4.4 optimization buffalo bills ranch, north platte, nebraska created by greg kelly, hanford high...
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4.4 Optimization
Buffalo Bill’s Ranch, North Platte, NebraskaCreated by Greg Kelly, Hanford High School, Richland, WashingtonRevised by Terry Luskin, Dover-Sherborn HS, Dover, Massachusetts
Photo 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 thearea at the endpoints = 0.
domain: x > 0 and 40 - 2x > 0 x < 20
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 the OPTIMIZED function, restate it in terms of one variable, determine a sensible domain.
2 Find the first derivative, set it equal to zero/ undefined, find function value at those critical points.
3 Find the function value at each end point.
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 can’s surface area.
22 2A r rh area of“lids”
lateralarea
To rewrite using one variable, we need another equation that relates r and h:
2V r h
31 L 1000 cm3 2
cm1000 r h
2
1000h
r
22
10 02
02A r r
r
2 20002A r
r
2
20004A r
r
2liter1 r h
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
If a domain endpoint could be the maximum or minimum, you have to evaluate the function at each endpoint, too.
Reminders:
If the function that you want to optimize has more than one variable, find a second connecting equation and substitute to rewrite the function in terms of one variable.
To confirm that the critical value you’ve found is a maximum or minimum, you should evaluate the function for that input value.