1. 3.7 Modeling and Optimization
Buffalo Bill’s Ranch, North Platte, Nebraska
Greg Kelly, Hanford High School, Richland, Washington
Photo by Vickie Kelly, 1999

2. 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?
There must be a local maximum here, since the endpoints are minimums.

3. 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?

4. Procedure for Solving Optimization Problems
Assign symbols to all given quantities and quantities
to be determined.
Write a primary equation for the quantity to be
maximized or minimized.
Reduce the primary equation to one having ONE
independent variable. This may involve the use of
secondary equation.
Determine the domain. Make sure it makes sense.
Determine the max or min by differentiation. (Find the first derivative and set it equal to zero)
If the maximum or minimum could be at the endpoints, you have to check them.

5. We can minimize the material by minimizing the area.
Example:
What dimensions for a one liter cylindrical can will use the least amount of material?
Motor
Oil
We can minimize the material by minimizing the area.
We need another equation that relates r and h:
area of
ends
lateral
area

6. Example 5:
What dimensions for a one liter cylindrical can will use the least amount of material?
area of
ends
lateral
area

7. 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 maximum
volume?
Since the box has a square
base, its volume is
V = x2h h
Note: We call this the primary
equation because it gives a
formula for the quantity we
wish to optimize. x x
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, solve
x2 + 4xh = 108 for h.

8. 3x2 = 108 Substitute this into the Volume equation.
To maximize V we find the derivative and it’s C.N.’s.
3x2 = 108
We can conclude that V is a maximum when x = 6 and
the dimensions of the box are 6 in. x 6 in. x 3 in.

9. Find the points on the graph of y = 4 – x2 that are closest
to the point (0,2).
What is the distance from the
point (x,y) and (0,2)? 4 3 2 1
(x,y)
(x,y)
We want this dist. to be minimized.

10. = 2x(2x2 – 3) = 0 C.N.’s imaginary dec. dec. inc. inc. 1st der. test
dec.
dec.
inc.
inc.
1st der. test
minimum
minimum
Two closest points.