4.4 Modeling and Optimization
Buffalo Bill’s Ranch, North Platte, Nebraska
Photo by Vickie Kelly, 1999
Greg Kelly, Hanford High School, Richland, Washington
Optimization
• Applying Calculus and its methods for finding
minimums and maximums in real life
situations such as
What is the largest area I can enclose with a given
amount of fence?
What is the least amount of material needed to
build a cylinder of given volume?
What dimensions will maximize the volume of a
rectangular prism with given surface area?
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?
A  x  40  2 x 
A  40 x  2 x
x
x
40  2 x
wx
l  40  2 x
w  1 0 ft
l  2 0 ft
A  40  4 x
0  40  4 x
4 x  40
x  10
2
There must be a
local maximum
here, since the
endpoints are
minimums. (and
the function is a
downward
parabola)

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?
A  x  40  2 x 
A  40 x  2 x
x
x
40  2 x
wx
l  40  2 x
w  1 0 ft
l  2 0 ft
A  40  4 x
0  40  4 x
4 x  40
x  10
2
A  1 0  4 0  2 1 0 
A  10  20 
A  200 ft
2

General procedures for finding solutions to optimization
problems:
1 Draw a picture and appropriately label the
important parts.
2. Write an equation (function) for the quantity you
want to optimize in terms of one variable.
3 Find and test all critical numbers. (answers are
usually fairly obvious.)

What is the largest open top box I can make by cutting
squares from each corner of a 6 by 6 sheet of metal?
V(x) = x(6 - 2x)(6 - 2x) = 36x – 24x2 + 4x3
V’(x) = 36 – 48x + 12x2
V(x) = x(6 - 2x)(6 - 2x)
12x2 – 48x + 36 = 0
V(1) = (1)(6 - 2)(6 - 2)
x2 – 4x + 3 = 0
(x – 3)(x – 1) = 0
x = 3 or x = 1
V(1) = 16 cubic units
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 surface
area.
2
We need another
A  2 r  2 rh
equation that
relates r and h:
area of
ends
V r h
lateral
area
2
1 L  1000 cm
1000   r h
2
A  2 r  2 r 
2
3

A  2 r 
2
10 0 0
r
2000
r
1000
r
2
h
A   4 r 
2000
r
2
2
Example 5:
What dimensions for a one liter cylindrical can will
use the least amount of material?
A  2 r  2 rh
V r h
2
2
1 L  1000 cm 
3
1000   r h
area of
ends
lateral
area
2
1000
r
2
2000
A  2 r  2 r 
2
h
A  2 r 
2
r
10 0 0
r
 4 r
2
2000  4 r
3
2
500

2000
r
3
r
1000
  5.42 
2
h
h  1 0 .8 3 cm
A   4 r 
2000
r
0  4 r 
2
2000
r
r 
3
500

r  5 .4 2 cm
2

To summarize:
Write a function for the amount that you want to optimize.
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.
If the end points could be the maximum or minimum,
you have to check. (This is rare)
