3.7 Optimization Problems
One of the most common applications of calculus involves the determination of minimum and maximum values.
Consider how frequently you hear or read terms such as…

3.7 Optimization Problems
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?
 2
V x h

Primary Equa tio n
S
 x
2 
4 xh

108

Secondary Equ atio n function of one variabl e
(use the second ary equation t o write V x
2 
4 xh

4 xh

108
 h

108
4
 x x x
2
108
2
V

  x
2 



27 x

i
 n terms of x ) .
x
2 x 

 x
4
3 x x
Solution
h

3.7 Optimization Problems

27 x

Feasible Domain (real-world)?
x must be positive A b
 x
2
is less than 108 x
3
4 dV dx x
0 108

27
 h
3 x
2

0
4
3
3 x
4 x
2
2


27
108 x x
2 
36 x
 
6 dV dx

?
V
2 d y dx
2
(6)
108
 x

6 in
?
3
 
2
9 0
Is this the maximum volume?
If the domain was a closed interval, extrema at its endpoints.
Diagram

3.7 Optimization Problems
 2
V x h  x

6, V

108, Solve for h
 2
108 (6) h h

3 h x x
 x x
Diagram is not drawn to scale

3.7 Optimization Problems
y x
2
d
  x
2
 x
1
  y
2
 y
1

2
The quantity to be minimized is distance: d
  x

0
  y

2

2
Work

3.7 Optimization Problems d
 x
2 
( y

2)
2

Primary Equa tio n y 4 x
2  at o
Use the secondary equation to write the primary equation in terms of one variable.
 d
 x
2 
( 4
 x
2 
2 )
2

 x
2 x
2   x
2 2
(2 )
(4 4 x
2  x
4
)
 x
4 
3 x
2 
4
Diagram

3.7 Optimization Problems
Find the minima of d
 x
4 
3 x
2 
4

we simply need the minima of f x
 x
4 
3 x
2 
4 f x

4 x
3 
6 x
 x x
2 
3)

2 x
2  
0 x
 
3
2
Table

INT.
TEST
F’
3.7 Optimization Problems
-2
( ,
3
)
2
-1
(

3
2
, 0)
1
3
(0, )
2
2
3
( , )
2

NEG.
POS.
NEG.
POS.
CONCL.
DECR.
INCR.
DECR.
INCR.
Min. @


3 Max. @ (0,4)
2
Closest points are

 
3 5
, & ,
2 2
 
 
3 5
2 2

 y x
2 f x

2 x

2 x
2 
3

Critical #’s

3.7 Optimization Problems
f x

4 x
3 
6 x
Min. @

  f x
 x
2 
6
3 5
, & ,
2 2
 
 
3 5
2 2

 Max. @ (0,4) f ''





3
2 




12

0 f '' 



3
2





12

0 f
 
6 0

Closest points are

 
3 5
, & ,
2 2
 
 
3 5
2 2



3.7 Optimization Problems


3 5
,
2 2


3 5
,
2 2



Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post.
Where should the stake be placed to use the least wire?
The quantity to be minimized is length. From the diagram you can between 0 and 30.
Domain?

3.7 Optimization Problems
Write y and z in terms of x.
W y z Primary Equa tio n y
2  x
2 
12
2 z
2 
(30
 x )
2 
28
2  x
2 
60 x

1684 y
 x
2 
144 z


(30
 x ) 2 
28 2
900 60 x
 x 2 
784
Secondary Equations!

3.7 Optimization Problems
W y z Primary Equation
Use the secondary equations to write the

W
 x
2 
144
 x
2 
60 x

1684 x in (0, 30)

( x
2 
144)
1
2

( x
2 
60 x

1684)
1
2 dw
 dx
1
2 x


1 x
  x


1
(
2
144) (2 )
1
(
2
60 1684) (2 60)
2 x x

 x
( x
2

144)
1
2

( 2 x

60 )
2 ( x
2

60 x

1684) 2
1

3.7 Optimization Problems w


 x x
2 
144
 x x
2 
144
 x x
2 
144

( x

30) x
2 
60 x

1684
( x

30) x
2 
60 x

1684
30
 x x
2 
60 x

1684

0 dw

0 dx
It could be the proportion from &$##!
 x x
2 
60 x

168 4

(30
 x ) x
2 
144
 2 2 x x

60 x

1684)

(30
 x
2 2
) ( x

144)

3.7 Optimization Problems
2 2 x x

60 x

1684)

(30
 x
2 2
) ( x

144)
 x
4 
60 x
3 
1684 x
2   x
 x
2
)( x
2 
144)
 x
4 
60 x
3 
1684 x
2 
900 x
2 
60 x
3  x
4   x

144 x
2
Put in Quadratic Form!
Combine Like Terms

1684 x
2 
900 x
2   x

144 x
2

1684 x
2 
1044 x
2 
8640 x

129600

x
2 
x


Obviously, 320 is a common factor

320(2 x
2 
27 x

405)

0

320(2 x

45)( x

9)

0  x
Factorable
9, 22.5

3.7 Optimization Problems w (9)

9
2 
144

9
2  

50
Conclusion :
The wire should be staked 9 feet from the 12 foot pole.

3.7 Optimization Problems
Remember, an extreme value can also occur at the endpoints of an interval.
4 ' of wire is to be used to form a square and/or a circle.
How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum total area?

A
 x
2   r
2
Primary Equa tio n
Solution

3.7 Optimization Problems
Remember, an extreme value can also occur at the endpoints of an interval.
4 ' of wire is to be used to form a square and/or a circle.
How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum total area?
A
 x
2   r
2 
Primary Equation
4

4 x

2
 r
 r

4

4 x
2


Secondary Equation
2

2 x

A
 x x
2
2 



2(1
 x )


4(1
 x )
2

2
2(1
 x )


1




1
 x
2 
 x
2
4 (1
 x )
2


4 (1 2 x
 x
2
)

1

(
 x
2
4 8 x

4 x
2
)
(4
 
) x
2 
8 x

4

3.7 Optimization Problems
x in [0,1]
You could use all or none of the wire for the square.
The perimeter of the square could be as little as zero or as much as 4.
A (0)

1.273

1

(4
 
) x
2

8 x
8 x
 dA dx
2

 x 8
 

0 x

8


4
0
4
A (
4

A


(1) 1
)

0.560
The maximum area occurs when x

0. That is, when all the wire is x

8 2
 
8 x
Diagram

8


8 2 4
4
  used for the circle.

3.7 Optimization Problems
You must expect that real-life applications often involve equations that are at least as complicated as the primary equations seen in today's examples.
Remember, one of the main goals of this course is to learn to use calculus to analyze equations that initially seem formidable.
HW 3.7 pp. 210-212/1-25 odd, 35