INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences 2011...
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Transcript of INTRODUCTORY MATHEMATICAL ANALYSIS For Business, Economics, and the Life and Social Sciences 2011...
INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
2011 Pearson Education, Inc.
Chapter 1 Chapter 1 Applications and More AlgebraApplications and More Algebra
2011 Pearson Education, Inc.
• To model situations described by linear or quadratic equations.
• To solve linear inequalities in one variable and to introduce interval notation.
• To model real-life situations in terms of inequalities.
• To solve equations and inequalities involving absolute values.
• To write sums in summation notation and evaluate such sums.
Chapter 1: Applications and More Algebra
Chapter ObjectivesChapter Objectives
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Chapter 1: Applications and More Algebra
Chapter OutlineChapter Outline
Applications of Equations
Linear Inequalities
Applications of Inequalities
Absolute Value
Summation Notation
1.6) Sequence
1.1)
1.2)
1.3)
1.4)
1.5)
2011 Pearson Education, Inc.
• Modeling: Translating relationships in the problems to mathematical symbols.
Chapter 1: Applications and More Algebra
1.1 Applications of Equations1.1 Applications of Equations
A chemist must prepare 350 ml of a chemical
solution made up of two parts alcohol and three
parts acid. How much of each should be used?
Example 1 - Mixture
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Solution:Let n = number of milliliters in each part.
Each part has 70 ml. Amount of alcohol = 2n = 2(70) = 140 mlAmount of acid = 3n = 3(70) = 210 ml
705
350
3505
35032
n
n
nn
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 1 - Mixture
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• Fixed cost is the sum of all costs that are independent of the level of production.
• Variable cost is the sum of all costs that are dependent on the level of output.
• Total cost = variable cost + fixed cost
• Total revenue = (price per unit) x (number of units sold)
• Profit = total revenue − total cost
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
2011 Pearson Education, Inc.
The Anderson Company produces a product for which the variable cost per unit is $6 and the fixed cost is $80,000. Each unit has a selling price of $10. Determine the number of units that must be sold for the company to earn a profit of $60,000.
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit
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Solution:
Let q = number of sold units.
variable cost = 6q
total cost = 6q + 80,000
total revenue = 10q
Since profit = total revenue − total cost
35,000 units must be sold to earn a profit of $60,000.
q
q
000,35
4000,140
000,80610000,60
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 3 – Profit
2011 Pearson Education, Inc.
A total of $10,000 was invested in two business ventures, A and B. At the end of the first year, A and B yielded returns of 6%and 5.75 %, respectively, on the original investments. How was the original amount allocated if the total amount earned was $588.75?
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment
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Solution:
Let x = amount ($) invested at 6%.
$5500 was invested at 6%
$10,000−$5500 = $4500 was invested at 5.75%.
5500
75.130025.0
75.5880575.057506.0
75.588000,100575.006.0
x
x
xx
xx
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 5 – Investment
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
A real-estate firm owns the Parklane Garden Apartments, which consist of 96 apartments. At $550 per month, every apartment can be rented. However, for each $25 per month increase, there will be three vacancies with no possibility of filling them. The firm wants to receive $54,600 per month from rent. What rent should be charged for each apartment?
2011 Pearson Education, Inc.
Solution 1:
Let r = rent ($) to be charged per apartment.
Total rent = (rent per apartment) x (number of apartments rented)
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
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Solution 1 (Con’t):
Rent should be $650 or $700.
256756
500,224050
32
000,365,13440504050
0000,365,140503
34050000,365,1
25
34050600,54
25
165032400600,54
25
550396600,54
2
2
r
rr
rr
rr
rr
rr
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
2011 Pearson Education, Inc.
Solution 2:
Let n = number of $25 increases.
Total rent = (rent per apartment) x (number of apartments rented)
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
2011 Pearson Education, Inc.
Solution 2 (Con’t):
The rent charged should be either
550 + 25(6) = $700 or
550 + 25(4) = $650.
4 or 6
046
02410
0180075075
75750800,52600,54
39625550600,54
2
2
2
n
nn
nn
nn
nn
nn
Chapter 1: Applications and More Algebra
1.1 Applications of Equations
Example 7 – Apartment Rent
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities1.2 Linear Inequalities
• Supposing a and b are two points on the real-number line, the relative positions of two points are as follows:
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• We use dots to indicate points on a number line.
• Suppose that a < b and x is between a and b.
• Inequality is a statement that one number is less than another number.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
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• Rules for Inequalities:
1. If a < b, then a + c < b + c and a − c < b − c.
2. If a < b and c > 0, then ac < bc and a/c < b/c.
3. If a < b and c < 0, then a(c) > b(c) and a/c > b/c.
4. If a < b and a = c, then c < b.
5. If 0 < a < b or a < b < 0, then 1/a > 1/b .
6. If 0 < a < b and n > 0, then an < bn.
If 0 < a < b, then .
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
nn ba
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
• Linear inequality can be written in the form
ax + b < 0where a and b are constants and a 0
• To solve an inequality involving a variable is to find all values of the variable for which the inequality is true.
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Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
Example 1 – Solving a Linear Inequality
Solve 2(x − 3) < 4.
Solution:Replace inequality by equivalent inequalities.
5 2
10
2
2
102
64662
462
432
x
x
x
x
x
x
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
1.2 Linear Inequalities
Example 3 – Solving a Linear Inequality
Solve (3/2)(s − 2) + 1 > −2(s − 4).
7
20
207
16443
442232
422122
32
42122
3
s
s
ss
ss
s-s
ss
The solution is ( 20/7 ,∞).
Solution:
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Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities1.3 Applications of Inequalities
Example 1 - Profit
• Solving word problems may involve inequalities.
For a company that manufactures aquarium heaters, the combined cost for labor and material is $21 per heater. Fixed costs (costs incurred in a given period, regardless of output) are $70,000. If the selling price of a heater is $35, how many must be sold for the company to earn a profit?
2011 Pearson Education, Inc.
Solution:
profit = total revenue − total cost
5000
000,7014
0000,702135
0 cost total revenue total
q
q
Let q = number of heaters sold.
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 1 - Profit
2011 Pearson Education, Inc.
After consulting with the comptroller, the president of the Ace Sports Equipment Company decides to take out a short-term loan to build up inventory. The company has current assets of $350,000 and current liabilities of $80,000. How much can the company borrow if the current ratio is to be no less than 2.5? (Note: The funds received are considered as current assets and the loan as a current liability.)
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio
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Solution:
Let x = amount the company can borrow.
Current ratio = Current assets / Current liabilities
We want,
The company may borrow up to $100,000.
x
x
xx
x
x
000,100
5.1000,150
000,805.2000,350
5.2000,80
000,350
Chapter 1: Applications and More Algebra
1.3 Applications of Inequalities
Example 3 – Current Ratio
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• On real-number line, the distance of x from 0 is called the absolute value of x, denoted as |x|.
DEFINITIONThe absolute value of a real number x, written |x|, is defined by
0 if ,
0 if ,
xx
xxx
Chapter 1: Applications and More Algebra
1.4 Absolute Value1.4 Absolute Value
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 1 – Solving Absolute-Value Equations
a. Solve |x − 3| = 2
b. Solve |7 − 3x| = 5
c. Solve |x − 4| = −3
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Solution:a. x − 3 = 2 or x − 3 = −2 x = 5 x = 1
b. 7 − 3x = 5 or 7 − 3x = −5 x = 2/3 x = 4
c. The absolute value of a number is never negative. The solution set is .
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 1 – Solving Absolute-Value Equations
2011 Pearson Education, Inc.
Absolute-Value Inequalities
• Summary of the solutions to absolute-value inequalities is given.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 3 – Solving Absolute-Value Equations
a. Solve |x + 5| ≥ 7
b. Solve |3x − 4| > 1
Solution:a.
We write it as , where is the union symbol.
b.
We can write it as .
2 12
75 or 75
xx
xx
,212,
3
5 1
143 or 143
xx
xx
,3
51,
2011 Pearson Education, Inc.
Properties of the Absolute Value
• 5 basic properties of the absolute value:
• Property 5 is known as the triangle inequality.
baba
aaa
abba
b
a
b
a
baab
.5
.4
.3
.2
.1
Chapter 1: Applications and More Algebra
1.4 Absolute Value
2011 Pearson Education, Inc.
Chapter 1: Applications and More Algebra
1.4 Absolute Value
Example 5 – Properties of Absolute Value
323251132 g.
222 f.
5
3
5
3
5
3 e.
3
7
3
7
3
7 ;
3
7
3
7
3
7 d.
77 c.
24224 b.
213737- a.
-
xxx
xx
Solution:
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Chapter 1: Applications and More Algebra
1.5 Summation Notation1.5 Summation Notation
DEFINITION
The sum of the numbers ai, with i successively taking on the values m through n is denoted as
nmmm
n
mii aaaaa
...21
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Evaluate the given sums.
a. b.
Solution:
a.
b.
Chapter 1: Applications and More Algebra
1.5 Summation Notation
Example 1 – Evaluating Sums
7
3
25n
n
6
1
2 1j
j
115
3328231813
275265255245235257
3
n
n
97
3726171052
1615141312111 2222226
1
2
j
j
2011 Pearson Education, Inc.
• To sum up consecutive numbers, we have
where n = the last number
2
1
1
nni
n
i
Chapter 1: Applications and More Algebra
1.5 Summation Notation
6
)12(1
1
2
nnni
n
i
2011 Pearson Education, Inc.
Evaluate the given sums.
a. b. c.
Solution:
a.
b.
c.
550,2510032
1011005 3535
100
1
100
1
100
1
kkk
kk
300,180,246
401201200999
200
1
2200
1
2
kk
kk
Chapter 1: Applications and More Algebra
1.5 Summation Notation
Example 3 – Applying the Properties of Summation Notation
2847144471
1
100
30
ij
100
1
35k
k
200
1
29k
k
100
30
4j
2011 Pearson Education, Inc.
1.6 Sequence
• Arithmetic sequenceAn arithmetic sequence is a sequence (bk) defined
recursively by b1=a and, for each positive integer k, bk+1= d + bk
Example
1.5, 1.5+0.7 , 1.5+2*0.7, 1.5 +3*0.7, 1.5+ 4*0.7, 1.5+5*0.7
1.5, 2.2, 2.9, 3.6, 4.3, 5.0
• Geometric sequenceA geometric sequence is a sequence (ck) defined
recursively by
c1=a and, for each positive integer k, ck+1= ck*r
Example
2 2*3 , 2*3*3, 2*3*3*3, 2*3*3*3*3
2, 6, 18, 48, 144
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Sums of sequences
• Sum of an arithmetic sequence - first n term
First term – a, common difference – d
• Sum of an geometric sequence
First term – a, common ratio – r
- Sum to first n term
- Sum of an infinite geometric sequence
for
)2)1((2
adnn
sn
1,1
)1(
rforr
ras
n
n
1
1
1,1
i
i
r
aarr
2011 Pearson Education, Inc.
• Example 1
A rich woman would like to leave $100,000 a year, starting now, to be divided equally among all her direct descendants. She puts no time limit on his bequeathment and is able to invest for this long-term outlay of funds at 2% compounded annually. How much must she invest now to meet such a long-term commitment?
2011 Pearson Education, Inc.
• Solution:
Let us write R=100,000, set the clock to 0 now, and measure times is years from now. With these conventions we are to account for payments for R payments of R at times 0,1,2…..k,.. By making a single investment now. Then the investment must equal to the sum
....)02.1(......)02.1()02.1( 21 kRRRR
First term=a=R=100,000
Common ratio=r=(1.02)-1. Since |r|<1, we can evaluate the required investment as
000,100,5
02.11
1
000,100
1
ra