When the number of years increases, the ability to
isolate 𝑖 becomes more and more difficult.
Intro example 2: find an expression for the
internal rate of return of the following project, and
then attempt to isolate 𝑖.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $10,000 0 1 0 $3000 2 0 $4000 3 0 $5000
Solution:
𝑁𝑃𝑉 = −10000 +3000
1 + 𝑖 1+
4000
1 + 𝑖 2+
5000
1 + 𝑖 3
Set this equal to zero to find an expression for the
internal rate of return:
0 = −10000 +3000
1 + 𝑖 1+
4000
1 + 𝑖 2+
5000
1 + 𝑖 3
Now, to isolate 𝑖, the −10000 will be first moved
to the other side:
10000 =3000
1 + 𝑖 1+
4000
1 + 𝑖 2+
5000
1 + 𝑖 3
Then to be able to add the fractions on the right, a
common denominator is needed. The common
denominator is 1 + 𝑖 3, so change the form of the
other fractions:
10000 =3000 1 + 𝑖 2
1 + 𝑖 3+
4000 1 + 𝑖
1 + 𝑖 3+
5000
1 + 𝑖 3
Simplifying:
10000 =3000 1 + 𝑖 2 + 4000 1 + 𝑖 + 5000
1 + 𝑖 3
175
Rearranging and expanding the brackets would
take a long time, and in the end, a cubic function
would need to be solved. Try this for yourself and
see if you can get:
0 = 10000𝑖3 + 27000𝑖2 + 20000𝑖 − 2000
Solving this is difficult, but a program can be used.
This above example is an exact method of finding
the internal rate of return. However, as the
number of years increases to 4, the exact solution
will involve solving something with 𝑖4 which is
difficult; for 5 years, the solution will involve 𝑖5, so
it gets more and more difficult.
However, there is a much easier method which
gives an approximate answer.
Theory: the method of finding the approximate
internal rate of return is split into two stages:
STAGE 1: find two interest rates; one which gives a
positive NPV, and one which gives a negative NPV.
This step is a trial-and-error method. The thought
process is:
Try a random (low) interest rate and find the NPV.
If it is positive, use another (higher) interest rate to
get a negative NPV. If the initial (low) interest rate
had a negative NPV, use another (lower) interest
rate to find a positive NPV.
This theory is demonstrated in the following
example.
Example 1: find two interest rates, one with a
positive NPV and one with a negative NPV, given
the following investment project:
Plan: find the NPV of the project using a relatively
low interest rate, then increase the interest rate
by, say 2%, to find one with a negative NPV.
Solution: use a low interest rate to begin with. This
can be anything you choose, so let’s use 3%.
𝑁𝑃𝑉3% = −8000 +5000
1.03 1+
2000
1.03 2+
2000
1.03 3
𝑁𝑃𝑉3% = $569.84
The low interest rate should give a positive NPV. If
it does not, lower the interest rate further. Now,
since the 3% gave a positive NPV, to get a negative
NPV, we use a higher interest rate, say 5%:
𝑁𝑃𝑉5% = −8000 +5000
1.05 1+
2000
1.05 2+
2000
1.05 3
𝑁𝑃𝑉5% = $303.64
This is still a positive NPV, so increase the interest
rate to say 10%:
𝑁𝑃𝑉10% = −8000 +5000
1.10 1+
2000
1.10 2+
2000
1.10 3
𝑁𝑃𝑉10% = −299.02
Both 3% and 5% gave a positive NPV, but for part
two of this approximations, the interest rate that
gives a NPV closest to zero is used, so in this case,
the 5% has a NPV closer to zero, than the 3%.
Theory:
STAGE 2: having two interest rates (relatively close
to one another) with one giving a positive NPV and
the other giving a negative NPV, apply the formula:
𝐼𝑅𝑅 ≈𝑖2𝑁𝑃𝑉1 − 𝑖1𝑁𝑃𝑉2
𝑁𝑃𝑉1 − 𝑁𝑃𝑉2
It does not matter which interest rate-NPV
combination you define with a subscript as 1 or 2,
as the formula works either way.
This method is a linear approximation of a curve.
The closer the positive and negative NPV’s are to
being zero, the better the approximation.
Example 1 cont: given the following pairs of NPV
and 𝑖, find the approximate IRR:
𝑖1 = 5%, 𝑁𝑃𝑉1 = 303.64
𝑖2 = 10%, 𝑁𝑃𝑉2 = −299.02
Plan: apply the IRR formula.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $8000 0 1 0 $5000 2 0 $2000 3 0 $2000
176
Solution: substitute this into the formula:
𝐼𝑅𝑅 ≈ 0.10 303.64 − 0.05 −299.02
303.64 − −299.02
𝐼𝑅𝑅 ≈45.31
602.66≈ 0.0752
Thus the internal rate of return of this project is
approximately 7.52%.
Theory: the IRR will always be between the two
interest rates which gave the positive and negative
NPV’s, and will be closer to the interest rate which
has its NPV closer to zero.
An internal rate of return higher than the market
interest rate means the project is worthwhile.
Conversely, an internal rate of return lower than
the market interest rate means the project is not
worthwhile.
Example 2: determine the internal rate of return
for the project shown below.
If the market rate is 9%, is the project
worthwhile?
Plan: use a low interest rate to find the NPV; then
if that gives a positive NPV, use another higher
interest rate to find a negative NPV. If the first
interest rate gives a negative NPV, use a lower
interest rate to find a positive NPV. Apply the
results to the IRR formula, and compare it to the
market rate.
Solution: finding the NPV at, say, 5%:
𝑁𝑃𝑉5% = −7000 +3000
1.05+
2500
1.052+
1500
1.053+
1500
1.054
𝑁𝑃𝑉5% = 654.53
Since this is a positive NPV, a higher interest rate
will give a negative NPV. So try 10%:
𝑁𝑃𝑉10% = −7000 +3000
1.1+
2500
1.12+
1500
1.13+
1500
1.14
𝑁𝑃𝑉10% = −55.12
Having two interest rates, one which gives a
positive NPV and one which gives a negative NPV,
apply the IRR formula:
𝐼𝑅𝑅 ≈𝑖2𝑁𝑃𝑉1 − 𝑖1𝑁𝑃𝑉2
𝑁𝑃𝑉1 − 𝑁𝑃𝑉2
Many students screw up with the substitution, so
it is best to rewrite the known variables, with
appropriate subscripts:
𝑖1 = 0.05, 𝑁𝑃𝑉1 = 654.53
𝑖2 = 0.10, 𝑁𝑃𝑉2 = −55.12
Substitute this into the formula:
𝐼𝑅𝑅 ≈ 0.10 654.53 − 0.05 −55.12
654.53 − −55.12
𝐼𝑅𝑅 ≈ 0.0961
Thus the internal rate of return is 9.61%. Since the
IRR is greater than the market rate of 9%, the
project is worthwhile.
Note that the IRR is closer to 10% than to 5%, as
the NPV of the 10% is closer to zero (i.e. −55.12)
than the NPV of the 5% (i.e. 654.53).
Exercises: 1. For the following data, find an approximation for
the internal rate of return.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $7,000 0 1 0 $3000 2 0 $2500 3 0 $2800
2. Given a market interest rate of 8.5%, determine if the following project is worthwhile using the IRR method.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $12,430 0 1 0 $7800 2 0 $5100 3 0 $1100
3. In your own words, describe what internal rate of return measures.
𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤 0 $7000 0 1 0 $3000 2 0 $2500 3 0 $1500 4 0 $1500
177
chapter eight summary
Index numbers represent values in terms of a base value:
𝐼𝑛𝑑𝑒𝑥𝑡 =𝑉𝑎𝑙𝑢𝑒𝑡
𝑉𝑎𝑙𝑢𝑒0
× 100
The base year is given an index value of 100. The percentage difference between two time periods in an index is found using:
%∆𝑡→𝑡+1=𝐼𝑡+1 − 𝐼𝑡
𝐼𝑡× 100
𝐴𝑟𝑖𝑡𝑚𝑒𝑡𝑖𝑐 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑦𝑒𝑎𝑟𝑠
𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑐𝑎𝑛𝑔𝑒
100+ 1
𝑛
− 1
A series is list of numbers with a constant pattern. Each individual entry of a series is important but so is the sum of the series.
An arithmetic series has a constant difference between terms.
A geometric series has a constant ratio between terms.
The 𝑛𝑡 term of a sequence is given by: 𝑇𝑛 = 𝑎 + 𝑛 − 1 𝑑
The sum of an arithmetic series is given by:
𝑆𝑛 =𝑛
2 2𝑎 + 𝑛 − 1 𝑑
The 𝑛𝑡 term of a geometric progression is found using the formula:
𝑇𝑛 = 𝑎𝑟𝑛−1 To find the ratio, divide a term by the one before it:
𝑟 =𝑇𝑛
𝑇𝑛−1
When 𝑝 terms are missing, the ratio is found by:
𝑟 = 𝑙𝑎𝑡𝑒𝑟 𝑡𝑒𝑟𝑚
𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑡𝑒𝑟𝑚
𝑝 +1
The sum of a geometric progression is given by:
𝑆𝑛 =𝑎 𝑟𝑛 − 1
𝑟 − 1
The simple interest formula is: 𝐼𝑡 = 𝑃0 ∙ 𝑟 ∙ 𝑡
The total value of an investment is: 𝑃𝑡 = 𝑃0 + 𝐼𝑡
For compound interest, the value of an investment after 𝑡 years is given by:
𝑉𝑡 = 𝑉0 1 + 𝑟 𝑡 The interest earned from compound interest is:
𝐼𝑡 = 𝑉𝑡 − 𝑉0 For an annual interest rate 𝑟, compounding 𝑚 times per year:
𝑉𝑡 = 𝑉0 1 +𝑟
𝑚
𝑚𝑡
Whenever finding when an investment will exceed a certain amount, always round UP to the nearest given time period.
For continuous compounding (or very short time period compounding) the 𝑒 formula is used:
𝑉𝑡 = 𝑉0𝑒𝑟𝑡
The annual interest rate is lower than the realised interest rate if the principal is compounded more than once per year. The realised interest rate is found using:
𝑟𝑟𝑒𝑎𝑙𝑖𝑠𝑒𝑑 = 1 +𝑟
𝑚
𝑚
− 1
Net Present Value is a way of finding the value of income in the future, in terms of today’s dollars. The formula for the NPV of a single future cash flow is:
𝑁𝑃𝑉 =𝑉𝑡
1 + 𝑟 𝑡
If there are multiple cash-flows at different years, the NPV of all the cash flows is the sum of the NPVs of the individual cash flows.
𝑁𝑃𝑉 = 𝑁𝑃𝑉𝑛
𝑛
Any project with a positive net present value is accepted, as it is profitable.
The internal rate of return (IRR) is the interest rate that would give a net present value of zero, and is unrelated to the market interest rate. To find an approximation to the IRR: STAGE 1: find two interest rates; one which gives a positive NPV, and one which gives a negative NPV. This step is a trial-and-error method. STAGE 2: having these two interest rates, apply the formula:
𝐼𝑅𝑅 ≈𝑖2𝑁𝑃𝑉1 − 𝑖1𝑁𝑃𝑉2
𝑁𝑃𝑉1 − 𝑁𝑃𝑉2
The IRR will always be between the two interest rates which gave the positive and negative NPV’s, and will be closer to the interest rate which has its NPV closer to zero. An internal rate of return higher than the market interest rate means the project is worthwhile. An internal rate of return lower than the market interest rate means the project is not worthwhile.
chapter eight questions
1. Given the following Consumer Price Index table: 𝑌𝑒𝑎𝑟 𝐶𝑃𝐼 1980 89.9 1981 92.4 1982 93.5 1983 − 1984 100 1985 103.1 1986 107.2
1987 −
Determine: a) The percentage change from 1980 to 1981. b) The percentage change from 1982 to 1984. c) The arithmetic average from 1982 to 1984. d) The geometric average from 1982 to 1984. e) If the geometric average from 1982 to the
missing value of 1987 is 3.32%, determine the
178
missing index entry. Hint: find 𝐼𝑡→𝑡+𝑛 then substitute into the geometric average formula.
2. The gold index has the following historic data: 𝑌𝑒𝑎𝑟 𝐺𝑜𝑙𝑑 𝐼𝑛𝑑𝑒𝑥 1971 150.2 1972 168.4 1973 172.3 1974 163.0 1975 154.0 1976 142.3 1977 133.1
Determine: a) The percentage change from 1971 to 1973. b) The percentage change from 1971 to 1977. c) The arithmetic average from 1971 to 1973. d) The geometric average from 1971 to 1973. e) The arithmetic average from 1971 to 1977. f) The geometric average from 1971 to 1977. g) Explain how the average percentage change over
the years 1971 to 1973 is positive, however if the time period is extended slightly from 1971 to 1977, the average percentage change is negative.
3. Given the following series: 3,3.75,4.5,5.25,6,6.75, …
Determine: a) The 13𝑡 term. b) The 21𝑠𝑡 term. c) The sum of the first 9 terms. d) The sum of the first 16 terms.
4. Given the following series: 1331,484,176,64, …
Determine:
a) The 15𝑡 term.
b) The 27𝑡 term. c) The sum of the first 11 terms. d) The sum of the first 16 terms.
5. An investment firm is offered one of two investments, each which has the following returns: Investment 1: $6000 in the first year, and this increases by $250 every year after. Investment 2: $4350 in the first year, and this increases by 13% every year after. If the investment is for 8 years, determine: a) Which of the two investments has a higher
return in the fifth year. b) Which of the two investments has a higher
return up to the fifth year. c) Which of the investments has a higher overall
return. 6. Mark lends David $1430 for an airfare home. The
deal is that David pays 3% simple interest per month, for the four months it takes to earn the money. Assuming David repays Mark in a lump-sum, determine: a) The interest Mark has earned over the four
months. b) The total amount David has to repay Mark after
the four months. 7. A father opens a savings account for his son, and
deposits $20,000. If the interest rate is 5% p.a.
(compounded annually), but the son takes out the interest at the beginning of each year, determine: a) How much money the son has taken out after 7
years. b) How much money the son has taken out after 11
years. 8. A firm saves $5600 at an interest rate of 4.65% p.a.
compounded monthly. Determine: a) The value of the account after 4 months. b) The value of the account after 5.5years. c) When the value of the account will first exceed
$10,000. d) The value of the account after 6 years, given that
after 3years, the firm withdrew $2000. 9. Having won $11,457 in the lottery, you save the total
amount at 7.1% p.a. compounded monthly. Determine: a) The value of the account in 5 years. b) When the value of the account will first exceed
$20,000. c) When the value of the account will first exceed
$20,000 given that after 2 years, you withdraw $5,000.
10. A student needs to have $5000 to pay his university fees in three years time. If the prevailing interest rate is set at 6.6%p.a. compounded semi-annually, determine how much the student needs to save today to be able to pay the university bill.
11. A firm wanting to save a future cash-flow has the option of two bank accounts:
Bank A: 5.1% p.a. compounded yearly
Bank B: 5% p.a. compounded weekly Determine which of the banks is the better deal.
12. A firm is looking to borrow money from one of three banks:
Universal Bank: 8.1%p.a. compounded quarterly
Sun Bank: 8%p.a. compounded weekly
Bank of Vark: 8.2% compounded semi-annually Determine which of the banks is the better deal.
13. Given the following investment project: 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤
0 $60,000 0 1 0 $48,200 2 0 $21,500 3 0 $9,100
For a market rate of 4%, determine: a) The NPV b) The internal rate of return c) If the project is worthwhile.
14. Given the following potential project: 𝑌𝑒𝑎𝑟 𝐶𝑎𝑠 − 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝐶𝑎𝑠 − 𝑖𝑛𝑓𝑙𝑜𝑤
0 $12,000 0 1 0 $5000 2 0 $5000 3 0 $2000
4 $4000 $2500 5 0 $4000
For a market rate of 9%, determine: a) The NPV and the IRR b) If the project is worthwhile.
179
180
chapter one solutions
Question 1: a) 15 b) −4 c) 50 d) −6 Question 2:
a) 1
6
b) 1
12
c) 1.5
d) −8
15
e) 21
49
f) 27
g) 36
13
h) 40
9
Question 3:
a) 5
6
b) 19
24
c) 25
12 𝑜𝑟 2
1
12
d) −7
22
e) 23
65
f) 5
12
g) 187
210
h) 68
21 𝑜𝑟 3
5
21
Question 4:
a) −25
42
b) 11
30
c) 3+𝑥
3𝑥
d) −5
6
e) 9
10
f) 25𝑥+12
12𝑥
g) 7𝑥
h) 3𝑦+2𝑥
3𝑥𝑦
Question 5: a) 𝑥3𝑦6 b) 216𝑥9𝑦1.5
c) 1
𝑥4𝑦8
d) 2
3𝑥2
e) 243𝑥4𝑦8 f) 1024𝑥2𝑦0.125
g) 𝑦7/6
6𝑥9/2
h) –𝑦0.125
25
Question 6: a) 34 b) 24 c) 9 d) −33𝑥 e) −186 f) −38 g) −64 h) −2𝑥 + 2𝑦 + 6 Question 7: a) 𝑥2 + 4𝑥 + 4 b) 𝑥2 − 6𝑥 + 9 c) 4𝑥2 + 4𝑥 + 1 d) 1 − 2𝑥 + 𝑥2 e) 𝑥2 + 8𝑥 + 5 f) −4𝑥2 + 7𝑥 − 1
g) 3𝑥2 − 6𝑥 + 4
h) – 𝑥2 − 2𝑥 − 4 Question 8: a) 𝑥 = 5 b) 𝑥 = 2 c) 𝑥 = 1
d) 𝑥 =11
3
e) 𝑥 =13
2
f) 𝑥 =1
21
g) 𝑥 = −9
10
h) 𝑥 = 3 −𝑦
3
i) 𝑥 =𝑦+1
2
j) 𝑥 =3𝑦−9
4
Question 9: a) 2 𝑥 + 𝑥𝑦 + 1 b) 2𝑥 𝑥 + 3𝑦 − 4𝑦2 c) 𝑦 13𝑥 − 12𝑦 + 1 d) 3𝑥2 1 + 3𝑥 + 4𝑥3 e) 𝑃𝑄 4 + 𝑟 − 𝑥𝑦
f) 1
6 2𝑥2 + 2𝑥4 + 1
g) 5𝑥 3𝑥2 + 5 − 10𝑥3
h) 1
𝑥 1 +
1
𝑥−
1
𝑥2 𝑜𝑟
1
𝑥3 𝑥2 + 𝑥 − 1
Question 10: a) 𝑥 𝑥 − 1 = 0
𝑥 = 0 𝑜𝑟 𝑥 = 1 b) 𝑥 𝑥 + 3 = 0
𝑥 = 0 𝑜𝑟 𝑥 = −3 c) 2𝑥 1 − 3𝑥 = 0
𝑥 = 0 𝑜𝑟 𝑥 =1
3
d) 4𝑥 𝑥 + 1 = 0 𝑥 = 0 𝑜𝑟 𝑥 = −1
e) 𝑥2 𝑥 − 1 = 0 𝑥 = 0 𝑜𝑟 𝑥 = 1
f) 𝑥3 3𝑥 − 5 = 0
𝑥 = 0 𝑜𝑟 𝑥 =5
3
Question 11: a) 4 𝑥2 + 3𝑥 + 3 b) 6 2 − 𝑥
c) 7𝑥2 +1
𝑥2
d) –𝑥2
2− 2.25𝑥 + 16
e) 𝑥2 4𝑥2 − 1 f) 𝑥 −5𝑥4 + 3𝑥2 − 5𝑥 + 3
Question 12: a) 𝑥 = 3 𝑜𝑟 𝑥 = 2 b) 𝑥 = −7 𝑜𝑟 𝑥 = 4 c) 𝑥 = 2 𝑜𝑟 𝑥 = 3
d) 𝑥 = −1
3 𝑜𝑟 𝑥 = −
15
6
e) 𝑥 = 0 𝑜𝑟 𝑥 = 2 f) 𝑥 = 1 Question 13: a) 3 b) 4 c) 1 d) 15 e) 0.1 f) 1
chapter two solutions
Question 1: a) 𝑦 = 𝑥 − 1
b) 𝑦 = −𝑥
2+
3
2
c) 𝑦 = 𝑥 + 1 d) 𝑦 = 2𝑥 − 4
e) 𝑦 =5 14𝑥+3
7 𝑜𝑟
𝑦 =70𝑥 + 15
7
f) 𝑦 = 2𝑥 −1
2
g) 𝑦 =5 𝑥−12
3 𝑜𝑟
𝑦 =5𝑥 − 60
3 𝑜𝑟
𝑦 =5𝑥
3− 20
h) 𝑦 = 5𝑥 − 12
i) 𝑦 = 9 −5𝑥
3
Question 2: a) 𝐷, 𝐹, 𝐴, 𝐵, 𝐶, 𝐸 b) 𝐴, 𝐵, 𝐹, 𝐶, 𝐷, 𝐸 c) 𝐴, 𝐹, 𝐵, 𝐶, 𝐷, 𝐸
Question 4:
a) 𝑦 =6
5𝑥 − 6
b) 𝑦 = −3𝑥 − 3 Question 5: a) 𝑦 = 6𝑥 − 3 b) 𝑦 = 5𝑥 − 9
c) 𝑦 =4
3𝑥
d) 𝑦 = −𝑥 + 6
e) 𝑦 = −5
4𝑥 − 8
1
4
Question 6: a) 𝑦 = 2𝑥 − 1 b) 𝑦 = 3𝑥 + 2 c) 𝑦 = 𝑥 d) 𝑦 = −2𝑥 − 12 e) 𝑦 = −2.5𝑥 − 6.6 Question 7: a) −6, −9 b) 6,15 c) −1,12
d) 331
7, −36
5
7
e) 9,4.25 Question 8: 𝑄 = 18, 𝑃 = 80 Question 9: 𝑄 = 5, 𝑃 = 22 Question 10:
𝜀𝑑 = −62
27 meaning this
good is elastic in demand, as |𝜀𝑑 | > 1. Question 11:
𝜀𝑠 =25.8
21=
129
105 meaning this
good is elastic in supply, as 𝜀𝑠 > 1. Question 12:
a) 𝑄 = 28
30𝑡𝑜𝑢𝑠𝑎𝑛𝑑𝑠,
𝑃 = 22182
300≈ 22.61
b) 𝜀𝑑 = −3391
68≈ −49.72
meaning this good is very elastic in demand.
c) 𝜀𝑠 =3391
1054≈ 3.22 which
means this good is elastic in supply.
Question 13:
a) 𝑄 =732
155≈ 4.723𝑚𝑖𝑙𝑙𝑖𝑜𝑛
𝑃 =3592
155≈ 23.17
b) 𝜀𝑑 = −8980
3111≈ −2.89
which is elastic.
c) 𝜀𝑠 = −898
61≈ −14.72
meaning this good is elastic in supply.
Note; the supply curve is backward bending in this case.
181
chapter three solutions
Question 1: a) −3𝑥 + 𝑦 = −1
2𝑥 + 𝑦 = −17 b) 𝑥 + 𝑦 = 3
3𝑥 + 𝑦 = −15 c) −𝑥 + 17𝑦 = 14
−3𝑥 + 2𝑦 = 1 d) 11𝑥 − 1.4𝑦 = 3
11𝑥 − 1.5𝑦 = 3 e) 3𝑥 − 3𝑦 − 𝑧 = −4
−15𝑥 + 4𝑦 = 13 −𝑥 − 𝑦 + 1.4𝑧 = 0
f) 𝑎 − 𝑏 + 𝑐 = −4 𝑎 − 𝑏 − 𝑐 = −1 3𝑎 − 3𝑏 − 𝑐 = 1
Question 2: a) 3, −1 b) 2,2 c) 3.75,2.75 d) 13, −11,1 e) 2,1,5 Question 3: a) 1,1 b) 2,1 c) 3, −3 d) 2, −1 e) 13,19 f) −8,12 g) −1.75,1.75 h) 3.25,18.75 Question 4: a) 3,2 b) 1,7 c) Infinite solutions d) 1,3 e) −1, −3 f) 20,4 g) −0.5,8.5
h) −0.5,8.5 i) −0.5,8.5 j) −0.5,2 k) 2.4,4.2 l) 4.4,7.6 m) 7.1, −1.9 Question 5: a) 𝑃 = 45.5 𝑎𝑛𝑑 𝑄 = 8.5
b) 𝑃 = 465
11 𝑎𝑛𝑑 𝑄 = 8
19
22
c) 𝑃 = 46 𝑎𝑛𝑑 𝑄 = 19
d) 𝑃 = 71
3 𝑎𝑛𝑑 𝑄 = 1
e) 𝑃 = 7.21 𝑎𝑛𝑑 𝑄 = 8.43 Question 6: a) −1,1,3 b) 3,4,7 c) 2,1,12 d) −1,2, −3 Question 7:
a) 1
6, 3
13
18, −3
17
18
b) 1.5,3.5,2 c) 1.75,1.25,3.25 d) 3.1,4.4,5.9 e) 5, −1.5,2.5 f) 3.1,2,1.6 g) −2.1, −2.2, −2.3 h) 2,4,6 Question 8: a) Unique solution b) Infinite solutions c) Unique solution d) Unique solution e) Infinite solutions f) No solutions Question 9: a) −11 b) 52
c) −10 d) 8 e) 0 f) −6.25 g) 3.4
h) −3
7
i) −34
315
j) −2.79 k) 0 l) 0 Question 10: a) 8 b) −30 c) 42 d) 56 e) −280 f) −2294 Question 11: a) 2,2,2 b) 2,8,4 c) 1,3,7 d) 0.2,0.4,0.6 Question 12: a) 𝑀: 2853 = 7𝑃 + 19𝑆 + 14𝐿
𝐼: 2873 = 9𝑃 + 21𝑆 + 11𝐿 𝑇: 3594 = 4𝑃 + 20𝑆 + 25𝐿
b) 𝑃 = 111, 𝑆 = 40 𝑎𝑛𝑑 𝐿 = 94 Question 13: 𝐶 = 4860, 𝐼 = 540 𝑎𝑛𝑑 𝑌 = 6000 Question 14: 𝐶 = 1827.91, 𝐼 = 180.78 𝑎𝑛𝑑 𝑌
= 2608.70 Question 15: 𝐶 = 800 − 800𝑠 , 𝐼 = 800𝑠 𝑎𝑛𝑑 𝑌
= 1000
chapter four questions
Question 1: Functions: 𝐴, 𝐶, 𝐷 Not Functions: 𝐵, 𝐸 Question 2: a) 𝑎 = −5, 𝑏 = −4, 𝑐 = 3 b) 𝑎 = −1, 𝑏 = 1, 𝑐 = 6 c) 𝑎 = 1, 𝑏 = 0, 𝑐 = −4 d) 𝑎 = 1, 𝑏 = −2, 𝑐 = −2 Question 3: 𝑦 = −3𝑥2 + 27𝑥 − 54: black
𝑦 = −𝑥2 + 18𝑥 − 45: blue 𝑦 = 𝑥2 − 6𝑥: red 𝑦 = 𝑥2 − 25𝑥 + 150: green Question 4:
a) 𝑥 =−3± 33
2
b) 𝑥 = 1.5 𝑜𝑟 𝑥 = −1.5 c) 𝑥 = 1 𝑜𝑟 𝑥 = −3 d) 𝑥 = −4
e) 𝑥 =36± 396
18
f) 𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
g) 𝑥 =−1± 69
2
Question 6: a) Two solutions b) Two solutions c) No solutions d) Two solutions e) Two solutions f) One solution Question 7: 𝑦 = −3𝑥3 − 3: blue 𝑦 = 2𝑥3 − 3: red 𝑦 = −𝑥3 + 5𝑥2 + 48𝑥 − 252:grey 𝑦 = 𝑥3 + 12𝑥2 + 11𝑥 − 168: black
Question 8: 𝑦 = 1.5𝑥 : blue 𝑦 = 3 2𝑥 :red 𝑦 = 2−𝑥 + 2: black Question 10: 𝑦 = 4 log 𝑥: black 𝑦 = log 𝑥2: blue
𝑦 = 2 log 𝑥 − 2 + 5: red Question 11:
𝑦 =2
𝑥−5+ 5: red
𝑦 =1
𝑥+4− 6: blue
𝑦 = −1
𝑥−2− 4: black
Question 12:
a) 𝑥 =12
log 4
b) 𝑥 = −13
2 log 3
c) 𝑥 =7
log 2−3
d) 𝑥 =2
log 2−4
e) 𝑥 = 4
f) 𝑥 =18
4 ln 2
g) 𝑥 = −log 8
182+log 8
h) 𝑥 =8
1+log 2
i) 𝑥 = −25−ln 3
ln 6+ 1
Question 13:
a) 𝑥 =ln 15
ln 2
b) 𝑥 =ln
18
4
ln 3
c) 𝑥 =ln 1
ln 4+ 1
d) 𝑥 =4± 24
2
e) 𝑥 =2 ln 3+ln 2
ln 2+3 ln 3 𝑜𝑟
𝑥 =ln 18
ln 54
f) No solutions
g) 𝑥 =ln 14
14 ln 2
Question 14:
a) 𝑡140 =ln
14
12
0.011≈ 14.01𝑦
b) 𝐺𝐷𝑃10 = 𝑈𝑆$828.75𝑏
c) 𝑡1000 =ln
100
55
0.041≈ 14.58𝑦
d) 𝑡 =ln
120
55
0.03= 26.01𝑦
182
Question 15: a) 𝑝 5 = 1.457
b) 𝑡 =ln
61
42 + ln 1.05
ln 1.03+ln 1.05 𝑜𝑟
𝑡 =ln 1.525
ln 1.0815 ≈ 5.39𝑦
c) 𝑡 =ln 2.05
ln 1.0815 ≈ 9.16𝑦
Question 16: a) 472.16 b) 15,170.89 c) $17.918𝑚𝑖𝑙𝑙𝑖𝑜𝑛
d) 13𝑒𝑚𝑝𝑙𝑜𝑦𝑒𝑒𝑠
chapter five solutions
Question 1: 𝐻, 𝐺, 𝐷, 𝐶, 𝐸, 𝐴, 𝐵, 𝐹 Question 2: a) 𝑓 ′ = 18 b) 𝑓 ′ = 2𝑥 c) 𝑓 ′ = 2𝑥 + 7 d) 𝑓 ′ = 3𝑥2 + 8𝑥
e) 𝑓 ′ = −5
𝑥2
f) 𝑓 ′ = −2
𝑥3
Question 3: a) 𝑦′ = 6𝑥 b) 𝑦′ = 30𝑥 + 1 c) 𝑦′ = 2𝑥 + 3𝑥2
d) 𝑦′ =51
2𝑥2 + 10𝑥
e) 𝑦′ = −32𝑥−5 + 3𝑥2 f) 𝑦′ = −24𝑥−3 − 3𝑥−4 + 3𝑥2
g) 𝑦′ = −2
𝑥3
h) 𝑦′ = −10
𝑥3 + 2𝑥
i) 𝑦′ = −1
𝑥2 −2
𝑥3 +3
𝑥4
Question 4: a) 𝑦′ = 2 𝑥 + 1 b) 𝑦′ = 15 3𝑥 − 4 4 c) 𝑦′ = 12 𝑥 + 5 5 d) 𝑦′ = −144 1 − 4𝑥 2 e) 𝑦′ = 300 2𝑥 + 1 𝑥2 + 𝑥 3
f) 𝑦′ = −48
5 4𝑥 − 1 3
Question 5: a) 𝑦′ = 3𝑥2 15𝑥 + 2 + 15𝑥3 b) 𝑦′ = 3𝑥2 𝑥9 + 1 + 9𝑥8 𝑥3 − 1 c) 𝑦′ = 60𝑥3 1 − 𝑥3 − 45𝑥6 𝑜𝑟
𝑦′ = 60𝑥3 − 105𝑥6 d) −28𝑥−8 𝑥2 + 𝑥4 + 2𝑥 + 4𝑥3 4𝑥−7
e) 𝑦′ = −𝑥2 + 4𝑥3 𝑥15 + 15𝑥14 −1
3𝑥3 + 𝑥4
f) 𝑦′ = 1.5𝑥−7 + 2 −1 − 𝑥−3 + 3𝑥−4 𝑥−6
−4+ 2𝑥
g) 𝑦′ = −36𝑥−7 𝑥6 − 6𝑥 + 6 6𝑥5 − 6 𝑥−6 − 6 Question 6:
a) 𝑦′ = −2
𝑥−1 2
b) 𝑦′ =4
1−2𝑥 2
c) 𝑦′ = 3𝑥2−4𝑥+3
1−𝑥2 2
d) 𝑦′ = −4𝑥
𝑥2−1 2
e) 𝑦′ =−130𝑥−3+39𝑥−2+1
5−𝑥 2
f) 𝑦′ =−𝑥4−4𝑥+6𝑥2
𝑥2−2 2
g) 𝑦′ =−39𝑥−4−3𝑥2+13𝑥−8−7𝑥−2
1+𝑥−4 2
h) 𝑦′ =6𝑥 𝑥2+1
2+1−18𝑥3 𝑥2 +1
2 +3𝑥2+6𝑥 𝑥2+1
3
1−3𝑥2 2
Question 7: a) 𝑦′ = 2𝑒2𝑥
b) 𝑦′ = 5𝑒5𝑥−1 c) 𝑦′ = −4𝑒1−4𝑥
d) 𝑦′ = 2𝑥𝑒𝑥2−1
e) 𝑦′ = 13 − 2𝑥 𝑒13𝑥−𝑥2
f) 𝑦′ = 2 4𝑥3 − 3𝑥−4 𝑒𝑥4−1+𝑥−3
g) 𝑦′ = −48
𝑥2 𝑒4/𝑥
h) 𝑦′ = −2
7 −
3
𝑥4 −1
2 𝑒𝑥−3−
𝑥
2
Question 8:
a) 𝑦′ =1
𝑥
b) 𝑦′ =2
𝑥
c) 𝑦′ =10
5𝑥−1
d) 𝑦′ =−3 2𝑥−1
𝑥2−𝑥
e) 𝑦′ =4
3 −
3
𝑥4+1
𝑥−3+𝑥 𝑜𝑟
𝑦′ =−
4𝑥4 +
43
𝑥−3 + 𝑥
f) 𝑦′ =2 2𝑥+3𝑥−4−𝑥−2
𝑥2−𝑥−3+1
𝑥
g) 𝑦′ =0.04 8+4𝑥−3
8𝑥−2𝑥−2
Question 9: a) 𝑦′ = 12𝑥2 2𝑥3 − 1 3𝑥 + 5 3 + 9 3𝑥 + 5 2 2𝑥3 − 1 2 𝑜𝑟
𝑦′ = 2𝑥3 − 1 3𝑥 + 5 2 54𝑥3 + 60𝑥2 − 9 b) 𝑦′ = 5𝑒−5𝑥−1 𝑥 − 1 4 + 4 𝑥 − 1 3 1 − 𝑒−5𝑥−1 𝑜𝑟
𝑦′ = 𝑥 − 1 3 5𝑥𝑒−5𝑥−1 − 9𝑒−5𝑥−1 + 4 c) 𝑦′ = 0.6𝑥2 + 1
d) 𝑦′ =−4𝑥𝑒𝑥2
1+𝑒𝑥2
2
e) 𝑦′ =180
𝑥4 12−𝑥−3
f) 𝑦′ =−5𝑥4 1+ln 𝑥
𝑥5−1 2 +1
𝑥 𝑥5−1
g) 𝑦′ = 10𝑥𝑒3𝑥−1 + 15𝑥2𝑒3𝑥−1 𝑜𝑟 𝑦′ = 5𝑥𝑒3𝑥−1 2 + 3𝑥
h) 𝑦′ = −𝑒𝑥 2𝑥−3−𝑥4 + 6𝑥−4+4𝑥3 𝑒𝑥 +1
3 2𝑥−3−𝑥4 2
i) 𝑦′ = 42𝑥2 2𝑥3 − 1 6 1 + ln 𝑒𝑥2+ 1 +
2𝑥𝑒𝑥2
𝑒𝑥2+1
2𝑥3 − 1 7 𝑜𝑟
𝑦′ = 2𝑥 2𝑥3 − 1 6 21𝑥 1 + ln 𝑒𝑥2+ 1
+𝑒𝑥2
2𝑥3 − 1
𝑒𝑥2+ 1
j) 𝑦′ = 7𝑒7𝑥−1+4𝑥3 ln 𝑥5 +𝑥−7 −
5
𝑥−7𝑥−8 𝑒7𝑥−1+𝑥4
ln 𝑥5 +𝑥−7 2+
72𝑥3 −7𝑥6
𝑥7−1
Question 10: a) 𝑦′ = 4𝑥3 + 45𝑥2
𝑦′′ = 12𝑥2 + 90𝑥
183
b) 𝑦′ = −68𝑥−5 + 𝑥−1 𝑦′′ = 340𝑥−6 − 𝑥−2
c) 𝑦′ = 0.27𝑥8 + 8𝑥 𝑥2 − 1 3 𝑦′ ′ = 2.16𝑥7 + 8 𝑥2 − 1 3 + 48𝑥2 𝑥2 − 1 2
d) 𝑦′ = 𝑒2 −4𝑥3 + 2𝑥𝑒𝑥2−1 1 − 𝑥 − 𝑒2 1 − 𝑥2 +
𝑒𝑥2−1 𝑜𝑟
𝑦′ = 𝑒2 −4𝑥3 + 2𝑥𝑒𝑥2−1 + 4𝑥4 − 2𝑥2𝑒𝑥2−1 − 1
+ 𝑥2 − 𝑒𝑥2−1
𝑦′′ = 𝑒2 −12𝑥2 + 2𝑒𝑥2−1 + 4𝑥2𝑒𝑥2−1 + 16𝑥3
− 6𝑥𝑒𝑥2−1 − 4𝑥3𝑒𝑥2−1 + 2𝑥
e) 𝑦′ =3𝑥2−1−72𝑥
3𝑥2+1 2
𝑦′′
= 6𝑥 − 72 3𝑥2 + 1 2 − 12𝑥 3𝑥2 + 1 3𝑥2 − 1 − 72𝑥
3𝑥2 + 1 4
f) 𝑦′ =10𝑥+1
5𝑥2+𝑥
𝑦′′ =10 5𝑥2 + 𝑥 − 10𝑥 + 1 2
5𝑥2 + 𝑥 2
Question 11:
𝑓 ′ =0.25𝑥−0.75 + 4
𝑥0.25 + 4𝑥
𝑓 ′ 10 = 0.0968 4𝑑. 𝑝. 𝑓 ′ 35 = 0.0282 4𝑑. 𝑝. Question 14:
a) 𝑀𝐶 =𝑑𝑇𝐶
𝑑𝑄= 3𝑄2 − 30𝑄 + 75
b) 𝑄 = 5 c) 𝑀𝐶 ′ 15 = 60 Question 15:
a) 𝑀𝐶 =𝑑𝑇𝐶
𝑑𝑄=
30𝑥
0.1𝑥2+2+ 15
b) 𝑀𝐶 ′ =60−3𝑥2
0.1𝑥2+2 2
𝑀𝐶 ′ 30 = −0.312 3𝑑. 𝑝. 𝑀𝐶 ′ 60 = −0.082 3𝑑. 𝑝.
chapter six solutions
Question 1: Local minima: 𝐶, 𝐸 Local maxima: 𝐷, 𝐹 Global minimum: 𝐴 Global maximum: 𝐵 Question 2: a) 2𝑥 + 3 = 0 b) 45𝑥2 − 4 1 − 𝑥 3 = 0
c) 24𝑥𝑒1−𝑥2− 24𝑥3𝑒1−𝑥2
= 0 𝑜𝑟
24𝑥𝑒1−𝑥2 1 − 𝑥2 = 0
d) 4 𝑥 − 1 𝑥3 − 1 2 + 4 𝑥3 − 1 𝑥 − 1 2 = 0 𝑜𝑟 4 𝑥 − 1 𝑥3 − 1 𝑥3 + 𝑥 − 2 = 0
e) 2
𝑥+1 2 = 0
Question 3: a) −1, −4 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 b) 2, −52 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
−5,193 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 c) 1, −2𝑒4 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 d) 3,0 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
5,4
𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
e) 1
2,
1
14.5 1
2
𝑒3.5 + 19 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
− 1
2, −
1
14.5 12
𝑒3.5 + 19
𝑚𝑖𝑛𝑖𝑚𝑢𝑚
f) 1 + 2, − 2
4+2 2 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
1 − 2, 2
4 − 2 2 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
Question 4: a) No stationary points
1
3, 83
25
27 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡
b) 4+ 244
6, −27.719 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
4 − 244
6, 41.127 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
1
3, 13
13
27 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡
c) −3,0 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
−1
3, 28
4
9 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
−10
6, 14
2
9 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡
d) 1, −𝑒3 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 2, −2𝑒2 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡
e) 0,0 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
2,4
𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
2 + 2, 6 + 4 2 𝑒−1− 2 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡
2 − 2, 6 − 4 2 𝑒−1+ 2 𝑖𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡
f) 3
8,
23
16 𝑚𝑖𝑛𝑖𝑚𝑢𝑚
No inflection point Question 5: a) 𝑄 = 5 𝑎𝑛𝑑 𝑄 = 95 b) 𝑄 = 50 c) 𝜋 = 202,500 Question 6: a) 𝑇𝑅 = 350𝑄 b) 𝜋 = 350𝑄 − 0.2𝑄2 − 40
c) 𝑄𝐵𝐸 = 875 ± 122468
0.4 𝑜𝑟
𝑄𝐵𝐸 ≈ 0 𝑜𝑟 𝑄𝐵𝐸 ≈ 1750 d) 𝑄𝑚𝑎𝑥 = 875 e) 𝜋 = 153,085 Question 7: a) 𝑇𝑅 = 10 + 0.25𝑄 b) 𝜋 = 10 + 0.25𝑄 − 0.2𝑒0.01𝑄+4
c) 𝑄 =ln 125−4
0.01≈ 82.8314
d) 𝑃 = 0.371 3𝑑. 𝑝. e) 𝜋 = 5.708 3𝑑. 𝑝. Question 8: a) 𝐴𝜋 = −𝑄 + 20
𝑀𝜋 = −2𝑄 + 20
184
b) 𝐴𝐶 = 0.1𝑞 + 2.5 +10
𝑞
𝑀𝐶 = 0.2𝑞 + 2.5
c) 𝐴𝑅 =𝑄+1
𝑄0.5
𝑀𝑅 = 1.5𝑄0.5 +0.5
𝑄0.5
d) 𝐴𝐶 = 0.01𝑥 ln 𝑥 + 1
𝑀𝐶 = 0.02𝑥 ln 𝑥 + 1 +0.01𝑥2
𝑥 + 1
e) 𝐴𝜋 =−0.1 𝑒𝑄−5+𝑒5−𝑄 +10
𝑄
𝑀𝜋 = −0.1𝑒𝑄−5 + 0.1𝑒5−𝑄 f) 𝐴𝜋 = 𝑒−𝑞+1
𝑀𝜋 = 𝑒−𝑞+1 1 − 𝑞 Question 9: 𝑄 = 0 𝑜𝑟 𝑄 = 12.5 Question 10:
𝐿 = 5
2
23
≈ 1.842 3𝑑. 𝑝.
Question 11: a) 𝐴𝜋 = −1.5𝑄2 + 50𝑄
𝑀𝜋 = −4.5𝑄2 + 100𝑄
b) 𝑄 = 0 𝑜𝑟 𝑄 = 331
3
c) 𝑄 = 222
9
d) 𝜋𝑚𝑎𝑥 = 8230.45 2𝑑. 𝑝.
e) 𝑄 = 162
3
Question 12:
𝐿 = 1
20
4
≈ 0.473 3𝑑. 𝑝.
Question 13:
𝜀𝑑 = −𝑒1.96 + 180
1.68𝑒1.96≈ −15.69
Since 𝜀𝑑 > 1 the good is elastic at this price. Question 14:
𝜀𝑠 =3 100 ln 6 + 18
154≈ 3.84
Since 𝜀𝑠 > 1 this good must be elastic in supply. Question 15:
%∆𝑇𝑅 = −0.25 so 𝑇𝑅 will decrease if the owner Question 16:
𝜀𝑑 = −27
107≈ 0.252
%∆𝑄 = −633
107%
%∆𝑇𝑅 = 1874
107%
Since demand is inelastic, the increase in price will outweigh the decrease in quantity sold, so total revenue will increase. Question 17: a) 𝑇𝐶 = 2500𝑄 + 1500 b) 𝑇𝑅 = −0.5𝑄2 + 2500𝑄 c) 𝜋 = −0.5𝑄2 + 2250𝑄 − 1500 d) 𝑄𝐵𝐸 ≈ 1 𝑎𝑛𝑑 𝑄𝐵𝐸 ≈ 4499 e) 𝑄𝑚𝑎𝑥 = 2250 f) 𝜋𝑚𝑎𝑥 = 2,529,750 g) 𝑃 = 1375 h) 𝑀𝜋 = −𝑄 + 2250
i) 𝐴𝜋 = −0.5𝑄 + 2250 −1500
𝑄
j) 𝑄 = 30002
3 ≈ 208.01 k) 𝜀𝑑 = −4 ∴ 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 l) %∆𝑇𝑅 = −30% Question 18: a) 𝑇𝐶 = 80𝑄 + 200 b) 𝑇𝑅 = −0.4𝑄2 + 4276 c) 𝜋 = −0.4𝑄2 + 4196𝑄 − 200 d) 𝑄𝐵𝐸 ≈ 0 𝑎𝑛𝑑 𝑄𝐵𝐸 ≈ 10489 e) 𝑄𝑚𝑎𝑥 = 5245 f) 𝜋𝑚𝑎𝑥 = 11,003,810 g) 𝑃 = 2178 h) 𝑀𝜋 = −0.8𝑄 + 4196
i) 𝐴𝜋 = −0.4𝑄 + 4196 −200
𝑄
j) 𝑄 = 5002
3 ≈ 63.00
k) 𝜀𝑑 = −1269
400 which is elastic in demand.
l) %∆𝑇𝑅 = −1613
16%
chapter seven solutions
Question 1: a) 𝑧𝑥 = 2𝑦 + 3 + 8𝑥𝑦3
𝑧𝑦 = 2𝑥 − 1 + 12𝑥2𝑦2
b) 𝑓𝑥 = 48𝑥3 + 𝑒𝑦 𝑓𝑦 = 𝑥𝑒𝑦
c) 𝑝𝑟 = 4.5𝑟−0.7𝑚0.7 + 15𝑚 𝑝𝑚 = 10.5𝑟0.3𝑚−0.3 + 15𝑟
d) 𝑧𝑥 = 43.2𝑥1.4𝑦4.8 − 3𝑦 + 2𝑥 𝑧𝑦 = 86.4𝑥2.4𝑦3.8 − 3𝑥
e) 𝑧𝑥 = 𝑥 + 𝑦 4 + 4 𝑥 + 𝑦 3 𝑥 − 𝑦 𝑜𝑟 𝑧𝑥 = 𝑥 + 𝑦 3 5𝑥 − 3𝑦 𝑧𝑦 = − 𝑥 + 𝑦 4 + 4 𝑥 + 𝑦 3 𝑥 − 𝑦 𝑜𝑟
𝑧𝑦 = 𝑥 + 𝑦 3 3𝑥 − 5𝑦
f) 𝑓𝑥 = 𝑦 3𝑥 − 4𝑦2 4 18𝑥 − 4𝑦2 𝑓𝑦 = 𝑥 3𝑥 − 4𝑦2 4 −44𝑦2 + 3𝑥
g) 𝑧𝑥 =2𝑥−𝑥2−2𝑥𝑦−𝑦3
1−𝑥−𝑦 2
𝑧𝑦 =−3𝑦2 − 3𝑦2𝑥 + 2𝑦3 + 𝑥2
1 − 𝑥 − 𝑦 2
h) 𝑧𝑥 =11𝑦2
1−𝑥𝑦 2
𝑧𝑦 =24𝑥𝑦 − 1 − 12𝑥2𝑦2
1 − 𝑥𝑦 2
i) 𝑧𝑥 =−2𝑥𝑦2+1
1−𝑥2𝑦2+𝑥
𝑧𝑦 =−2𝑦𝑥2
1 − 𝑥2𝑦2 + 𝑥
j) 𝑧𝑥 = 4𝑒4𝑥−𝑦 𝑥3 + 4𝑥𝑦 + 3𝑥2 + 4𝑦 𝑒4𝑥−𝑦 𝑧𝑦 = −𝑒4𝑥−𝑦 𝑥3 + 4𝑥𝑦 + 4𝑥𝑒4𝑥−𝑦
k) 𝑧𝑥 = 2𝑥𝑦2𝑒4𝑥𝑦 + 4𝑦𝑒4𝑥𝑦 𝑥2𝑦2 − 1 𝑧𝑦 = 2𝑦𝑥2𝑒4𝑥𝑦 + 4𝑥𝑒4𝑥𝑦 𝑥2𝑦2 − 1
l) 𝑧𝑥 =9𝑦 𝑥𝑦−1−𝑥
𝑥𝑦−1 2 −1
15−𝑥
𝑧𝑦 =9𝑥 𝑥𝑦 − 1 − 9𝑦
𝑥𝑦 − 1 2
m) 𝑧𝑥 =−
𝑦
1+𝑥𝑦 1−𝑒𝑥𝑦 +𝑦𝑒𝑥𝑦 1−ln 1+𝑥𝑦
1−𝑒𝑥𝑦 2
185
𝑧𝑦 =−
𝑥1 + 𝑥𝑦
1 − 𝑒𝑥𝑦 + 𝑥𝑒𝑥𝑦 1 − ln 1 + 𝑥𝑦
1 − 𝑒𝑥𝑦 2
Question 2: a) 𝑧𝑥𝑥 = 2, 𝑧𝑦𝑦 = 2, 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = −3
b) 𝑧𝑥𝑥 = 2𝑦, 𝑧𝑦𝑦 = 4𝑥, 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 4𝑦 + 2𝑥
c) 𝑧𝑥𝑥 = 6𝑥𝑦4 , 𝑧𝑦𝑦 = 12𝑥3𝑦2 , 𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 12𝑥2𝑦3 + 2
d) 𝑓𝑥𝑥 = 192𝑥2𝑦8 , 𝑓𝑦𝑦 = 896𝑥4𝑦5 + 36𝑥,
𝑓𝑥𝑦 = 𝑓𝑦𝑥 = 512𝑥3𝑦7 + 36𝑦 − 13
e) 𝑧𝑥𝑥 = 4𝑒2𝑥2+𝑦 + 16𝑥2𝑒2𝑥2+𝑦
𝑧𝑦𝑦 = 𝑒2𝑥2+𝑦
𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 4𝑥𝑒2𝑥2+𝑦
f) 𝑧𝑥𝑥 = 2𝑒𝑥2+𝑦2+ 4𝑥2𝑒𝑥2+𝑦2
𝑧𝑦𝑦 = 2𝑒𝑥2+𝑦2+ 4𝑦2𝑒𝑥2+𝑦2
− 4𝑥
𝑧𝑥𝑦 = 𝑧𝑦𝑥 = 4𝑥𝑦𝑒𝑥2+𝑦2− 4𝑦
g) 𝑓𝑥𝑥 =−8𝑥2+4𝑦3
2𝑥2+𝑦3 2
𝑓𝑦𝑦 =12𝑦𝑥2 − 3𝑦4
2𝑥2 + 𝑦3 2
𝑓𝑥𝑦 = 𝑓𝑦𝑥 = −12𝑥𝑦2
2𝑥2 + 𝑦3 2
h) 𝑧𝑥𝑥 =26+2𝑥𝑦−2𝑥2−𝑦2
13−𝑥𝑦 +𝑥2 2 + 30
𝑧𝑦𝑦 = −𝑥2
13 − 𝑥𝑦 + 𝑥2 2+ 30
𝑧𝑥𝑦 = 𝑧𝑦𝑥 =−13 + 𝑥2
13 − 𝑥𝑦 + 𝑥2 2− 30
Question 3: a) Stationary point at 0,0 . SOC gives a possible
maximum. Delta test confirms this. b) Stationary point at −1.75, −0.25 . SOC gives
possible maximum. Delta test confirms this. c) Two stationary points: 𝑃𝑜𝑖𝑛𝑡 1: 0,0
𝑃𝑜𝑖𝑛𝑡 2: 1
5184
3, 24
1
5184
2
3 . SOC says the first point
is a saddle point. SOC says the second point is a possible minimum. Delta test confirms this.
d) Stationary point at 5
11,
9
11 . SOC gives a possible
maximum. Delta test confirms this.
e) Stationary point at 5
4
3,
5
3
3 . SOC gives possible
maximum. Delta test confirms this.
f) Stationary point at 8
5,
8
5 . SOC says it is an inflection
point. Question 4: a) 𝑇𝑅 = 1 + 20𝐾0.5𝐿0.6 b) 𝜋 = 20𝐾0.5𝐿0.6 − 5𝐾 − 50𝐿 − 399 c) Maximum profit at 𝐿 = 10.486, 𝐾 = 26.214. SOC
provides a possible maximum, delta test confirms this. The profit is 7.2144.
d) 𝑃 = 20.08 e) 𝑄 = 13.107 Question 5: There is a stationary point at 𝐾 = 12.8601, 𝐿 = 12.0563. The SOC gives a possible maximum, and the delta test confirms this. Question 6: 𝑑𝑧 ≈ 1.7579 and the actual change in 𝑧 is 1.7583 4𝑑. 𝑝. Question 7: 𝑑𝑅 = 2421.64 however the actual change in 𝑅 is 2409.80. Question 8: a) 𝑑𝑄 = −0.016𝐾0.4𝐿0.5 + 0.0064𝐿0.5𝐾0.4 b) Exact change in output is −1.0874 4𝑑. 𝑝. Question 9:
Maximum profit is obtained at 𝐿 =1
4, 𝐾 =
1
6. The SOC
indicates a maximum, and the delta test proves this. The maximum profit attainable is 1.198 3𝑑. 𝑝. . Question 10: a) 𝑇𝑅 = 300𝑇 + 150𝑅 b) 𝜋 = 300𝑇 + 150𝑅 − 3𝑇2 − 4𝑅2 − 𝑅𝑇 − 400
c) 𝑇 = 4741
47 𝑎𝑛𝑑 𝑅 = 12
36
47 with the SOC indicating a
maximum, and the delta test proving this. d) 𝜋𝑚𝑎𝑥 = 7738.30 Question 11: a) 𝑇𝑅 = 1100 − 90𝐿2 − 135𝐾2 + 112.5𝐿𝐾 + 360𝐿 +
540𝐾 b) 𝑇𝐶 = 50𝐿 + 35𝐾 + 550 c) 𝜋 = 550 − 90𝐿2 − 135𝐾2 + 112.5𝐿𝐾 + 310𝐿 +
505𝐾
d) 𝐿 = 3581
639 𝑎𝑛𝑑 𝐾 = 3
319
639 for maximum profit. SOC
indicates a possible maximum, and delta test confirms this.
e) 𝜋𝑚𝑎𝑥 = 2039.48 2𝑑. 𝑝. Question 12: a) 𝑇𝑅𝑅 = 𝑅2 − 4𝐴2 + 𝐴𝑅 + 1400𝑅
𝑇𝑅𝐴 = 𝐴2 − 6𝑅2 + 3𝐴𝑅 + 1200𝐴 b) 𝑇𝐶 = 650𝑅 + 350𝐴 + 10,000 c) 𝜋 = 750𝑅 + 850𝐴 − 5𝑅2 − 3𝐴2 + 4𝐴𝑅 − 10000
d) 𝑅 = 1796
11 𝑎𝑛𝑑 𝐴 = 261
4
11 with the SOC giving a
possible maximum, and the delta test confirming this. e) 𝑃𝐴 𝑑𝑜𝑚𝑒𝑠𝑡𝑖𝑐 = 1259.96 𝑎𝑛𝑑 𝑃𝑅 𝑑𝑜𝑚 𝑒𝑠𝑡𝑖𝑐 = 319.04 f) 𝜋𝑚𝑎𝑥 = 168,409.09 g) The prices don’t make sense as the cost of the
refrigerator is greater than the sales price, however due to the interdependence of the two goods, then it does make sense. That is take a small loss on one good to make a big profit on another.
chapter eight solutions
Question 1: a) 2.781% b) 6.952% c) 3.476% d) 3.418% e) 110.1
Question 2: a) 14.714% b) −11.385% c) 7.357% d) 7.105% e) −1.897%
186
f) −1.994% g) A big drop in price in between. Question 3: a) 12 b) 18 c) 54 d) 138 Question 4:
a) 𝑇15 = 1331 4
11
14
b) 𝑇21 = 1331 4
11
26
c) 𝑆11 =1331
4
11
11−1
−7
11
≈ 2091.54
d) 𝑆16 =1331
4
11
16−1
−7
11
≈ 2091.57
Question 5: a) Investment 2 (7092.56) is better than investment 1
(7000) in the fifth year. b) Investment 1 (32,500) is better than investment 2
(28,189.18) up to the fifth year. c) Investment 2 has a return of 55,494.09 whilst
investment 1 has a return of 55,000. Question 6: a) 171.60 b) 1601.60 Question 7: a) 6000
b) 10,000 Question 8: a) 5687.31 b) 7228.43 c) 12𝑦 6 𝑚𝑜𝑛𝑡𝑠 d) 5099.35 Question 9: a) 16322.67 b) 7𝑦 11𝑚𝑜𝑛𝑡𝑠 c) 14𝑦 8𝑚𝑜𝑛𝑡𝑠 Question 10: 4114.98 Question 11: Bank 2 is a better deal at the effective rate 5.125% is better than the 5.1% offered by Bank 1. Question 12: Bank of Vark has the best effective rate of 8.368%, which is better than Sun Bank (8.322%) and Universal Bank (8.349%). Question 13: a) 𝑁𝑃𝑉 = 14,313.98 b) 𝐼𝑅𝑅 ≈ 20.5% c) The project is worthwhile as 𝑁𝑃𝑉 > 0 and
𝐼𝑅𝑅 > 𝑚𝑎𝑟𝑘𝑒𝑡 𝑟𝑎𝑡𝑒 Question 14: a) 𝑁𝑃𝑉 = −122.99 and 𝐼𝑅𝑅 ≈ 8.51% b) The project is not worthwhile, as more could be
earned from the market interest rate.
187
Index A Absolute values (26) Adding fractions (8) Annual interest rates (170) Arithmetic average (162) Arithmetic series (163) Averages
arithmetic (162) geometric (162) standard (130)
B BIMDAS (18) Break-even (129)
C Compound interest (168) Cubic functions (80)
D Determinant (65-66) Differentiation
multiple variable (140) partial
applications (147,155) complex (144) simple (142) second order (146)
rules Chain (104) 𝑒 (109) ln (110) power (103) product (105) quotient (108)
single variable elasticity (133) first principles (99) introduction (97) second derivative (111)
total (148) Dividing fractions (7)
E 𝑒 (85) Elasticity
differentiation (133) interpreting (44) introduction (42) total revenue (135)
Equations (21) Exponential functions (82)
F Factorisation (25) Fractions (7-12)
G Geometric average (162) Geometric series (165) Gradients
function (112) negative (33)
H Hyperbolic functions (90)
I Index numbers (161) Indices (14) Inequalities (26) Inflection point (123) Interest rates
annual (170) compound (168) simple (167)
Internal rate of return (174) Intersecting lines (38)
J Jacobian Determinant (68)
L Linear Equations,
graphing (34) introduction (31) macroeconomic applications (40) main features (32) obtaining equations (36)
Logarithm applications (91) functions (84) graphs (88) rules (84)
M Marginal value (130) Matrix
applications (62) determinant (65-66) introduction (54) notes (61) solving 2 × 2 (55) solving 3 × 3 (59)
Multiple variable differentiation (140) Multiplying fractions (7)
N Negative numbers (6) Net present value (172) Non-linear function (74)
O Optimal point, nature (120) Optimisation
single variable application (124) graphical (118) mathematical (118)
multiple variable (151)
P Partial differentiation Profit (126)
Q Quadratic Functions
defining (75) graph (76) sketching (77)
S Simple interest (167) Simultaneous equations
introduction (50) two (50) three (52)
T Total Revenue
definition (127) elasticity (135)
V Variables, defining (13)
Z Zero (27)
