2. Optimization Problems
1. Draw an appropriate figure and label the quantities
relevant to the problem.
2. Write a primary equation that relates the given and
unknown quantities.
3. If necessary, reduce the primary equation to 1 variable
(use a secondary equation if necessary).
4. Determine the desired max/min using the derivative(s).
5. Check solutions with possible values (domain).
3. Example: An open box is to be made from a 16” by 30” piece of
cardboard by cutting out squares of equal size from the four
corners and bending up the sides. What size should the squares
be to obtain a box with the largest volume?
V = l×w×h
l =30-2x
w =16-2x
h = x
V = 30-2x
( )× 16-2x
( )× x
V = 30x -2x2
( )× 16-2x
( )= 480x -60x2
-32x2
+ 4x3
V = 4x3
-92x2
+480x
V ' =12x2
-184x+480 = 0
V ' = 4 3x2
- 46x +120
( )= 0
V ' = 4 x -12
( ) 3x-10
( )= 0
x
x
16
30
0 £ x £8
x =
10
3
and x =12
x =
10
3
4. Example: An offshore oil well located at a point W that is 5
km from the closest point A on a straight shoreline. Oil is to
be piped from W to a shore point B that is 8 km from A by
piping it on a straight line under water from W to some shore
point P between A and B and then on to B via pipe along the
shoreline. If the cost of laying pipe is $1 million under water
and $½ million over land, where should the point P be
located to minimize the coast of laying the pipe?
5 km
A P B
x 8 – x
8 km
W
WP = x2
+25
PB =8- x
C =1× x2
+25 +
1
2
× 8- x
( )
0 £ x £8
C' =
1
2 x2
+25
×2x -
1
2
= 0
x
x2
+25
=
1
2 x2
+25 = 2x
x2
+25= 4x2 3x2
= 25
x =
5
3
x = -
5
3
C
5
3
æ
è
ç
ö
ø
÷ » 8.330127
C(8) » 9.433981 abs max
5. Example: You have 200 feet of fencing to enclose two adjacent
rectangular corrals. What dimensions should be used to maximize the
area?
A = 2x×
200- 4x
3
=
400x -8x2
3
P = 4x+3y = 200
y =
200- 4x
3
A' =
3 400-16x
( )
9
= 0
x = 25
0 £ x £ 200
y =
100
3
6. Example: Find the radius and height of the right circular
cylinder of largest volume that can be inscribed in a right
circular cone with radius 4 inches and height 10 inches.
r
h
V = pr2
h
10-h
10-h
r
=
10
4
h =10-
5
2
r
V = pr2
10-
5
2
r
æ
è
ç
ö
ø
÷
V =10pr2
-
5p
2
r3
V ' = 20pr -
15p
2
r2
= 0
5pr 4-
3
2
r
æ
è
ç
ö
ø
÷ = 0
r =
8
3
0 £ r £ 4
r(0) = 0
r
8
3
æ
è
ç
ö
ø
÷ » 74.42962
r(4) = 0
abs max
h =
10
3
