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18. (a)
The base measures 10
2
2
x
in. by
}
15
2
2
2
x
}
in., so the
volume formula is
V
(
x
)
55
2
x
3
2
25
x
2
1
75
x
.
(b)
We require
x
.
0, 2
x
,
10, and 2
x
,
15. Combining
these requirements, the domain is the interval (0, 5).
[0, 5] by [
2
20, 80]
(c)
[0, 5] by [
2
20, 80]
The maximum volume is approximately 66.02 in
3
when
x
<
1.96 in.
(d)
V
9
(
x
)
5
6
x
2
2
50
x
1
75
The critical point occurs when
V
9
(
x
)
5
0, at
x
}
50
6
1
ˇ
2
7
w
0
w
0
w
}
5
}
25
6
6
5
ˇ
7
w
}
, that is,
x
<
1.96 or
x
<
6.37. We discard
the larger value because it is not in the domain. Since
V
0
(
x
)
5
12
x
2
50, which is negative when
x
<
1.96,
the critical point corresponds to the maximum volume.
The maximum volume occurs when
x
5
}
25
2
6
5
ˇ
7
w
}
<
1.96, which confirms the result in (c).
19. (a)
The “sides” of the suitcase will measure 24
2
2
x
in. by
18
2
2
x
in. and will be 2
x
in. apart, so the volume
formula is
V
(
x
)
5
2
x
(24
2
2
x
)(18
2
2
x
)
5
8
x
3
2
168
x
2
1
864
x
.
(b)
We require
x
.
0, 2
x
,
18, and 2
x
,
24. Combining
these requirements, the domain is the interval (0, 9).
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 Spring '08
 GERMAN
 Critical Point

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