HW11 Solutions
Math 1A Spring 2009
with Professor Stankova
§
4.7 #52, 54, 56, 58, 62
52.
We want to maximize the derivative of
y
= 1 + 40
x
3

3
x
5
. So, to maximize
y
0
=
120
x
2

15
x
4
, ﬁrst take the derivative of
y
0
, and ﬁnd the critical numbers.
y
00
= 240
x

60
x
3
= 60
x
(4

x
2
) = 60
x
(2

x
)(2 +
x
) = 0
So, critical numbers are
x
= 0
,

2
,
2. We do not have a closed interval here, so if we check
to see where
y
0
is increasing and decreasing (by looking at the sign of
y
00
= (
y
0
)
0
), we see that
the only possible places for
y
0
to be a maximum are at
x
=

2
,
2.
interval
(
∞
,

2)
(

2
,
0)
(0
,
2)
(2
,
∞
)
sign of (
y
0
)
0
+

+

Now
y
0
(2) =
y
0
(

2) = 240, so we get
y
0
at a maximum for both
x
values. To ﬁnd the ac
tual points on the curve,
y
(2) = 1+40(2)
3

3(2)
5
= 225, and
y
(

2) = 1+40(

2)
3

3(

2)
5
=

223, so the points are (2
,
225) and (

2
,

223).
54. (a)
If
P
(
x
) =
R
(
x
)

C
(
x
) is at a maximum, then
P
0
(
x
) =
R
0
(
x
)

C
0
(
x
) = 0, so
R
0
(
x
) =
C
0
(
x
)
(b)
P
(
x
) =
xp
(
x
)

C
(
x
) = 1700
x

7
x
2

16000 + 500
x

1
.
6
x
2
+ 0
.
004
x
3
. To maximize
this, ﬁrst ﬁnd
P
0
(
x
) and determine critical numbers.
P
0
(
x
) = 1700

14
x

500+3
.
2
x

0
.
012
x
2
=

0
.
012
x
2

10
.
8
x
+1200 To ﬁnd where
P
0
(
x
) = 0,
use the quadritic formula, and we see the critical numbers are
x
= 100
,

1000. We only take
values of
x
≥
0 since
x
is the amount produced. So make a table to check the sign of
P
0
(
x
):
interval
(0
,
100)
(100
,
∞
)
sign of
P
0
(
x
)
+

From this, we can conclude that a production level of
x
= 100 does give the maximum
proﬁt.
56. (a)
For $10 each, Terry sells 20 necklaces per day and for $11 each, he can sell 18. So
p
(20) = 10 and
p
(18) = 11. If
p
=
mx
+
b
, then
m
=
10

11
20

18
=

1
2
. Plug in values to ﬁnd
b
:
10 =

1
2
20 +
b
=

10 +
b