CHAPTER 3

SECTION 3.7

OPTIMIZATION PROBLEMS

Applying Our Concepts

• We know about

max and min …

• Now how can

we use those

principles?

Use the Strategy• What is the quantity to be optimized?

– The volume

• What are the measurements (in terms of

x)?

• What is the variable which will

manipulated to determine the optimum

volume?

• Now use calculus

principlesx

30”

60”

Guidelines for Solving Applied

Minimum and Maximum Problems

Optimization

Optimization

Maximizing or minimizing a quantity based on a given situation

Requires two equations:

Primary Equation

what is being maximized or minimized

Secondary Equation

gives a relationship between variables

To find the maximum (or minimum) value of a function:

1 Write it in terms of one variable.

2 Find the first derivative and set it equal to zero.

3 Check the end points if necessary.

Ex. 1 A manufacturer wants to design an open box

having a square base and a surface area of 108 in2.

What dimensions will produce a box with maximum

volume?

x

x

h

Since the box has a square

base, its volume is

V = x2h

Note: We call this the primary

equation because it gives a

formula for the quantity we

wish to optimize.

The surface area = the area of the base + the area of the 4 sides.

S.A. = x2 + 4xh = 108 We want to maximize the volume,

so express it as a function of just

one variable. To do this, solve

x2 + 4xh = 108 for h.

x

xh

4

108 2Substitute this into the Volume equation.

x

xxhxV

4

108 222

427

3xx

To maximize V we find the derivative and it’s C.N.’s.

04

327

2x

dx

dV3x2 = 108 6.'. xsNC

We can conclude that V is a maximum when x = 6 and

the dimensions of the box are 6 in. x 6 in. x 3 in.

2. Find the point on that is closest to (0,3).2f x x

2 20 3d x y

2. Find the point on that is closest to (0,3).

Primary Secondary

Intervals: 52

, 52

0,Test values: 3 1d ’(test pt)

d(x) dec increl min

52

x

2f x x

2y x

2 23d x y

***The value of the root will be smallest

when what is inside the root is smallest.

2 23d x y

22 2 3d x x x2 4 26 9d x x x x

4 25 9d x x x3' 4 10d x x x

34 10 0x x22 2 5 0x x

20 2 5 0x x52

0x x

52

0,

1

decrel max

52,

3

increl min

52

x

5 52 2, 5 5

2 2,

Minimize distance

2. A rectangular page is to contain 24 square inches of print. The

margins at the top and bottom are 1.5 inches. The margins on each side are

1 inch. What should the dimensions of the print be to use the least paper?

2. A rectangular page is to contain 24 square inches of print. The

margins at the top and bottom are 1.5 inches. The margins on each side are

1 inch. What should the dimensions of the print be to use the least paper?

Primary Secondary

2 3A x y 24xy

11

1.5

1.5

224 in

x

2x

y3y24

xy24( ) 2 3

xA x x

4824 3 6x

x

13 48 30x x

2

48'( ) 3x

A x

crit #'s: 0, 4xIntervals: 0,4 4,24Test values: 1 10

A ’(test pt)

A(x) incdecrel min

4x

Smallest

(x is near zero)

Largest

(y is near zero)

24x0x

4

4

2y

6y

Print dimensions: 6 in x 4 in

Page dimensions: 9 in x 6 in

1. Find two positive numbers whose sum is 36 and

whose product is a maximum.

P xy

1. Find two positive numbers whose sum is 36 and

whose product is a maximum.Primary Secondary

36x y

36y x36P x x x

'( ) 36 2P x x

36 2 0x

18x

36 18 18y

Intervals: 0,18 18,36Test values: 1 20P ’(test pt)

P(x) inc decrel max

18x

18,18

A Classic Problem

You have 40 feet of fence to enclose a rectangular garden

along the side of a barn. What is the maximum area that

you can enclose?

x x

40 2x

40 2A x x

240 2A x x

40 4A x

0 40 4x

4 40x

10x40 2l x

w x 10 ftw

20 ftl

There must be a

local maximum

here, since the

endpoints are

minimums.

A Classic Problem

You have 40 feet of fence to enclose a rectangular garden

along the side of a barn. What is the maximum area that

you can enclose?

x x

40 2x

40 2A x x

240 2A x x

40 4A x

0 40 4x

4 40x

10x

10 40 2 10A

10 20A

2200 ftA40 2l x

w x 10 ftw

20 ftl

Example 5: What dimensions for a one liter cylindrical can will

use the least amount of material?

We can minimize the material by minimizing the area.

22 2A r rh

area of

ends

lateral

area

We need another

equation that

relates r and h:

2V r h

31 L 1000 cm

21000 r h

2

1000h

r

2

2

10 02

02A r r

r

2 20002A r

r

2

20004A r

r

Example 5: What dimensions for a one liter cylindrical can will

use the least amount of material?

22 2A r rh

area of

ends

lateral

area

2V r h

31 L 1000 cm

21000 r h

2

1000h

r

2

2

10 02

02A r r

r

2 20002A r

r

2

20004A r

r

2

20000 4 r

r

2

20004 r

r

32000 4 r

3500r

3500

r

5.42 cmr

2

1000

5.42h

10.83 cmh

If the end points could be the maximum or minimum,

you have to check.

Notes:

If the function that you want to optimize has more than

one variable, use substitution to rewrite the function.

If you are not sure that the extreme you’ve found is a

maximum or a minimum, you have to check.

Example #1

• A company needs to construct a cylindrical

container that will hold 100cm3. The cost

for the top and bottom of the can is 3 times

the cost for the sides. What dimensions are

necessary to minimize the cost.r

h 2

2

22 rrhSA

hrV

Minimizing Cost

hr

hr

2

2

100

100

2

2

2

2200

2100

2

rr

SA

rr

rSA 26200

)( rr

rC

rr

rC 12200

)(2

2

2

22 rrhSA

hrV

Domain: r>0

Minimizing Cost

rr

12200

02

rr

rC 12200

)(2

3

12200

3

2

12200

12200

r

r

rr

744.13350r

0)744.1(

12100

)(3

C

rrC

Concave up – Relative min

464.10

1002

h

hr

The container will have a radius of

1.744 cm and a height of 10.464 cm

1.7440

- - - - - - + + + + +

C' changes from neg. to pos. Rel. min