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?
x
x
V = l × w× h
l = 30 - 2x
w =16 - 2x
16
0£ x £8
h=x
30
V = (30 - 2x ) × (16 - 2x ) × x
V = (30x - 2x 2 ) × (16 - 2x ) = 480x - 60x 2 - 32x 2 + 4x 3
V = 4x 3 - 92x 2 + 480x
V ' =12x 2 -184x + 480 = 0
V ' = 4 (3x 2 - 46x +120) = 0
10
x=
and x =12
V ' = 4 ( x -12) (3x -10) = 0
3
10
x=
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?
W
0£ x £8
5 km
C' =
A
P
B
1
1
× 2x - = 0
2
2 x 2 + 25
8–x
x
8 km
WP = x 2 + 25
PB = 8- x
1
C =1× x 2 + 25 + × (8 - x )
2
x
x + 25
2
=
1
2
x 2 + 25 = 2x
3x 2 = 25
x 2 + 25 = 4x 2
æ 5 ö
C ç ÷ » 8.330127
è 3ø
C(8) » 9.433981 abs max
x=
5
3
x=-
5
3
5. Example: You have 200 feet of fencing to enclose two adjacent
rectangular corrals. What dimensions should be used to maximize the
area?
0 £ x £ 200
P = 4x + 3y = 200
A = 2x ×
200 - 4x 400x - 8x
=
3
3
3( 400 -16x )
=0
9
100
x = 25
y=
3
A' =
2
y=
200 - 4x
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.
10 - h 10
2
=
V = pr h
r
4
5 ö
2æ
5
V = p r ç10 - r ÷
h =10 - r
è
2 ø
2
5p 3
r
2
15p 2
V ' = 20p r r =0
2
æ
3 ö
5p r ç 4 - r ÷ = 0
è 2 ø
8
r=
r(0) = 0
3
10
æ8ö
h=
r ç ÷ » 74.42962
3
è 3ø
r(4) = 0
V =10p r 2 -
0£r£4
10-h
r
h
abs max
