1. The problem aims to find the dimensions of a cylindrical container that will minimize cost for a given volume of 100cm3. 2. The cost of the sides is represented by a function of the radius, while the cost of the tops and bottom is three times the cost of the sides. 3. By setting up the volume and surface area equations and substituting one for the other, an equation for cost is derived as a function of just the radius. 4. Taking the derivative and setting it equal to zero finds the relative minimum, giving an optimal radius of 1.744cm, and substituting back gives the corresponding optimal height of 10.464cm.

1. CHAPTER 3
SECTION 3.7
OPTIMIZATION PROBLEMS

2. Applying Our Concepts
• We know about
max and min …
• Now how can
we use those
principles?

4. Use the Strategy
• What is the quantity to be optimized?
– The volume
• What are the measurements (in terms of
x)?
• What is the variable which will
manipulated to determine the optimum
volume?
60”
• Now use calculus
x
principles
30”

5. Guidelines for Solving Applied
Minimum and Maximum Problems

6. Optimization

7. Optimization
Maximizing or minimizing a quantity based on a given situation
Requires two equations:
Primary Equation
what is being maximized or minimized
Secondary Equation
gives a relationship between variables

8. To find the maximum (or minimum) value of a function:
1
Write it in terms of one variable.
2
Find the first derivative and set it equal to zero.
3
Check the end points if necessary.

9. Ex. 1 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
h base, its volume is
V = x 2h
x
x
Note: We call this the primary
equation because it gives a
formula for the quantity we
wish to optimize.
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.

10. h
V
108 x 2
4x
2
xh
Substitute this into the Volume equation.
108 x
x
4x
2
2
27 x
x3
4
To maximize V we find the derivative and it’s C.N.’s.
dV
dx
27
3x 2
4
0
3x2 = 108
C.N.' s x
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.
6

11. 2.
Find the point on f x
x 2 that is closest to (0,3).

12. x 2 that is closest to (0,3).
Find the point on f x
2.
Minimize distance
Secondary
Primary
d
x 0
d
x
2
2
y 3
y 3
y x2
2
2
***The value of the root will be smallest
when what is inside the root is smallest.
d
d x
d x
x
2
y 3
2
2
2
Intervals:
2
x
x 3
x 2 x 4 6x 2 9
d x x 4 5x 2 9
d ' x 4x3 10x
4x 3 10x 0
2x 2x 2 5 0
x 0
2x 2 5 0
x 0
x
5
2
Test values:
5
2
,
5
2
,0
0,
5
2
5
2
,
3
1
1
3
dec
inc
dec
inc
d ’(test pt)
d(x)
rel max
rel min
x
5
2
5 5
2 2
,
rel min
x
5
2
5 5
2 2
,

13. 2.
A rectangular page is to contain 24 square inches of print. The
margins at the top and bottom are 1.5 inches. The margins on each side are
1 inch. What should the dimensions of the print be to use the least paper?

14. 2.
A rectangular page is to contain 24 square inches of print. The
margins at the top and bottom are 1.5 inches. The margins on each side are
1 inch. What should the dimensions of the print be to use the least paper?
Primary
A
A( x )
Secondary
xy 24
x 2 y 3
x 2
24
x
48
x
3x 48 x
1
A '( x ) 3
y
3
24 3 x
1.5
6
30
48
x2
y
y
2
24
x
24 in
1
24
4
y 3
y
1
x
1.5
6
x 2
Smallest
Largest
(x is near zero)
x 0
crit #'s: x 0, 4
(y is near zero)
x 24
Page dimensions: 9 in x 6 in
0,4
4,24
Test values:
Print dimensions: 6 in x 4 in
Intervals:
1
10
dec
inc
A ’(test pt)
A(x)
rel min
x 4

15. 1.
Find two positive numbers whose sum is 36 and
whose product is a maximum.

16. 1.
Find two positive numbers whose sum is 36 and
whose product is a maximum.
Primary
P
P x
Secondary
xy
x 36 x
y
P '( x ) 36 2x
36 2x 0
x 18
Intervals:
Test values:
0,18
1
18,36
inc
dec
20
P ’(test pt)
P(x)
rel max
x 18
x y 36
y 36 x
36 18 18
18,18

17. 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 2x
x
x
A 40 x 2 x 2
A
40 2x
w
l
x
40 2x
0 40 4x
w 10 ft
l
40 4x
20 ft
4x 40
x 10
There must be a
local maximum
here, since the
endpoints are
minimums.

18. 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 2x
x
x
A 40 x 2 x 2
A
40 2x
w
l
x
40 2x
A 10 40 2 10
0 40 4x
w 10 ft
l
40 4x
20 ft
4x 40
x 10
A 10 20
A 200 ft 2

19. 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 area.
We need another
equation that
relates r and h:
V
r 2h
3
1 L 1000 cm
1000
1000
r2
r 2h
A 2 r 2 2 rh
area of
ends
A 2 r
lateral
area
1000
2 r
r2
2
A 2 r
2
h
A
4 r
2000
r
2000
r2

20. Example 5:
What dimensions for a one liter cylindrical can will
use the least amount of material?
r 2h
V
A 2 r 2 2 rh
3
1 L 1000 cm
1000
1000
r2
r 2h
lateral
area
1000
2 r
r2
A 2 r2
h
A 2 r
1000
5.42
area of
ends
2
h
h 10.83 cm
A
0
2
4 r
4 r
2000
r2
4 r
2000 4 r 3
500
2000
r
2000
r2
2000
r2
r3
500
r
3
r
5.42 cm

21. Notes:
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.

22. Example #1
• A company needs to construct a cylindrical
container that will hold 100cm3. The cost
for the top and bottom of the can is 3 times
the cost for the sides. What dimensions are
necessary to minimize the cost.
2
r
V
h
r h
SA 2 rh 2 r
2

23. Minimizing Cost
V
r 2h
100
SA 2 rh 2 r 2
r
SA 2 r
SA
r 2h
100
100
r
2
2 r
200
2 r2
r
Domain: r>0
2
C (r )
C (r )
h
2
200
6 r2
r
200
r
2
12 r

24. Minimizing Cost
C (r )
0
200
r2
200
r
200
r2
2
3
r
C (1.744 ) 0
12
Concave up – Relative min
------ +++++
12 r
1.744
0
C' changes from neg. to pos.
200 12 r 3
3
200
12
r
12 r
12 r
100
C (r )
r
3 50
3
1.744
100
r
2
Rel. min
h
h 10.464
The container will have a radius of
1.744 cm and a height of 10.464 cm