Contributed by:

In this section, we will learn:

How to solve problems involving maximization and minimization of factors.

How to solve problems involving maximization and minimization of factors.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

The methods we have learned in this

chapter for finding extreme values have

practical applications in many areas of life.

A businessperson wants to minimize costs and

maximize profits.

A traveler wants to minimize transportation time.

Fermat’s Principle in optics states that light follows

the path that takes the least time.

The methods we have learned in this

chapter for finding extreme values have

practical applications in many areas of life.

A businessperson wants to minimize costs and

maximize profits.

A traveler wants to minimize transportation time.

Fermat’s Principle in optics states that light follows

the path that takes the least time.

3.
APPLICATIONS OF DIFFERENTIATION

4.7

Optimization Problems

In this section, we will learn:

How to solve problems involving

maximization and minimization of factors.

4.7

Optimization Problems

In this section, we will learn:

How to solve problems involving

maximization and minimization of factors.

4.
OPTIMIZATION PROBLEMS

In this section (and the next), we solve

such problems as:

Maximizing areas, volumes, and profits

Minimizing distances, times, and costs

In this section (and the next), we solve

such problems as:

Maximizing areas, volumes, and profits

Minimizing distances, times, and costs

5.
OPTIMIZATION PROBLEMS

In solving such practical problems, the

greatest challenge is often to convert the word

problem into a mathematical optimization

problem—by setting up the function that is

to be maximized or minimized.

In solving such practical problems, the

greatest challenge is often to convert the word

problem into a mathematical optimization

problem—by setting up the function that is

to be maximized or minimized.

6.
OPTIMIZATION PROBLEMS

Let’s recall the problem-solving

principles discussed in Chapter 1 and

adapt them to this situation.

Let’s recall the problem-solving

principles discussed in Chapter 1 and

adapt them to this situation.

7.
STEPS IN SOLVING OPTIMIZATION PROBLEMS

Thus, there are six steps involved in

solving optimization problems.

These are as follows.

Thus, there are six steps involved in

solving optimization problems.

These are as follows.

8.
1. UNDERSTAND THE PROBLEM

Read the problem carefully until it is

clearly understood.

Ask yourself:

What is the unknown?

What are the given quantities?

What are the given conditions?

Read the problem carefully until it is

clearly understood.

Ask yourself:

What is the unknown?

What are the given quantities?

What are the given conditions?

9.
2. DRAW A DIAGRAM

In most problems, it is useful to draw

a diagram and identify the given and

required quantities on the diagram.

In most problems, it is useful to draw

a diagram and identify the given and

required quantities on the diagram.

10.
3. INTRODUCE NOTATION

Assign a symbol to the quantity that

is to be maximized or minimized.

Let’s call it Q for now.

Assign a symbol to the quantity that

is to be maximized or minimized.

Let’s call it Q for now.

11.
3. INTRODUCE NOTATION

Also, select symbols (a, b, c, . . . , x, y)

for other unknown quantities and label

the diagram with these symbols.

It may help to use initials as suggestive symbols.

Some examples are: A for area, h for height,

and t for time.

Also, select symbols (a, b, c, . . . , x, y)

for other unknown quantities and label

the diagram with these symbols.

It may help to use initials as suggestive symbols.

Some examples are: A for area, h for height,

and t for time.

12.
4. EXPRESS Q IN TERMS OF THE VARIABLES

Express Q in terms of

some of the other symbols

from Step 3.

Express Q in terms of

some of the other symbols

from Step 3.

13.
5. EXPRESS Q IN TERMS OF ONE VARIABLE

If Q has been expressed as a function of

more than one variable in Step 4, use the

given information to find relationships—in the

form of equations—among these variables.

Then, use the equations to eliminate all but

one variable in the expression for Q.

If Q has been expressed as a function of

more than one variable in Step 4, use the

given information to find relationships—in the

form of equations—among these variables.

Then, use the equations to eliminate all but

one variable in the expression for Q.

14.
5. EXPRESS Q IN TERMS OF ONE VARIABLE

Thus, Q will be expressed as

a function of one variable x, say,

Q = f(x).

Write the domain of this function.

Thus, Q will be expressed as

a function of one variable x, say,

Q = f(x).

Write the domain of this function.

15.
6. FIND THE ABSOLUTE MAX./MIN. VALUE OF f

Use the methods of Sections 4.1 and 4.3

to find the absolute maximum or minimum

value of f.

In particular, if the domain of f is a closed interval,

then the Closed Interval Method in Section 4.1

can be used.

Use the methods of Sections 4.1 and 4.3

to find the absolute maximum or minimum

value of f.

In particular, if the domain of f is a closed interval,

then the Closed Interval Method in Section 4.1

can be used.

16.
OPTIMIZATION PROBLEMS Example 1

A farmer has 2400 ft of fencing and wants

to fence off a rectangular field that borders

a straight river. He needs no fence along

the river.

What are the dimensions of the field that

has the largest area?

A farmer has 2400 ft of fencing and wants

to fence off a rectangular field that borders

a straight river. He needs no fence along

the river.

What are the dimensions of the field that

has the largest area?

17.
OPTIMIZATION PROBLEMS Example 1

In order to get a feeling for what

is happening in the problem, let’s

experiment with some special cases.

In order to get a feeling for what

is happening in the problem, let’s

experiment with some special cases.

18.
OPTIMIZATION PROBLEMS Example 1

Here are three

possible ways of

laying out the 2400 ft

of fencing.

Here are three

possible ways of

laying out the 2400 ft

of fencing.

19.
OPTIMIZATION PROBLEMS Example 1

We see that when we try shallow, wide

fields or deep, narrow fields, we get

relatively small areas.

It seems plausible that there is some intermediate

configuration that produces the largest area.

We see that when we try shallow, wide

fields or deep, narrow fields, we get

relatively small areas.

It seems plausible that there is some intermediate

configuration that produces the largest area.

20.
OPTIMIZATION PROBLEMS Example 1

This figure

the general case.

We wish to maximize the area A of

the rectangle.

Let x and y be the depth and width of the rectangle

(in feet).

Then, we express A in terms of x and y: A = xy

This figure

the general case.

We wish to maximize the area A of

the rectangle.

Let x and y be the depth and width of the rectangle

(in feet).

Then, we express A in terms of x and y: A = xy

21.
OPTIMIZATION PROBLEMS Example 1

We want to express A as a function of

just one variable.

So, we eliminate y by expressing it in terms of x.

To do this, we use the given information that

the total length of the fencing is 2400 ft.

Thus, 2x + y = 2400

We want to express A as a function of

just one variable.

So, we eliminate y by expressing it in terms of x.

To do this, we use the given information that

the total length of the fencing is 2400 ft.

Thus, 2x + y = 2400

22.
OPTIMIZATION PROBLEMS Example 1

From that equation, we have:

y = 2400 – 2x

This gives:

A = x(2400 – 2x) = 2400x - 2x2

Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0).

From that equation, we have:

y = 2400 – 2x

This gives:

A = x(2400 – 2x) = 2400x - 2x2

Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0).

23.
OPTIMIZATION PROBLEMS Example 1

So, the function that we wish to maximize

is: A(x) = 2400x – 2x2 0 ≤ x ≤ 1200

The derivative is: A’(x) = 2400 – 4x

So, to find the critical numbers, we solve: 2400 – 4x = 0

This gives: x = 600

So, the function that we wish to maximize

is: A(x) = 2400x – 2x2 0 ≤ x ≤ 1200

The derivative is: A’(x) = 2400 – 4x

So, to find the critical numbers, we solve: 2400 – 4x = 0

This gives: x = 600

24.
OPTIMIZATION PROBLEMS Example 1

The maximum value of A must

occur either at that critical number or

at an endpoint of the interval.

A(0) = 0; A(600) = 720,000; and A(1200) = 0

So, the Closed Interval Method gives the maximum

value as:

A(600) = 720,000

The maximum value of A must

occur either at that critical number or

at an endpoint of the interval.

A(0) = 0; A(600) = 720,000; and A(1200) = 0

So, the Closed Interval Method gives the maximum

value as:

A(600) = 720,000

25.
OPTIMIZATION PROBLEMS Example 1

Alternatively, we could have observed that

A’’(x) = –4 < 0 for all x

So, A is always concave downward

and the local maximum at x = 600 must be

an absolute maximum.

Alternatively, we could have observed that

A’’(x) = –4 < 0 for all x

So, A is always concave downward

and the local maximum at x = 600 must be

an absolute maximum.

26.
OPTIMIZATION PROBLEMS Example 1

Thus, the rectangular field should

600 ft deep

1200 ft wide

Thus, the rectangular field should

600 ft deep

1200 ft wide

27.
OPTIMIZATION PROBLEMS Example 2

A cylindrical can is to be made to

hold 1 L of oil.

Find the dimensions that will minimize

the cost of the metal to manufacture the can.

A cylindrical can is to be made to

hold 1 L of oil.

Find the dimensions that will minimize

the cost of the metal to manufacture the can.

28.
OPTIMIZATION PROBLEMS Example 2

Draw the diagram as in

this figure, where

r is the radius and h the

height (both in

Draw the diagram as in

this figure, where

r is the radius and h the

height (both in

29.
OPTIMIZATION PROBLEMS Example 2

To minimize the cost of

the metal, we minimize

the total surface area of

the cylinder (top, bottom,

To minimize the cost of

the metal, we minimize

the total surface area of

the cylinder (top, bottom,

30.
OPTIMIZATION PROBLEMS Example 2

We see that the sides are made from

a rectangular sheet with dimensions

2πr and h.

We see that the sides are made from

a rectangular sheet with dimensions

2πr and h.

31.
OPTIMIZATION PROBLEMS Example 2

So, the surface

area is:

A = 2πr2 +

So, the surface

area is:

A = 2πr2 +

32.
OPTIMIZATION PROBLEMS Example 2

To eliminate h, we use the fact that

the volume is given as 1 L, which we take

to be 1000 cm3.

Thus, πr2h = 1000

This gives h = 1000/(πr2)

To eliminate h, we use the fact that

the volume is given as 1 L, which we take

to be 1000 cm3.

Thus, πr2h = 1000

This gives h = 1000/(πr2)

33.
OPTIMIZATION PROBLEMS Example 2

Substituting this in the expression for A gives:

2 1000 2 2000

A 2 r 2 r 2

2 r

r r

So, the function that we want to minimize is:

2 2000

A(r ) 2 r r 0

r

Substituting this in the expression for A gives:

2 1000 2 2000

A 2 r 2 r 2

2 r

r r

So, the function that we want to minimize is:

2 2000

A(r ) 2 r r 0

r

34.
OPTIMIZATION PROBLEMS Example 2

To find the critical numbers, we differentiate:

3

2000 4( r 500)

A '(r ) 4 r 2

2

r r

Then, A’(r) = 0 when πr3 = 500

So, the only critical number is: r 3 500 /

To find the critical numbers, we differentiate:

3

2000 4( r 500)

A '(r ) 4 r 2

2

r r

Then, A’(r) = 0 when πr3 = 500

So, the only critical number is: r 3 500 /

35.
OPTIMIZATION PROBLEMS Example 2

As the domain of A is (0, ∞), we can’t use the

argument of Example 1 concerning endpoints.

However, we can observe that A’(r) < 0 for r 3 500 /

3

and A’(r) > 0 for r 500 /

So, A is decreasing for all r to the left of the critical

number and increasing for all r to the right.

Thus, r 3 500 / must give rise to an absolute

minimum.

As the domain of A is (0, ∞), we can’t use the

argument of Example 1 concerning endpoints.

However, we can observe that A’(r) < 0 for r 3 500 /

3

and A’(r) > 0 for r 500 /

So, A is decreasing for all r to the left of the critical

number and increasing for all r to the right.

Thus, r 3 500 / must give rise to an absolute

minimum.

36.
OPTIMIZATION PROBLEMS Example 2

Alternatively, we could

argue that A(r) → ∞

as r → 0+ and A(r) → ∞

as r → ∞.

So, there must be

a minimum value of A(r),

which must occur at

the critical number.

Alternatively, we could

argue that A(r) → ∞

as r → 0+ and A(r) → ∞

as r → ∞.

So, there must be

a minimum value of A(r),

which must occur at

the critical number.

37.
OPTIMIZATION PROBLEMS Example 2

The value of h corresponding to

3

r 500 / is:

1000 1000 500

h 2 23

2 3 2r

r (500 / )

The value of h corresponding to

3

r 500 / is:

1000 1000 500

h 2 23

2 3 2r

r (500 / )

38.
OPTIMIZATION PROBLEMS Example 2

Thus, to minimize the cost of

the can,

The radius should be r 3 500 / cm

The height should be equal to twice the radius—

namely, the diameter

Thus, to minimize the cost of

the can,

The radius should be r 3 500 / cm

The height should be equal to twice the radius—

namely, the diameter

39.
OPTIMIZATION PROBLEMS Note 1

The argument used in the example

to justify the absolute minimum is a variant

of the First Derivative Test—which applies

only to local maximum or minimum values.

It is stated next for future reference.

The argument used in the example

to justify the absolute minimum is a variant

of the First Derivative Test—which applies

only to local maximum or minimum values.

It is stated next for future reference.

40.
FIRST DERIV. TEST FOR ABSOLUTE EXTREME VALUES

Suppose that c is a critical number of a

continuous function f defined on an interval.

a. If f’(x) > 0 for all x < c and f’(x) < 0 for all x > c,

then f(c) is the absolute maximum value of f.

b. If f’(x) < 0 for all x < c and if f’(x) > 0 for all x > c,

then f(c) is the absolute minimum value of f.

Suppose that c is a critical number of a

continuous function f defined on an interval.

a. If f’(x) > 0 for all x < c and f’(x) < 0 for all x > c,

then f(c) is the absolute maximum value of f.

b. If f’(x) < 0 for all x < c and if f’(x) > 0 for all x > c,

then f(c) is the absolute minimum value of f.

41.
OPTIMIZATION PROBLEMS Note 2

An alternative method for solving

optimization problems is to use implicit

Let’s look at the example again to illustrate

the method.

An alternative method for solving

optimization problems is to use implicit

Let’s look at the example again to illustrate

the method.

42.
IMPLICIT DIFFERENTIATION Note 2

We work with the same equations

A = 2πr2 + 2πrh πr2h = 100

However, instead of eliminating h,

we differentiate both equations implicitly

with respect to r :

A’ = 4πr + 2πh + 2πrh’ 2πrh + πr2h’ = 0

We work with the same equations

A = 2πr2 + 2πrh πr2h = 100

However, instead of eliminating h,

we differentiate both equations implicitly

with respect to r :

A’ = 4πr + 2πh + 2πrh’ 2πrh + πr2h’ = 0

43.
IMPLICIT DIFFERENTIATION Note 2

The minimum occurs at a critical

So, we set A’ = 0, simplify, and arrive at

the equations

2r + h + rh’ = 0 2h + rh’ = 0

Subtraction gives: 2r - h = 0 or h = 2r

The minimum occurs at a critical

So, we set A’ = 0, simplify, and arrive at

the equations

2r + h + rh’ = 0 2h + rh’ = 0

Subtraction gives: 2r - h = 0 or h = 2r

44.
OPTIMIZATION PROBLEMS Example 3

Find the point on the parabola

y2 = 2x

that is closest to the point (1, 4).

Find the point on the parabola

y2 = 2x

that is closest to the point (1, 4).

45.
OPTIMIZATION PROBLEMS Example 3

The distance between

the point (1, 4) and 2 2

d ( x 1) ( y 4)

the point (x, y) is:

However, if (x, y) lies on

the parabola, then x = ½ y2.

So, the expression for d

becomes:

d ( 12 y 2 1) 2 ( y 4) 2

The distance between

the point (1, 4) and 2 2

d ( x 1) ( y 4)

the point (x, y) is:

However, if (x, y) lies on

the parabola, then x = ½ y2.

So, the expression for d

becomes:

d ( 12 y 2 1) 2 ( y 4) 2

46.
OPTIMIZATION PROBLEMS Example 3

Alternatively, we could have

substituted y 2 x to get d in terms

of x alone.

Alternatively, we could have

substituted y 2 x to get d in terms

of x alone.

47.
OPTIMIZATION PROBLEMS Example 3

Instead of minimizing d, we minimize

its square:

2 2

d f ( y ) y 1 y 4

2 1

2

2

You should convince yourself that the minimum of d

occurs at the same point as the minimum of d2.

However, d2 is easier to work with.

Instead of minimizing d, we minimize

its square:

2 2

d f ( y ) y 1 y 4

2 1

2

2

You should convince yourself that the minimum of d

occurs at the same point as the minimum of d2.

However, d2 is easier to work with.

48.
OPTIMIZATION PROBLEMS Example 3

Differentiating, we obtain:

f '( y ) 2 y 1 y 2( y 4) y 8

1

2

2 3

So, f’(y) = 0 when y = 2.

Differentiating, we obtain:

f '( y ) 2 y 1 y 2( y 4) y 8

1

2

2 3

So, f’(y) = 0 when y = 2.

49.
OPTIMIZATION PROBLEMS Example 3

Observe that f’(y) < 0 when y < 2 and f’(y) > 0

when y > 2.

So, by the First Derivative Test for Absolute

Extreme Values, the absolute minimum

occurs when y = 2.

Alternatively, we could simply say that, due to

the geometric nature of the problem, it’s obvious that

there is a closest point but not a farthest point.

Observe that f’(y) < 0 when y < 2 and f’(y) > 0

when y > 2.

So, by the First Derivative Test for Absolute

Extreme Values, the absolute minimum

occurs when y = 2.

Alternatively, we could simply say that, due to

the geometric nature of the problem, it’s obvious that

there is a closest point but not a farthest point.

50.
OPTIMIZATION PROBLEMS Example 3

The corresponding value

of x is:

x = ½ y2 = 2

Thus, the point on y2 = 2x

closest to (1, 4) is (2, 2).

The corresponding value

of x is:

x = ½ y2 = 2

Thus, the point on y2 = 2x

closest to (1, 4) is (2, 2).

51.
OPTIMIZATION PROBLEMS Example 4

A man launches his boat

from point A on a bank of

a straight river, 3 km

and wants to reach point

(8 km downstream on

the opposite bank) as

quickly as possible.

A man launches his boat

from point A on a bank of

a straight river, 3 km

and wants to reach point

(8 km downstream on

the opposite bank) as

quickly as possible.

52.
OPTIMIZATION PROBLEMS Example 4

He could proceed in

of three ways:

Row his boat directly across

the river to point C and then

run to B

Row directly to B

Row to some point D

between

C and B and then run to B

He could proceed in

of three ways:

Row his boat directly across

the river to point C and then

run to B

Row directly to B

Row to some point D

between

C and B and then run to B

53.
OPTIMIZATION PROBLEMS Example 4

If he can row 6 km/h and

run 8 km/h, where should

he land to reach B as

soon as possible?

We assume that the speed

of

the water is negligible

compared with the speed at

which he rows.

If he can row 6 km/h and

run 8 km/h, where should

he land to reach B as

soon as possible?

We assume that the speed

of

the water is negligible

compared with the speed at

which he rows.

54.
OPTIMIZATION PROBLEMS Example 4

If we let x be the distance from C to D,

The running distance is: |DB| = 8 – x

The Pythagorean Theorem gives the rowing

distance as: |AD| = x 2 9

If we let x be the distance from C to D,

The running distance is: |DB| = 8 – x

The Pythagorean Theorem gives the rowing

distance as: |AD| = x 2 9

55.
OPTIMIZATION PROBLEMS Example 4

distance

We use the equation time=

rate

Then, the rowing time is: x 2 9 / 6

The running time is: (8 – x)/8

So, the total time T as a function of x is:

2

x 9 8 x

T ( x)

6 8

distance

We use the equation time=

rate

Then, the rowing time is: x 2 9 / 6

The running time is: (8 – x)/8

So, the total time T as a function of x is:

2

x 9 8 x

T ( x)

6 8

56.
OPTIMIZATION PROBLEMS Example 4

The domain of this function T is [0, 8].

Notice that if x = 0, he rows to C, and if x = 8,

he rows directly to B.

The derivative of T is: T '( x )

x 1

6 x2 9 8

The domain of this function T is [0, 8].

Notice that if x = 0, he rows to C, and if x = 8,

he rows directly to B.

The derivative of T is: T '( x )

x 1

6 x2 9 8

57.
OPTIMIZATION PROBLEMS Example 4

Thus, using the fact that x ≥ 0,

x 1

we have: T '( x) 0

6 x 9 8

2

4 x 3 x 2 9

2 2

16 x 9( x 9)

2 9

7 x 81 x

7

The only critical number is: 9 / 7

Thus, using the fact that x ≥ 0,

x 1

we have: T '( x) 0

6 x 9 8

2

4 x 3 x 2 9

2 2

16 x 9( x 9)

2 9

7 x 81 x

7

The only critical number is: 9 / 7

58.
OPTIMIZATION PROBLEMS Example 4

To see whether the minimum occurs at

this critical number or at an endpoint of

the domain [0, 8], we evaluate T at all three

points: T (0) 1.5

9 7

T 1 8 1.33

7

73

T (8) 1.42

6

To see whether the minimum occurs at

this critical number or at an endpoint of

the domain [0, 8], we evaluate T at all three

points: T (0) 1.5

9 7

T 1 8 1.33

7

73

T (8) 1.42

6

59.
OPTIMIZATION PROBLEMS Example 4

Since the smallest of

these values of T 9/ 7

occurs when x = ,

the absolute minimum

value of T must occur

The figure illustrates

this calculation by

showing the graph

of T.

Since the smallest of

these values of T 9/ 7

occurs when x = ,

the absolute minimum

value of T must occur

The figure illustrates

this calculation by

showing the graph

of T.

60.
OPTIMIZATION PROBLEMS Example 4

Thus, the man should

land 9/ 7

the boat at a point

(≈ 3.4 km) downstream

from his starting point.

Thus, the man should

land 9/ 7

the boat at a point

(≈ 3.4 km) downstream

from his starting point.

61.
OPTIMIZATION PROBLEMS Example 5

Find the area of the largest rectangle

that can be inscribed in a semicircle

of radius r.

Find the area of the largest rectangle

that can be inscribed in a semicircle

of radius r.

62.
OPTIMIZATION PROBLEMS E. g. 5—Solution 1

Let’s take the semicircle

to be the upper half of

the circle x2 + y2 = r2 with

center the origin.

Then, the word

inscribed means

that the rectangle

has two vertices

on the semicircle

and two vertices

on the x-axis.

Let’s take the semicircle

to be the upper half of

the circle x2 + y2 = r2 with

center the origin.

Then, the word

inscribed means

that the rectangle

has two vertices

on the semicircle

and two vertices

on the x-axis.

63.
OPTIMIZATION PROBLEMS E. g. 5—Solution 1

Let (x, y) be the vertex

that lies in the first

Then, the rectangle

has sides of

lengths 2x and y.

So, its area is:

A = 2xy

Let (x, y) be the vertex

that lies in the first

Then, the rectangle

has sides of

lengths 2x and y.

So, its area is:

A = 2xy

64.
OPTIMIZATION PROBLEMS E. g. 5—Solution 1

To eliminate y, we use the fact that (x, y)

lies on the circle x2 + y2 = r2.

2 2

So, y r x

2 2

Thus, A 2 x r x

To eliminate y, we use the fact that (x, y)

lies on the circle x2 + y2 = r2.

2 2

So, y r x

2 2

Thus, A 2 x r x

65.
OPTIMIZATION PROBLEMS E. g. 5—Solution 1

The domain of this function is 0 ≤ x ≤ r.

Its derivative is:

2 2 2

2 2 2x 2(r 2 x )

A ' 2 r x

2 2 2 2

r x r x

This is 0 when 2x2 = r2, that is x = r / 2 ,

(since x ≥ 0).

The domain of this function is 0 ≤ x ≤ r.

Its derivative is:

2 2 2

2 2 2x 2(r 2 x )

A ' 2 r x

2 2 2 2

r x r x

This is 0 when 2x2 = r2, that is x = r / 2 ,

(since x ≥ 0).

66.
OPTIMIZATION PROBLEMS E. g. 5—Solution 1

This value of x gives a maximum value of A,

since A(0) = 0 and A(r) = 0 .

Thus, the area of the largest inscribed

rectangle is: 2

r r 2 r 2

A 2 r r

2 2 2

This value of x gives a maximum value of A,

since A(0) = 0 and A(r) = 0 .

Thus, the area of the largest inscribed

rectangle is: 2

r r 2 r 2

A 2 r r

2 2 2

67.
OPTIMIZATION PROBLEMS Example 5

A simpler solution is possible

if we think of using an angle as

a variable.

A simpler solution is possible

if we think of using an angle as

a variable.

68.
OPTIMIZATION PROBLEMS E. g. 5—Solution 2

Let θ be the angle

shown here.

Then, the area of the

rectangle is:

A(θ) = (2r cos θ)(r sin θ)

= r2(2 sin θ cos

θ)

= r2 sin 2θ

Let θ be the angle

shown here.

Then, the area of the

rectangle is:

A(θ) = (2r cos θ)(r sin θ)

= r2(2 sin θ cos

θ)

= r2 sin 2θ

69.
OPTIMIZATION PROBLEMS E. g. 5—Solution 2

We know that sin 2θ has a maximum

value of 1 and it occurs when 2θ = π/2.

So, A(θ) has a maximum value of r2

and it occurs when θ = π/4.

We know that sin 2θ has a maximum

value of 1 and it occurs when 2θ = π/2.

So, A(θ) has a maximum value of r2

and it occurs when θ = π/4.

70.
OPTIMIZATION PROBLEMS E. g. 5—Solution 2

Notice that this trigonometric solution

doesn’t involve differentiation.

In fact, we didn’t need to use calculus at all.

Notice that this trigonometric solution

doesn’t involve differentiation.

In fact, we didn’t need to use calculus at all.

71.
APPLICATIONS TO BUSINESS AND ECONOMICS

Let us now look at

optimization problems in business

and economics.

Let us now look at

optimization problems in business

and economics.

72.
MARGINAL COST FUNCTION

In Section 3.7, we introduced the idea of

marginal cost.

Recall that if C(x), the cost function, is the cost of

producing x units of a certain product, then the marginal

cost is the rate of change of C with respect to x.

In other words, the marginal cost function is

the derivative, C’(x), of the cost function.

In Section 3.7, we introduced the idea of

marginal cost.

Recall that if C(x), the cost function, is the cost of

producing x units of a certain product, then the marginal

cost is the rate of change of C with respect to x.

In other words, the marginal cost function is

the derivative, C’(x), of the cost function.

73.
DEMAND FUNCTION

Now, let’s consider marketing.

Let p(x) be the price per unit that the company

can charge if it sells x units.

Then, p is called the demand function

(or price function), and we would expect it

to be a decreasing function of x.

Now, let’s consider marketing.

Let p(x) be the price per unit that the company

can charge if it sells x units.

Then, p is called the demand function

(or price function), and we would expect it

to be a decreasing function of x.

74.
REVENUE FUNCTION

If x units are sold and the price per unit

is p(x), then the total revenue is:

R(x) = xp(x)

This is called the revenue function.

If x units are sold and the price per unit

is p(x), then the total revenue is:

R(x) = xp(x)

This is called the revenue function.

75.
MARGINAL REVENUE FUNCTION

The derivative R’ of the revenue

function is called the marginal revenue

It is the rate of change of revenue with respect

to the number of units sold.

The derivative R’ of the revenue

function is called the marginal revenue

It is the rate of change of revenue with respect

to the number of units sold.

76.
MARGINAL PROFIT FUNCTION

If x units are sold, then the total profit

is P(x) = R(x) – C(x)

and is called the profit function.

The marginal profit function is P’,

the derivative of the profit function.

If x units are sold, then the total profit

is P(x) = R(x) – C(x)

and is called the profit function.

The marginal profit function is P’,

the derivative of the profit function.

77.
MINIMIZING COSTS AND MAXIMIZING REVENUES

In Exercises 53–58, you are asked to use

the marginal cost, revenue, and profit

functions to minimize costs and maximize

revenues and profits.

In Exercises 53–58, you are asked to use

the marginal cost, revenue, and profit

functions to minimize costs and maximize

revenues and profits.

78.
MAXIMIZING REVENUE Example 6

A store has been selling 200 DVD burners

a week at $350 each. A market survey

indicates that, for each $10 rebate offered to

buyers, the number of units sold will increase

by 20 a week.

Find the demand function and the revenue function.

How large a rebate should the store offer to maximize

its revenue?

A store has been selling 200 DVD burners

a week at $350 each. A market survey

indicates that, for each $10 rebate offered to

buyers, the number of units sold will increase

by 20 a week.

Find the demand function and the revenue function.

How large a rebate should the store offer to maximize

its revenue?

79.
DEMAND & REVENUE FUNCTIONS Example 6

If x is the number of DVD burners sold

per week, then the weekly increase in sales

is x – 200.

For each increase of 20 units sold, the price

is decreased by $10.

If x is the number of DVD burners sold

per week, then the weekly increase in sales

is x – 200.

For each increase of 20 units sold, the price

is decreased by $10.

80.
DEMAND FUNCTION Example 6

So, for each additional unit sold, the decrease

in price will be 1/20 x 10 and the demand

function is:

p(x) = 350 – (10/20)(x – 200)

= 450 – ½x

So, for each additional unit sold, the decrease

in price will be 1/20 x 10 and the demand

function is:

p(x) = 350 – (10/20)(x – 200)

= 450 – ½x

81.
REVENUE FUNCTION Example 6

The revenue function is:

R(x) = xp(x)

= 450x – ½x2

The revenue function is:

R(x) = xp(x)

= 450x – ½x2

82.
MAXIMIZING REVENUE Example 6

Since R’(x) = 450 – x, we see that

R’(x) = 0 when x = 450.

This value of x gives an absolute maximum

by the First Derivative Test (or simply by observing

that the graph of R is a parabola that opens

downward).

Since R’(x) = 450 – x, we see that

R’(x) = 0 when x = 450.

This value of x gives an absolute maximum

by the First Derivative Test (or simply by observing

that the graph of R is a parabola that opens

downward).

83.
MAXIMIZING REVENUE Example 6

The corresponding price is:

p(450) = 450 – ½(450) = 225

The rebate is: 350 – 225 = 125

Therefore, to maximize revenue, the store

should offer a rebate of $125.

The corresponding price is:

p(450) = 450 – ½(450) = 225

The rebate is: 350 – 225 = 125

Therefore, to maximize revenue, the store

should offer a rebate of $125.