Optimization

4.7

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 40 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

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.

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 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 ofends

lateralarea

We need another equation that relates r and h:

2V r h

31 L 1000 cm21000 r h

2

1000h

r

22

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 ofends

lateralarea

2V r h

31 L 1000 cm21000 r h

2

1000h

r

22

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

• A rectangular field, bounded on one side by a building, is to be fenced in on the other 3 sides. If 3,000 feet of fence is to be used, find the dimensions of the largest field that can be fenced in.

ww

3000-2w

lwAwwA )23000(

223000 wwA

wdw

dA43000

w430000

CV at w = 750

42

2

dw

Ad Always concave down

Max at: w = 750 and l = 1500

Max Area = 1,125,000 sq. ft.

• A physical fitness room consists of a rectangular region with a semicircle on each end. If the perimeter of the room is to be a 200 meter running track, find the dimensions that will make the area of the rectangular region as large as possible.

x

2rmP 200

xr 22200

We want to maximize the

area of the rectangle.

rxA 2

2 variables, so lets solve the above

equation for r

rx

2

2200

xx

A

2

22002

22200 xxA

x

dx

dA 4200

x4200

0

x50

4

2

2

dx

Ad

Concave down

50

,50When rx