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3.
The circumference of the original circle of radius 4 is 8
p
.
Thus, 0
#
x
#
8
p
.
[0, 8
p
] by [
2
10, 40]
4.
The maximum occurs at about
x
5
4.61. The maximum
volume is about
V
5
25.80.
5.
Start with
}
d
d
V
x
} 5 }
2
3
p
}
rh
}
d
d
x
r
} 1 }
p
3
}
r
2
}
d
d
h
x
}
. Compute
}
d
d
x
r
}
and
}
d
d
h
x
}
,
substitute these values in
}
d
d
V
x
}
, set
}
d
d
V
x
} 5
0, and solve for
x
to
obtain
x
5
}
8(3
2
3
ˇ
6
w
)
p
}
<
4.61.
Then
V
5
}
128
2
p
7
ˇ
3
w
}
<
25.80.
Quick Review 4.4
1.
y
95
3
x
2
2
12
x
1
12
5
3(
x
2
2)
2
Since
y
9$
0 for all
x
(and
y
9.
0 for
x
±
2),
y
is
increasing on (
2‘
,
‘
) and there are no local extrema.
2.
y
95
6
x
2
1
6
x
2
12
5
6(
x
1
2)(
x
2
1)
y
05
12
x
1
6
The critical points occur at
x
52
2 or
x
5
1, since
y
95
0
at these points. Since
y
0
(
2
2)
52
18
,
0, the graph has a
local maximum at
x
52
2. Since
y
0
(1)
5
18
.
0, the graph
has a local minimum at
x
5
1. In summary, there is a local
maximum at (
2
2, 17) and a local minimum at (1,
2
10).
3.
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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