43. continued
(b)
Note that
x
.
4.25, and let
f
(
x
)
5
L
2
5
}
2
x
2
2
x
3
8.5
}
. Since
y
#
11, the approximate domain of
f
is 5.20
#
x
#
8.5.
Then
f
9
(
x
)
55
For
x
.
5.20, the critical point occurs at
x
5 }
5
8
1
} 5
6.375 in., and this corresponds to a minimum
value of
f
(
x
) because
f
9
(
x
)
,
0 for 5.20
,
x
,
6.375
and
f
9
(
x
)
.
0 for
x
.
6.375. Therefore, the value of
x
that minimizes
L
2
is
x
5
6.375 in.
(c)
The minimum value of
L
is
!
}
2
§
(6
§
2
.3
§
(6
7
§
.
5
3
§
)
7
§
2
5
§
)
3
§
8
§
.5
}
§
<
11.04 in.
44.
Since
R
5
M
2
1
}
C
2
}
2 }
M
3
}
2
5 }
C
2
}
M
2
2 }
1
3
}
M
3
, we have
}
d
d
M
R
} 5
CM
2
M
2
. Let
f
(
M
)
5
CM
2
M
2
. Then
f
9
(
M
)
5
C
2
2
M
, and the critical point for
f
occurs at
M
5 }
C
2
}
.This value corresponds to a maximum because
f
9
(
M
)
.
0 for
M
, }
C
2
}
and
f
9
(
M
)
,
0 for
M
. }
C
2
}
. The value
of
M
that maximizes
}
d
d
M
R
}
is
M
5 }
C
2
}
.
45.
The profit is given by
P
(
x
)
5
(
n
)(
x
2
c
)
5
a
1
b
(100
2
x
)(
x
2
c
)
52
bx
2
1
(100
1
c
)
bx
1
(
a
2
100
bc
).
Then
P
9
(
x
)
2
bx
1
(100
1
c
)
b
5
b
(100
1
c
2
2
x
).
The critical point occurs at
x
5
}
100
2
1
c
} 5
50
1 }
2
c
}
, and this
value corresponds to the maximum profit because
P
9
(
x
)
.
0 for
x
,
50
1 }
2
c
}
and
P
9
(
x
)
,
0 for
x
.
50
1 }
2
c
}
.
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 Spring '08
 GERMAN
 Critical Point, 2m, 2 m, 2 2 CM, 2 1 3 M

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