Transcript CHAPTER 3 SECTION 3.7 OPTIMIZATION PROBLEMS

CHAPTER 3
SECTION 3.7
OPTIMIZATION PROBLEMS
Applying Our Concepts
• We know about
max and min …
• Now how can
we use those
principles?
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”
Guidelines for Solving Applied
Minimum and Maximum Problems
Optimization
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
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.

1.
An open box having a square base and a surface area of 108
square inches is to have a maximum volume. Find its dimensions.
1.
An open box having a square base and a surface area of 108
square inches is to have a maximum volume. Find its dimensions.
Primary
Secondary
V  x2y
 108  x 2 
V (x)  x 

4
x


3
108 x x


4
4
 27x  41 x3
2
V '( x )  27  34 x 2
27  34 x 2  0
x2  27   34 
x  6
SA  x 2  4xy
108  x 2  4xy
108  x 2
y
4x
2
108   6 
y
4 6
3y
y
x
x
Domain of x will range from x being as small as
possible to x as large as possible.
Largest
Smallest
(y is near zero)
2
(x is near zero)
x  108
x 0
 6,
Intervals:
0,6 
Test values:
1
V ’(test pt)


V(x)
inc
dec
Dimensions: 6 in x 6 in x 3 in
rel max
x6
108
10

2.
Find the point on f  x   x 2 that is closest to (0,3).
Find the point on f  x   x 2 that is closest to (0,3).
2.
Minimize distance
Secondary
Primary
d
 x  0   y  3
d
 x    y  3
2
2
yx
2
2
***The value of the root will be smallest
when what is inside the root is smallest.
d   x    y  3
2

2
Intervals:

d x  x  x  3
d  x   x2  x 4  6x 2  9
2
2
d  x   x 4  5x 2  9
d '  x   4x3  10x
4x 3  10x  0
2x  2x 2  5   0
x 0
x 0
2
2x 2  5  0
x   52
2
 ,    
Test values:
d ’(test pt)
d(x)
5
2
3

5
2
,0
1
dec

inc
dec
rel max

5
2
5 5
2 2
,
5
2
5
2
1

rel min
x
 0,  
 
,
3

inc
rel min
x
5
2
5 5
2 2
,


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?
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   x  2 y  3
A( x )   x  2  24x  3
 24  3x  48x  6
1
 3x  48x  30
A '( x )  3 
48
x2
2
Secondary
1.5
xy  24
y  24x
y
24 in
1
24
 4
Print dimensions: 6 in x 4 in
Page dimensions: 9 in x 6 in
y
1
x
1.5
y 6
3 x  48

x2
crit #'s: x  0, 4
y 3
2
x2
Smallest
Largest
(x is near zero)
(y is near zero)
x  24
x 0
Intervals:
0,4  4,24
Test values:
1
10
A ’(test pt)


A(x)
dec
inc
rel min
x4
1.
Find two positive numbers whose sum is 36 and
whose product is a maximum.
1.
Find two positive numbers whose sum is 36 and
whose product is a maximum.
Primary
P  xy
P  x   x 36  x 
P '( x )  36  2x
Test values:
0,18 18,36
20
1

P ’(test pt)

P(x)
inc
dec
rel max
x  18
x  y  36
y  36  x
y  36  18  18
18,18
36  2 x  0
x  18
Intervals:
Secondary
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
40  2x
w x
l  40  2 x
w  10 ft
l  20 ft
A  40 x  2 x 2
A  40  4 x
0  40  4x
4 x  40
x  10
There must be a
local maximum
here, since the
endpoints are
minimums.

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
40  2x
w x
l  40  2 x
w  10 ft
l  20 ft
A  40 x  2 x 2
A  40  4 x
0  40  4x
4 x  40
x  10
A  10  40  2 10
A  10  20
A  200 ft 2

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   r 2 h
1000
h
2
r
A  2 r 2  2 rh
area of
ends
lateral
area
1000
A  2 r  2 r 
 r2
2
2000
A  2 r 
r
2
2000
A  4 r  2
r
Example 5:
What dimensions for a one liter cylindrical can will
use the least amount of material?
A  2 r 2  2 rh
V   r 2h
3
1
L

1000
cm


1000   r 2 h
1000
h
2
r
1000
  5.42 
2
area of
ends
lateral
area
1000
2
A  2 r  2 r 
 r2
2000
 4 r
2
r
2000  4 r 3
500

2000
A  2 r 
r
2
h
h  10.83 cm
 r3
500
2000
A  4 r  2
r
r
2000
0  4 r  2
r
r  5.42 cm
3


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.

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.
V  r h
2
r
h
SA  2rh  2r
2
Minimizing Cost
V  r 2 h
100  r 2h
100
h
2
r
SA  2rh  2r 2
SA  2r
100
 2r
r
200
SA 
 2r 2
r
Domain: r>0
2
2
200
C (r ) 
 6r 2
r
C (r ) 
 200
r
2
 12r
Minimizing Cost
C (r ) 
 200
 12r
r2
 200
0  2  12r
r
200
r2
 12r
r 3
50
3
100
3
 12
r
C (1.744)  0
Concave up – Relative min
------ +++++
0
1.744
C' changes from neg. to pos.  Rel. min
200  12r 3
200
12 
C (r ) 
r
3
 1.744
100
h
r
h  10.464
The container will have a radius of
2
1.744 cm and a height of 10.464 cm