43.
Let
S
be the stiffness of the beam. Then,
S
=
kwh
3
,
where
k
is a constant of proportionality. Since
w
2
+
h
2
=
225
,
or
h
=
√
225
−
w
2
, S
can be
expressed as a function of
w,

3.5
Additional Applied Optimization
139
S(w)
=
kw(
225
−
w
2
)
3
/
2
which is the function to be maximized.
S (w)
=
k
w
·
3
2
(
225
−
w
2
)
1
/
2
(
−
2
w)
+
(
225
−
w
2
)
3
/
2
(
1
)
=
k(
225
−
w
2
)
1
/
2
−
3
w
2
+
225
−
w
2
=
k(
225
−
w
2
)
1
/
2
(
225
−
4
w
2
)
S (w)
=
0 when
w
=
15
2
(rejecting the solution
w
=
15
,
which is not possible given the diameter)
When 0
< w <
15
2
, S (w) >
0 so
C
is increasing
15
2
< w <
15
, S (x)) <
0 so
S
is decreasing.
So, the dimensions for maximum stiffness are
w
=
15
2
inches and
h
=
225
−
15
2
2
≈
13
.
0
inches.
45.
Let
x
be the number of miles from the house to
plant
A
. Then, 18
−
x
is its distance from plant
B
,
and 1
≤
x
≤
16
.
Let
P (x)
be the concentration of
particulate matter at the house. Then,
P (x)
=
80
x
+
720
18
−
x
which is the function to minimize.
P (x)
= −
80
x
2
+
0
−
(
720
)(
−
1
)
(
18
−
x)
2
P (x)
=
0 when
80
x
2
=
720
(
18
−
x)
2
2
x
2
+
9
x
−
81
=
0
or,
x
=
9
2
(rejecting negative solution)
P (
4
.
5
)
=
0
, P (
1
)
≈
122
.
4
, P (
16
)
=
365;
So, the total pollution is minimized when the house
is 4.5 miles from plant
A
.
47.
Let
C(N)
be the total cost of using
N
machines.
Now, the setup cost of
N
machines is
aN
and the
operating cost of
N
machines is
b
N
.
So,
C(N)
=
aN
+
b
N
which is the function to minimize.
C (N)
=
a
=
b
N
2
C (N)
=
0 when
a
=
b
N
2
,
or when
aN
=
b
N
(setup cost = operating cost)
C (N)
=
2
b
N
3
,
which is positive for all
N
in the domain
N
≥
1,
so there is an absolute minimum when setup cost
equals operating cost.
49.
Frank is right. In the cost function,
C(x)
=
5
(
900
)
2
+
x
2
+
4
(
3
,
000
−
x)
note where the distance downstream appears. Since it
is only part of the constant term in
C(x),
it drops out
when finding
C (x)
. So, the critical value is always
x
=
1
,
200 (as long as the distance downstream is at
least 1,200 meters).
When 0
≤
x <
1
,
200
, C (x) <
0 so
C
is decreasing
x >
1
,
200
, C (x)) >
0 so
C
is increasing
So, the absolute minimum cost is always when the
cable reaches the bank 1,200 meters downstream.
51. (a)
Let
x
be the number of machines and let
t
be the
number of hours required to produce
q
units.
The set up cost is
xs
and the operating cost is
pt
. Since each machine produces
n
units per
hour, then
q
=
xnt,
or
t
=
q
nx
. The total cost is
C(x)
=
xs
+
p
q
nx
which is the function to be minimized.

140
Chapter 3.
Additional Applications of the Derivative
C (x)
=
s
−
pq
nx
2
C (x)
=
0 when
s
=
pq
nx
2
or,
x
=
pq
ns
1
/
2
C (x)
=
2
pq
nx
3
,
so
C
pq
ns
1
/
2
>
0
and there is a relative minimum when
x
=
pq
ns
1
/
2
.
Further, since
C (x) >
0 for
all values of
x
in the domain
x
≥
1, it is the
absolute minimum.
(b)
The setup cost
xs,
at this minimum, becomes
xs
=
s
pq
ns
=
pqs
n
and the operating cost
pt
, at this minimum,
becomes
P
q
n
pq
ns
=
pq
pqn
s
=
pq
s
pqn
=
pqs
n
So, the setup cost equals the operating cost
when the total cost is minimized.
53. (a)
Let
x
be the number of units produced,
p(x)
the
price per unit,
t
the tax per unit, and
C(x)
the
total cost.