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?
V  x h  Primary Equation
2
S  x  4 xh  108  Secondary Equation
2
Express V as a function of one variable
(use the secondary equation to write V in terms of x).
x  4 xh  108
2
4 xh  108  x
108  x
h
4x
2
2
 108  x 
V x 

 4x 
2
2
h
3
x
V  x   27 x 
4
x
x
Solution
3.7 Optimization Problems
x3
V  x   27 x 
4
Feasible Domain (real-world)?
2
Ab  x is less than 108
x must be positive
2
d
y
dV
3 6

 9  0
?
0  x  108
2
2
dx
dx x  6
2
dV
3x
0
 27 
V (6)  ?
dx
4
3
3x 2
108 in
 27
4
Is this the maximum volume?
h
2
3x  108 If the domain was a closed interval,
x
x
x 2  36
x  6
we would also have to check V for
extrema at its endpoints.
Diagram
3.7 Optimization Problems
V x h
2
 x  6, V  108, Solve for h
108  (6) h
2
h3
h
x
x
 6" x 6" x 3"
Diagram is not drawn to scale
3.7 Optimization Problems
Which points on the graph of
2
y  4 x
are closest to the point (0,2)?
d
 x2  x1    y2  y1 
2
2
The quantity to be minimized is distance:
d
 x  0
2
  y  2
2
Work
3.7 Optimization Problems
d  x  ( y  2)
2
2
 Primary Equation
y  4  x  Secondary Equation
2
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
2 2
 x  (4  4 x  x )
2
2
4
 x  3x  4
4
2
Diagram
3.7 Optimization Problems
Find the minima of
d  x  3x  4
4
2
 we simply need the minima of
f ( x)  x  3x  4
4
2
f '( x)  4 x  6 x
3
 2 x(2 x  3)  0
2
 x  0 or 2 x  3  0
2
It's time for a table!
3
x
2
Table
3.7 Optimization Problems
INT.
3
)
2
(, 
3
( ,0)
2
(0,
3
)
2
(
3
, )
2
TEST
-2
-1
1
2
F’
NEG.
POS.
NEG.
POS.
CONCL.
DECR.
INCR.
DECR.
INCR.
3
Min. @ 
2
Max. @ (0,4)
 Closest points are
y  4  x2
f '( x)  2 x  2 x 2  3
 3 5
 3 5
, 
  ,  & 
 2 2
 2 2
Critical #’s
3.7 Optimization Problems
Double check with the 2nd Derivative Test:
f '( x)  4 x3  6 x
f ''( x)  12 x  6
2
 3 5
 3 5
Min. @   ,  & 
,  Max. @ (0,4)
 2 2
 2 2
 3
 3
f ''  
  12  0 f '' 
  12  0 f ''  0  6  0
 2
 2
 3 5
 3 5
 Closest points are   ,  & 
,
 2 2
 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
see that x varies
between 0 and 30.
Domain?
3.7 Optimization Problems
W  y  z  Primary Equation
y  x  12
2
2
2
y  x 2  144
Write y and z in terms of x.
z 2  (30  x)2  282
z  (30  x) 2  282
 900  60 x  x2  784
Secondary Equations!
 x2  60 x  1684
3.7 Optimization Problems
W  y  z  Primary Equation
Use the secondary equations to write the
primary equation in terms of x.
 W  x2  144  x2  60 x  1684
1
2
x in (0,30)
 ( x  144)  ( x  60 x  1684)
2
2
1
2
1
1
dw


1
1
 ( x 2  144) 2 (2 x)  ( x 2  60 x  1684) 2 (2 x  60)
dx
2
2
x

( x  144)
2
1
2
(2 x  60)

2( x  60 x  1684)
2
1
2
3.7 Optimization Problems
w


x
x 2  144
x
x  144
2
x
x 2  144



( x  30)
x 2  60 x  1684
( x  30)
x  60 x  1684
2
0
30  x
dw
0
dx
It could be the
proportion
from &$##!
x 2  60 x  1684
 x x 2  60 x  1684  (30  x) x 2  144
 x 2 ( x 2  60 x  1684)  (30  x)2 ( x 2  144)
3.7 Optimization Problems
x 2 ( x 2  60 x  1684)  (30  x)2 ( x 2  144)
 x  60 x  1684 x  (900  60 x  x )( x  144)
4
3
2
2
2
 x 4  60 x3  1684 x 2  900 x 2  60 x3  x 4  129600  8640 x  144 x 2
Put in Quadratic Form!
Combine Like Terms
 1684 x  900 x  129600  8640 x  144 x
2
2
 1684 x  1044 x  8640 x  129600
2
2
2
 640 x  8640 x  129, 600  0
Obviously, 320 is
a common factor
 320(2 x  27 x  405)  0
Factorable
2
2
 320(2 x  45)( x  9)  0
 x  9, 22.5
3.7 Optimization Problems
w(9)  9  144  9  60(9)  1684
2
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?
The quantity to
be maximized is
area.
A  x r .
2
2
 Primary Equation
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
 x 2  4(1  x)2


maximum total area?
A  x 2   r 2  Primary Equation
 x 2  4(1  2 x  x 2 ) 

1
 ( x 2  4  8 x  4 x 2 )


4  4 x  2 r  Secondary Equation
4  4x
2  2x 2(1  x)
r


2


 2(1  x) 
 A  x2   

 2 
4(1  x)
2
x 

1
2
1
2

 (4   ) x  8 x  4 

3.7 Optimization Problems
Feasible Domain? 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
A(0)  1.273
zero or as much as 4.
4
1
A(
)  0.560
2
 (4   ) x  8 x  4 
4

dA (8  2 ) x  8

0
dx

8 x  2 x  8  0
A(1)  1
The maximum area occurs
when x  0. That is,
8
4
x  8  2   8  x 

8

2

4


Diagram
when all the wire is
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