Lecture 5

Basic Calculus for Economists

Analyzing a Limit

We can examine what occurs at a particular point by the limit. Using the function f (x) = 2x – 1, let’s examine what happens near x = 2 through the following chart:

We see that as x approaches 2, f (x) approaches 3.

x 1.5 1.9 1.99 1.999 2 2.001 2.01 2.1 2.5

f (x) 2 2.8 2.98 2.998 ? 3.002 3.02 3.2 4

Limits

In limit notation we have

.312lim2

xx

Definition: We write

Lxfcx

)(lim

or as x c, then f (x) L,

if the functional value of f (x) is close to the single real number L whenever x is close to, but not equal to, c (on either side of c).

2

3

One-Sided Limits

■ We write

and call K the limit from the left (or left-hand limit) if f (x) is close to K whenever x is close to c, but to the left of c on the real number line.

■ We write

and call L the limit from the right (or right-hand limit) if f (x) is close to L whenever x is close to c, but to the right of c on the real number line.

■ In order for a limit to exist, the limit from the left and the limit from the right must exist and be equal.

Kxfcx

)(lim

Lxfcx

)(lim

Limit Properties

Let f and g be two functions, and assume that the following two limits exist and are finite:

Then the limit of the sum of the functions is equal to the sum of the limits. the limit of the difference of the functions is equal to the difference of the limits.

MxgLxfcxcx

)(limand)(lim

Limit Properties(continued)

the limit of a constant times a function is equal to the constant times the limit of the function.

the limit of the product of the functions is the product of the limits of the functions.

the limit of the quotient of the functions is the quotient of the limits of the functions, provided M 0.

the limit of the nth root of a function is the nth

root of the limit of that function.

Examples

13

8

13lim

2lim

13

2lim

4

4

4

x

x

x x

x

x

x

2 2

2 2 2lim 3 lim lim3 4 6 2x x x

x x x x

From these examples we conclude that

1.lim ( ) ( )

2.lim ( ) ( )x c

x c

f x f c

r x r c

f any polynomial function

r any rational function with a nonzero denominator at x = c

If and , then

is said to be indeterminate.

The term “indeterminate” is used because the limit may or may not exist.

Indeterminate Forms

0)(lim

xfcx )(

)(lim

xg

xfcx

0)(lim

xgcx

It is important to note that there are restrictions on some of the limit properties. In particular if 0)(lim

xr

cx

then finding may present difficulties, since the

denominator is 0.

( )lim

( )x c

f x

r x

Example

This example illustrates some techniques that can be useful for indeterminate forms.

2

2 2 2

4 ( 2)( 2)lim lim lim( 2) 4

2 2x x x

x x xx

x x

Algebraic simplification is often useful when the numerator and denominator are both approaching 0.

Let f (x) = 3x - 1. Find

Solution:

Difference Quotients

.)()(

lim0 h

afhafh

1331)(3)( hahahaf

13)( aaf

hafhaf 3)()(

.33

lim)()(

lim00

h

h

h

afhafhh

Definition of Continuity

A function f is continuous at a point x = c if

1.

2. f (c) exists

3.

A function f is continuous on the open interval (a,b) if it is continuous at each point on the interval.

If a function is not continuous, it is discontinuous.

)()(lim cfxfcx

exists)(lim xfcx

f (x) = x – 1 at x = 2.

1. The limit exists!

2. f(2) = 1

3.

Therefore this function is continuous at x = 2.

Example 1

)2(11lim2

fxx

2

1

11lim2

xx

Example 2

f (x) = (x2 – 9)/(x + 3) at x = -3

1.

The limit exists (reduce the fraction).

2. f (-3) = 0/0 is undefined!

3.

The function is not continuous at x = -3. (Graph should have an open circle there.)

63

9lim

2

3

x

xx

)3(3

9lim

2

3

f

x

xx

Example 3 f (x) = |x|/x at x = 0 and at x = 1.

1. Does not exist!

2. f (0) = 0/0 Undefined!

3.

The function is not continuous at x = 0.

This function is continuous at x = 1.

x

xx 0lim

)0(lim0

fx

xx

Limits in Mathematica

Continuity Properties

If two functions are continuous on the same interval, then their sum, difference, product, and quotient are continuous on the same interval, except for values of x that make the denominator 0.

Examples of Continuous Functions

A constant function is continuous for all x. For integer n > 0, f (x) = xn is continuous for all x. A polynomial function is continuous for all x. A rational function is continuous for all x, except those

values that make the denominator 0. For n an odd positive integer, is continuous

wherever f (x) is continuous. For n an even positive integer, is continuous

wherever f (x) is continuous and nonnegative.

n xf )(

n xf )(

Infinite LimitsThere are various possibilities under which

does not exist. For example, if the one-sided limits are different at x = a, then the limit does not exist.

Another situation where a limit may fail to exist involves functions whose values become very large as x approaches a. The special symbol (infinity) is used to describe this type of behavior.

)(lim xfax

To illustrate this case, consider the function f (x) = 1/(x-1), which is discontinuous at x = 1. As x approaches 1 from the right, the values of f (x) are positive and become larger and larger. That is, f (x) increases without bound. We write this symbolically as

Since is not a real number, the limit above does not actually exist. We are using the symbol (infinity) to describe the manner in which the limit fails to exist, and we call this an infinite limit.

1( ) as 1

1f x x

x

Limit 1x 1

, x 1, Direction 1

Limit 1x 1

, x Infinity0

Example(continued)As x approaches 1 from the left, the values of f (x) are negative and become larger and larger in absolute value. That is, f (x) decreases through negative values without bound. We write this symbolically as

1( ) as 1

1f x x

x

The graph of this function is as shown:

Note that does not exist. 1

1lim

1 xx

Limit 1x 1

, x 1, Direction 1

In[3]:= Limit 1x 1

, x 1, Direction 1Out[3]=

Infinite Limits and Vertical Asymptotes

Definition:

The vertical line x = a is a vertical asymptote for the graph of y = f (x) if f (x) or f (x) - as x a+ or x a–.

That is, f (x) either increases or decreases without bound as x approaches a from the right or from the left.

Note: If any one of the four possibilities is satisfied, this makes x = a a vertical asymptote. Most of the time, the limit will be infinite (+ or -) on both sides, but it does not have to be.

Vertical Asymptotesof Polynomials

How do we locate vertical asymptotes? If a function f is continuous at x = a, then

Since all of the above limits exist and are finite, f cannot have a vertical asymptote at x = a. In order for f to have a vertical asymptote at x = a, at least one of the limits above must be an infinite limit, and f must be discontinuous atx = a. We know that polynomial functions are continuous for all real numbers, so a polynomial has no vertical asymptotes.

)()(lim)(lim)(lim afxfxfxfaxaxax

Vertical Asymptotes of Rational FunctionsSince a rational function is discontinuous only at the zeros of its denominator, a vertical asymptote of a rational function can occur only at a zero of its denominator. The following is a simple procedure for locating the vertical asymptotes of a rational function:

If f (x) = n(x)/d(x) is a rational function, d(c) = 0 and n(c) 0, then the line x = c is a vertical asymptote of the graph of f.

However, if both d(c) = 0 and n(c) = 0, there may or may not be a vertical asymptote at x = c.

Example

Let

Describe the behavior of f at each point of discontinuity. Use and - when appropriate. Identify all vertical asymptotes.

1

22

2

x

xxxf

Example(continued)

Let

Describe the behavior of f at each point of discontinuity. Use and - when appropriate. Identify all vertical asymptotes.

Solution: Let n(x) = x2 + x - 2 and d(x) = x2 - 1. Factoring the denominator, we see that d(x) = x2 - 1 = (x+1)(x-1) has two zeros, x = -1 and x = 1. These are the points of discontinuity of f.

1

22

2

x

xxxf

Example(continued)

Since d(-1) = 0 and n(-1) = -2 0, the theorem tells us that the line x = -1 is a vertical asymptote.

Now we consider the other zero of d(x), x = 1. This time n(1) = 0 and the theorem does not apply. We use algebraic simplification to investigate the behavior of the function at x = 1:

Since the limit exists as x approaches 1, f does not have a vertical asymptote at x = 1. The graph of f is shown on the next slide. 2

3

)1)(1(

)2)(1(lim

1

2lim)(lim

1

2

2

11

xx

xxx

xxxf

x

xx

Example(continued)

2

2

2( )

1

x xf x

x

Vertical Asymptote

Point of discontinuity

We begin our consideration of limits at infinity by considering power functions of the form x p and 1/x p, where p is a positive real number.

If p is a positive real number, then x p increases as x increases, and it can be shown that there is no upper bound on the values of x p. We indicate this by writing

or

Limits at Infinity ofPower Functions

.lim

p

xx

xx p as

In[5] := fx : x2

gx : 1

x2

In[7] := Limitfx, x InfinityLimitgx, x Infinity

Out[7]=

Out[8]= 0

Since the reciprocals of very large numbers are very small numbers, it follows that 1/x p approaches 0 as x increases without bound. We indicate this behavior by writing

or

This figure illustrates this behavior for f (x) = x2 and g(x) = 1/x2.

Power Functions (continued)

xx p

as01

.01

lim px x

0)(lim

)(lim

xg

xf

x

x

Power Functions (continued)In general, if p is a positive real number and k is a nonzero real number, then

defined isit iflim

lim

0limlim

p

x

p

x

pxpx

kx

kx

x

k

x

k

Note: k and p determine whether the limit at is or -.

The last limit is only defined if the pth power of a negative number is defined. This means that p has to be an integer, or a rational number with odd denominator.

Limits at Infinity of Polynomial Functions

What about limits at infinity for polynomial functions?

As x increases without bound in either the positive or the negative direction, the behavior of the polynomial graph will be determined by the behavior of the leading term (the highest degree term). The leading term will either become very large in the positive sense or in the negative sense (assuming that the polynomial has degree at least 1). In the first case the function will approach and in the second case the function will approach -.

In mathematical shorthand, we write this asThis covers all possibilities.

)(lim xfx

Limits at Infinity andHorizontal Asymptotes

A line y = b is a horizontal asymptote for the graph of y = f (x) if f (x) approaches b as either x increases without bound or decreases without bound. Symbolically, y = b is a horizontal asymptote if

In the first case, the graph of f will be close to the horizontal line y = b for large (in absolute value) negative x. In the second case, the graph will be close to the horizontal line y = b for large positive x.

Note: It is enough if one of these conditions is satisfied, but frequently they both are.

bxfbxfxx

)(limor)(lim

Example

This figure shows the graph of a function with two horizontal asymptotes, y = 1 and y = -1.

There are three possible cases for these limits.

1. If m < n, then

The line y = 0 (x axis) is a horizontal asymptote for f (x).

2. If m = n, then

The line y = am/bn is a horizontal asymptote for f (x) .

3. If m > n, f (x) does not have a horizontal asymptote.

Horizontal Asymptotes of Rational Functions

0,0,)(01

11

011

1

nmnn

nn

mm

mm ba

bxbxbxb

axaxaxaxf

0)(lim

xfx

nn

mm

xx xb

xaxf

lim)(lim

If

then

m

m

x b

axf

)(lim

1000 500 500 1000

1 .49

1 .50

1 .51

1 .52

Horizontal Asymptotes of Rational Functions (continued)

Notice that in cases 1 and 2 on the previous slide that the limit is the same if x approaches or -. Thus a rational function can have at most one horizontal asymptote. (See figure). Notice that the numerator and denominator have the same degree in this example, so the horizontal asymptote is the ratio of the leading coefficients of the numerator and denominator.

2

2

3 5 9

2 7

x xy

x

y = 1.5

Example

Find the horizontal asymptotes of each function.

4 2

6

3 1a.) ( )

8 10

x xf x

x

5

3

2 1b.) ( )

7

xf x

x

Example Solution

Find the horizontal asymptotes of each function.

4 2

6

3 1a.) ( )

8 10

x xf x

x

Since the degree of the numerator is less than the degree of the denominator in this example, the horizontal asymptote is y = 0 (the x axis).

5

3

2 1b.) ( )

7

xf x

x

Since the degree of the numerator is greater than the degree of the denominator in this example, there is no horizontal asymptote.

Summary

An infinite limit is a limit of the form

(y goes to infinity). It is the same as a vertical asymptote (as long as a is a finite number).

A limit at infinity is a limit of the form

(x goes to infinity). It is the same as a horizontal asymptote (as long as L is a finite number).

)(limor,)(lim,)(lim xfxfxfaxaxax

Lxfx

)(lim

The Rate of Change

For y = f (x), the average rate of change from x = a to x = a + h is

0,)()(

hh

afhaf

The above expression is also called a difference quotient. See Chiang 6.1. It can be interpreted as the slope of a secant.

See the picture on the next slide for illustration.

h

f (a + h) – f (a)

slope

Visual Interpretation

P

Q h

afhaf )()(

Average rate of change = slope of the secant line through P and Q

Example 1

The revenue generated by producing and selling widgets is given by R(x) = x (75 – 3x) for 0 x 20.

What is the change in revenue if production changes from 9 to 12?

Example 1

The revenue generated by producing and selling widgets is given by R(x) = x (75 – 3x) for 0 x 20.

What is the change in revenue if production changes from 9 to 12?

R(12) – R(9) = $468 – $432 = $36.

Increasing production from 9 to 12 will increase revenue by $36.

Example 1 (continued)

The revenue is R(x) = x (75 – 3x) for 0 x 20.

What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12?

Example 1 (continued)

The revenue is R(x) = x (75 – 3x) for 0 x 20.

What is the average rate of change in revenue (per unit change in x) if production changes from 9 to 12?

To find the average rate of change we divide the change in revenue by the change in production:

Thus the average change in revenue is $12 when production is increased from 9 to 12.

Like Change in Y over change in X.

123

36

912

)9()12(

RR

The Instantaneous Rate of ChangeConsider the function y = f (x) only near the point P = (a, f (a)).

The difference quotient

gives the average rate of change of f over the interval [a, a+h].

If we make h smaller and smaller, in the limit we obtain the instantaneous rate of change of the function at the point P:

h

afhafh

)()(lim

0

0,)()(

hh

afhaf

h

f (a + h) – f (a)

Tangent

Visual Interpretation

P

Let h approach 0

h

afhaf

h

)()(

0

lim

Slope of tangent =instantaneous rate of change.

Q

Given y = f (x), the instantaneous rate of change at x = a is

provided that the limit exists. It can be interpreted as the slope of the tangent at the point (a, f (a)). See illustration on previous slide.

h

afhafh

)()(lim

0

Instantaneous Rate of Change

The Derivative

For y = f (x), we define the derivative of f at x, denoted f ’ (x), to be

h

xfhxfxf

h

)()(lim)('

0

if the limit exists.

If f ’(a) exists, we call f differentiable at a.

If f ’(x) exist for each x in the open interval (a, b), then f is said to be differentiable over (a, b).

Interpretations of the Derivative

If f is a function, then f ’ is a new function with the following interpretations:

■ For each x in the domain of f ’, f ’ (x) is the slope of the line tangent to the graph of f at the point (x, f (x)).

■ For each x in the domain of f ’, f ’ (x) is the instantaneous rate of change of y = f (x) with respect to x.

■ If f (x) is the position of a moving object at time x, then v = f ’ (x) is the velocity of the object at that time.

Finding the Derivative

To find f ‘ (x), we use a four-step process:

Step 1. Find f (x + h)

Step 2. Find f (x + h) – f (x)

Step 3. Find

Step 4. Find

h

xfhxf )()(

h

xfhxfh

)()(lim

0

Find the derivative of f (x) = x 2 – 3x.

Step 1. f (x + h) = (x + h)2 – 3(x + h) = x2 + 2xh + h2 – 3x – 3h

Step 2. Find f (x + h) – f (x) = 2xh + h2 – 3h

Step 3. Find

Step 4. Find

Example 2

3232lim)()(

lim00

xhx

h

xfhxfhh

3232)()( 2

hxh

hhxh

h

xfhxf

Example 3

Find the slope of the tangent to the graph of f (x) = x 2 – 3x at x = 0, x = 2, and x = 3.

Solution: In example 2 we found the derivative of this function at x to be

f ’ (x) = 2x – 3Hence

f ’ (0) = -3

f ’ (2) = 1, and

f ’ (3) = 3In[15]:= Limit x h 2 3x h x2 3x

h, h 0

Out[15]= 3 2 x

Example 4

We know that the derivative of f (x) = x 2 – 3x is f ’ (x) = 2x – 3.

Verify this for x = 2 using Mathematica.

D[] is the derivative function

tangent equation

Example 5

Find the derivative of f (x) = 2x – 3x2 using Mathematica with a symbolic differentiation routine.

In[21]:= D2x 3x2, xOut[21]= 2 6 x

Find the derivative of f (x) = 2x – 3x2 using the four-step process.

Step 1. f (x + h) = 2(x + h) – 3(x + h)2

Step 2. f (x + h) – f (x) = 2h – 6xh - 3h2

Step 3.

Step 4.

Example 6

hxh

hxhx

h

xfhxf362

362)()( 2

xhxh

62362lim0

Nonexistence of the Derivative

The existence of a derivative at x = a depends on the existence of the limit

h

afhafaf

h

)()(lim)('

0

If the limit does not exist, we say that the function is nondifferentiable at x = a, or f ’ (a) does not exist.

Nonexistence of the Derivative(continued)

Some of the reasons why the derivative of a function may not exist at x = a are

■ The graph of f has a hole or break at x = a, or

■ The graph of f has a sharp corner at x = a, or

■ The graph of f has a vertical tangent at x = a.

Summary■ For y = f (x), we defined the derivative of f at x, denoted

f ’ (x), to be

if the limit exists.

■ We have seen how to find the derivative algebraically, using the four-step process.

h

xfhxfxf

h

)()(lim)('

0

Derivative Notation

In the preceding section we defined the derivative of a function. There are several widely used symbols to represent the derivative. Given y = f (x), the derivative may be represented by any of the following:

■ f ’(x)

■ y’

■ dy/dx

Example 1 (continued)

What is the slope of a constant function?

The graph of f (x) = C is a horizontal line with slope 0, so we would expect f ’(x) = 0.

Theorem 1. Let y = f (x) = C be a constant function, then

y’ = f ’(x) = 0.

Power Rule

A function of the form f (x) = xn is called a power function. This includes f (x) = x (where n = 1) and radical functions (fractional n).

Theorem 2. (Power Rule) Let y = xn be a power function, then

y’ = f ’(x) = n xn – 1.

THEOREM 2 IS VERY IMPORTANT. IT WILL BE USED A LOT!

Example 2

Differentiate f (x) = x5.

Solution:

By the power rule, the derivative of xn is n xn–1.

In our case n = 5, so we get f ’(x) = 5 x4.

Differentiate

Solution:

Rewrite f (x) as a power function, and apply the power rule:

Example 3

3/1)( xxf

.)( 3 xxf

3 2

3/2

3

1

3

1)('

xxxf

Constant Multiple Property

Theorem 3. Let y = f (x) = k u(x) be a constant k times a function u(x). Then

y’ = f ’(x) = k u’(x).

In words: The derivative of a constant times a function is the constant times the derivative of the function.

Example 4

Differentiate f (x) = 7x4.

Solution:

Apply the constant multiple property and the power rule.

f ’(x) = 7(4x3) = 28 x3.

Sum and Difference Properties

Theorem 5. If y = f (x) = u(x) ± v(x),

then y’ = f ’(x) = u’(x) ± v’(x).

In words:■ The derivative of the sum of two differentiable functions is the sum of the derivatives.■ The derivative of the difference of two differentiable functions is the difference of the derivatives.

Differentiate f (x) = 3x5 + x4 – 2x3 + 5x2 – 7x + 4.

Solution:

Apply the sum and difference rules, as well as the constant multiple property and the power rule.

f ’(x) = 15x4 + 4x3 – 6x2 + 10x – 7.

Example 5

Applications

Remember that the derivative gives the instantaneous rate of change of the function with respect to x. That might be:

■ Instantaneous velocity.

■ Tangent line slope at a point on the curve of the function.

■ Marginal Cost. If C(x) is the cost function, that is, the total cost of producing x items, then C’(x) approximates the cost of producing one more item at a production level of x items. C’(x) is called the marginal cost.

Tangent Line Example

Let f (x) = x4 - 6x2 + 10.

(a) Find f ’(x)

(b) Find the equation of the tangent line at x = 1

Tangent Line Example (continued)Let f (x) = x4 - 6x2 + 10.

(a) Find f ’(x)

(b) Find the equation of the tangent line at x = 1

Solution:

(a) f ’(x) = 4x3 - 12x

(b) Slope: f ’(1) = 4(13) - 12(1) = -8.Point: If x = 1, then y = f (1) = 1 - 6 + 10 = 5. Point-slope form: y - y1 = m(x - x1)

y - 5 = -8(x -1) y = -8x + 13

T he other w ay to do this ... allow s y ou to fill in values for t he funct ion

In[23]:= fx : x4 6x2 10

In[24]:= f1Out[24]= 5

You can jus t use f P R IM E of x t o find the derivat ive. If y ou w ant t o evaluat e it at a p oint , y ou jus t rep lace t he x w ith t he value y ou are int eres t ed in .

In[25]:= f'xOut[25]= 12 x 4 x3

In[26]:= f'1Out[26]= 8

T hus , at a value of x= 1, f(x) = 5, and the s lop e (or f'(x) is -8 .

Application Example

The total cost (in dollars) of producing x portable radios per day is

C(x) = 1000 + 100x – 0.5x2

for 0 ≤ x ≤ 100.

1. Find the marginal cost at a production level of x radios.

Example (continued)

The total cost (in dollars) of producing x portable radios per day is

C(x) = 1000 + 100x – 0.5x2

for 0 ≤ x ≤ 100.

1. Find the marginal cost at a production level of x radios.

Solution: The marginal cost will be

C’(x) = 100 – x.

Example (continued)

2. Find the marginal cost at a production level of 80 radios and interpret the result.

Solution: C’(80) = 100 – 80 = 20.

It will cost approximately $20 to produce the 81st radio.

3. Find the actual cost of producing the 81st radio and compare this with the marginal cost.

Solution: The actual cost of the 81st radio will be

C(81) – C(80) = $5819.50 – $5800 = $19.50.

This is approximately equal to the marginal cost.

Example (continued): for you

4. If the price of the radio is $11.00, how many radios do you decide to produce?

5. Calculate the firm’s total profits.

6. Calculate the firm’s break-even point

7. Show and plot the firm’s profit function

Summary

If f (x) = C, then f ’(x) = 0

If f (x) = xn, then f ’(x) = n xn-1

If f (x) = ku(x), then f ’ (x) = ku’(x)

If f (x) = u(x) ± v(x), then f ’(x) = u’(x) ± v’(x).

Increments

In a previous section we defined the derivative of f at x as the limit of the difference quotient:

Increment notation will enable us to interpret the numerator and the denominator of the difference quotient separately.

h

xfhxfxf

h

)()(lim)('

0

ExampleLet y = f (x) = x3. If x changes from 2 to 2.1, then y will change from y = f (2) = 8 to y = f (2.1) = 9.261.

We can write this using increment notation. The change in x is called the increment in x and is denoted by x. is the Greek letter “delta”, which often stands for a difference or change. Similarly, the change in y is called the increment in y and is denoted by y.

In our example,

x = 2.1 – 2 = 0.1

y = f (2.1) – f (2) = 9.261 – 8 = 1.261.

Graphical Illustration of IncrementsFor y = f (x)

x = x2 - x1 y = y2 - y1

x2 = x1 + x = f (x2) – f (x1) = f (x1 + x) – f (x1)

(x1, f (x1))

(x2, f (x2))

x1 x2

x

y■ y represents the

change in y corresponding to a x change in x.

■ x can be either positive or negative.

Assume that the limit exists.

For small x,

Multiplying both sides of this equation by x gives us

y f ’(x) x.

Here the increments x and y represent the actual changes in x and y.

Differentials

x

yxf

x

0

lim)('

x

yxf

)('

One of the notations for the derivative is

If we pretend that dx and dy are actual quantities, we get

We treat this equation as a definition, and call dx and dy differentials.

Differentials (continued)

dx

dyxf )('

dxxfdy )('

x and dx are the same, and represent the change in x.

The increment y stands for the actual change in y resulting from the change in x.

The differential dy stands for the approximate change in y, estimated by using derivatives.

In applications, we use dy (which is easy to calculate) to estimate y (which is what we want).

Interpretation of Differentials

dxxfdyy )('

Example 1

Find dy for f (x) = x2 + 3x and evaluate dy for x = 2 and dx = 0.1.

Solution:

dy = f ’(x) dx = (2x + 3) dx

When x = 2 and dx = 0.1, dy = [2(2) + 3] 0.1 = 0.7.

Example 2 Cost-Revenue

A company manufactures and sells x transistor radios per week. If the weekly cost and revenue equations are

find the approximate changes in revenue and profit if production is increased from 2,000 to 2,010 units/week.

000,80

000,110)(

2000,5)(2

x

xxxR

xxC

The profit is

We will approximate R and P with dR and dP, respectively, using x = 2,000 and dx = 2,010 – 2,000 = 10.

Example 2 Solution

000,5000,1

8)()()(2

x

xxCxRxP

per week60$10)500

000,210(

)500

10()('

dxx

dxxRdR

per week40$10)500

000,28(

)500

8()('

dxx

dxxPdP

Marginal Cost

Remember that marginal refers to an instantaneous rate of change, that is, a derivative.

Definition:

If x is the number of units of a product produced in some time interval, then

Total cost = C(x)

Marginal cost = C’(x)

Marginal Revenue andMarginal Profit

Definition:If x is the number of units of a product sold in some time interval, then

Total revenue = R(x) Marginal revenue = R’(x)

If x is the number of units of a product produced and sold in some time interval, then

Total profit = P(x) = R(x) – C(x)Marginal profit = P’(x) = R’(x) – C’(x)

Marginal Cost and Exact Cost

Assume C(x) is the total cost of producing x items. Then the exact cost of producing the (x + 1)st item is

C(x + 1) – C(x).

The marginal cost is an approximation of the exact cost.

C’(x) ≈ C(x + 1) – C(x).

Similar statements are true for revenue and profit.

ExampleThe total cost of producing x electric guitars is

C(x) = 1,000 + 100x – 0.25x2.

1. Find the exact cost of producing the 51st guitar.

The exact cost is C(x + 1) – C(x).

C(51) – C(50) = 5,449.75 – 5375 = $74.75.

2. Use the marginal cost to approximate the cost of producing the 51st guitar.

The marginal cost is C’(x) = 100 – 0.5x

C’(50) = $75.

Marginal Average Cost

Definition:

If x is the number of units of a product produced in some time interval, then

Average cost per unit =

Marginal average cost =

x

xCxC

)()(

)()(' xCdx

dxC

If x is the number of units of a product sold in some time interval, then

Average revenue per unit =

Marginal average revenue =

If x is the number of units of a product produced and sold in some time interval, then

Average profit per unit =

Marginal average profit =

Marginal Average Revenue Marginal Average Profit

x

xRxR

)()(

)()(' xRdx

dxR

x

xPxP

)()(

)()(' xPdx

dxP

Warning!

To calculate the marginal averages you must calculate the average first (divide by x), and then the derivative. If you change this order you will get no useful economic interpretations.

STOP

Example 2

The total cost of printing x dictionaries is $20,000 to start your business, and there are $10 in variable costs: Thus,

C(x) = 20,000 + 10x

1. Find the average cost per unit if 1,000 dictionaries are produced.

Example 2 (continued)

The total cost of printing x dictionaries is

C(x) = 20,000 + 10x

1. Find the average cost per unit if 1,000 dictionaries are produced.

= $30

x

xCxC

)()(

)000,1(C000,1

000,10000,20

x

x10000,20

Example 2 (continued)

2. Find the marginal average cost at a production level of 1,000 dictionaries, and interpret the results.

Marginal average cost = )()(' xCdx

dxC

x

x

dx

dxC

1020000)('

21000

20000)1000('C

2

20000

x

02.0

This means that if you raise production from 1,000 to 1,001 dictionaries, the price per book will fall approximately 2 cents.

Example 2 (continued)

3. Use the results from above to estimate the average cost per dictionary if 1,001 dictionaries are produced.

Average cost for 1000 dictionaries = $30.00Marginal average cost = - 0.02

The average cost per dictionary for 1001 dictionaries would be the average for 1000, plus the marginal average cost, or

$30.00 + $(- 0.02) = $29.98

The price-demand equation and the cost function for the production of television sets are given by

where x is the number of sets that can be sold at a price of $p per set, and C(x) is the total cost of producing x sets.

1. Find the marginal cost.

Solution: The marginal cost is C’(x) = $30.

Example 3

xxCx

xp 30000,150)(and30

300)(

2. Find the revenue function in terms of x.

The revenue function is

3. Find the marginal revenue.

The marginal revenue is

4. Find R’(1500) and interpret the results.

At a production rate of 1,500, each additional set increases revenue by approximately $200.

Example 3 (continued)

30300)()(

2xxxpxxR

15300)('

xxR

200$15

1500300)1500(' R

Example 3 (continued)

5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point.

0 < y < 700,000

0 < x < 9,000

Example 3 (continued)

5. Graph the cost function and the revenue function on the same coordinate. Find the break-even point.

0 < y < 700,000

0 < x < 9,000

(600,168,000) (7500, 375,000)

Solution: There are two break-even points.

C(x)

R(x)

6. Find the profit function in terms of x.

The profit is revenue minus cost, so

7. Find the marginal profit.

8. Find P’(1500) and interpret the results.

At a production level of 1500 sets, profit is increasing at a rate of about $170 per set.

Example 3 (continued)

15000027030

)(2

xx

xP

15270)('

xxP

17015

1500270)1500(' P

6. Find Maximum Profits… without finding MR or MC

Example 3 (continued) 15000027030

)(2

xx

xP