Transcript Calculus 4.4

4.4
Modeling
Optimization
Ch 4.4
Modeling &and
Optimization
none
Local max (-2, 10); Local min (1,-8)
sin 
1, 3 ; 1, 3 
cos 
112 
cm
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?
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

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.

Example
Find two numbers whose sum is 20 and whose product is as
large as possible.
Example
Find two numbers whose sum is 20 and whose product is as
large as possible.
x + y = 20,
y = 20 - x,
0 < x  20
Px = x  y
P  x  = x  20  x 
= 20x - x 2
P  x  = 20 - 2x = 0 at x = 10
Test endpoints and critical point:
P 0 = 0
P 10  = 100
P  20  = 0
The 2 numbers are 10,10
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
An open-top box is to be made by cutting congruent squares of
side length x from the corners of a 20 by 25 inch sheet of tin and
bending up the sides. How large should the squares be to make
the box hold as much as possible? What is the maximum
volume?
x
25
x
20
Example
An open-top box is to be made by cutting congruent squares of
side length x from the corners of a 20 by 25 inch sheet of tin and
bending up the sides. How large should the squares be to make
the box hold as much as possible? What is the maximum
volume?
V =  25  2x  20  2x  x, 0 < x < 10
x
25
=
x
20
 500 - 90x + 4x  x
2
= 4x 3 - 90x 2 + 500x
V  x  = 12x 2 - 180x + 500 = 0 at x  3.68, 11.32
Test endpoints and critical points:
V  0 = 0
V'
+
3.68
V  3.68  = 820.53
Max volume at x = 3.68
V 10  = 0
Volume = 820.53 in 2