3.7 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

What’s happening atx = 10?

Rel. Maximum!

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 ftA 40 2l x

w x 10 ftw

20 ftl

One application of finding relative extrema is to optimize afunction. That is, to find the maximum or minimum values

in a particular situation.

Key words that might mean optimization:-greatest profit

-least cost-least time

-greatest strength-least size

-optimum size-greatest distance

etc…

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

1 Make a sketch, if helpful.2 Write down all the GIVEN quantities.3 Write down quantities that you want to find.4 Write down the primary equation for the quantity

that is to be Maximized or Minimized.5 Re-write the primary equation as a single variable,

using a secondary equation.6. Determine the Feasible Domain. (realistic values)7. Find the Max or Min by using Calculus.

Example 1: What dimensions for a one liter cylindrical can will use the least amount of material? (Assume can is closed)

We can minimize the material by minimizing its surface area.

22 2A r rh We need secondary 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

the only given: volume = 1 liter

Primary Eqtn:

(Substitute h into the primary eqtn.)

Example 1: What dimensions for a one liter cylindrical can will use the least amount of material?

22 2A r rh 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

If the end points could be the maximum or minimum, you have to check them as well.

Important Points:

If the function that you want to optimize has more than one variable, use substitution (with the secondary eqtn) to rewrite the function to be optimized in terms of one variable.

If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check it. (Use FDT.)

Example 2: Which points on the graph of are closestto the point (0, 2)?

24 xy

Use the distance formula:2

122

12 )()( yyxxd

22 )2()0( yxd 43)24( 24222 xxxx

Since d is smallest when the expression under the radical is smallest, we only need to minimize 43)( 24 xxxf

064)(' 3 xxxf3

0,2

x Possible critical numbers:

Using the FDT or SDT, you will see that x = 0 gives a rel. maxand both give rel. minimums.

2

3,

2

3 xx

So the closest points on the curve to (0,2) are:

2

5,

2

3

2

5,

2

3

Ex. 3 We want to construct a box whose base length is 3 timesits base width. The material used to construct the top and bottomof the box costs $10 per square foot. The material used to buildthe sides of the box costs $6 per sq. ft. If the box must have avolume of 50 ft3 find the dimensions that will minimize the costof the box.

wl = 3w

h

Primary Eqtn to Minimize:

Secondary Eqtn:

)22(6)(102 whlhlwC )26(660 2 whwhw

whw 4860 2 hwlwh 2350

Solve for a variable in the secondary to substitute into the primary:

23

50

wh

ww

wwwC

80060

3

504860 2

22

22 800

120800120'w

wwwC

Critical #’s would be at: 8821.1,0wUse SDT to verify there is a minimum at w=1.8821:

31600120" wC

0)8821.1(" C

So the cost is minimized when width is approx. 1.8821.

6463.5)8821.1(3 l

7050.4)8821.1(3

50

3

5022

wh

What would this minimum cost be?

60.637$)7050.4)(8821.1(48)8821.1(60 2 C