7.
Let
x
be the side length of the cutout square (0
,
x
,
4).
Then the base measures 8
2
2
x
in. by 15
2
2
x
in.,
and the volume is
V
(
x
)
5
x
(8
2
2
x
)(15
2
2
x
)
5
4
x
3
2
46
x
2
1
120
x
. Then
V
9
(
x
)
5
12
x
2
2
92
x
1
120
5
4(3
x
2
5)(
x
2
6). Then the
critical point (in 0
,
x
,
4) occurs at
x
5 }
5
3
}
. Since
V
9
(
x
)
.
0 for 0
,
x
, }
5
3
}
and
V
9
(
x
)
,
0 for
}
5
3
}
,
x
,
4, the
critical point corresponds to the maximum volume.
The maximum volume is
V
1
}
5
3
}
2
5 }
24
2
5
7
0
}
<
90.74 in
3
, and the
dimensions are
}
5
3
}
in. by
}
1
3
4
}
in. by
}
3
3
5
}
in.
Graphical support:
[0, 4] by [
2
25, 100]
8.
Note that the values
a
and
b
must satisfy
a
2
1
b
2
5
20
2
and so
b
5
ˇ
4
w
0
w
0
w
2
w
a
w
2
w
. Then the area is given by
A
5 }
1
2
}
ab
5 }
1
2
}
a
ˇ
4
w
0
w
0
w
2
w
a
w
2
w
for 0
,
a
,
20, and
}
d
d
A
a
} 5 }
1
2
}
a
1
}
2
ˇ
4
w
0
w
1
0
w
2
w
a
w
2
w
}
2
(
2
2
a
)
1 }
1
2
}
ˇ
4
w
0
w
0
w
2
w
a
w
2
w
55
}
ˇ
2
4
w
0
0
w
0
0
w
2
2
w
a
a
w
2
2
w
}
. The critical point occurs
when
a
2
5
200. Since
}
d
d
A
a
} .
0 for 0
,
a
,
ˇ
2
w
0
w
0
w
and
}
d
d
A
a
} ,
0 for
ˇ
2
w
0
w
0
w
,
a
,
20, this critical point corresponds
to the maximum area. Furthermore, if
a
5
ˇ
2
w
0
w
0
w
then
b
5
ˇ
4
w
0
w
0
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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
 Critical Point

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