3.7 Modeling and Optimization

Buffalo Bill’s Ranch, North Platte, NebraskaGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1999

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

Procedure for Solving Optimization Problems

1. Assign symbols to all given quantities and quantitiesto be determined.

2. Write a primary equation for the quantity to be maximized or minimized.

3. Reduce the primary equation to one having ONEindependent variable. This may involve the use ofsecondary equation.

4. Determine the domain. Make sure it makes sense.5. Determine the max or min by differentiation. (Find the first

derivative and set it equal to zero)6. If the maximum or minimum could be at the endpoints, you

have to check them.

Example: 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 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 maximumvolume?

x

x

h

Since the box has a squarebase, its volume isV = x2h

Note: We call this the primaryequation 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, solvex2 + 4xh = 108 for h.

x

xh

4

108 2 Substitute 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

2

x

dx

dV 3x2 = 108 6.'. xsNC

We can conclude that V is a maximum when x = 6 andthe dimensions of the box are 6 in. x 6 in. x 3 in.

Find the points on the graph of y = 4 – x2 that are closest to the point (0,2).

4

3

2

1

(x,y)(x,y)

What is the distance from the point (x,y) and (0,2)?

22 )2()0( yxd

We want this dist. to be minimized.

43 24 xx

222 )24( xx

21

24

3

432

64'

xx

xxd

21

24

3

432

64'

xx

xxd

imaginary

= 2x(2x2 – 3) = 0

C.N.’s 2

6,0

2

6 02

6

dec.

inc.

dec.

inc.

1st der. test

2

5,

2

6

minimum

2

5,

2

6

minimum

Two closest points.