7.
Let
x
be the side length of the cut-out 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
w
2
w
a
w
2
w
5
ˇ
2
w
0
w
0
w
, so the maximum area occurs
when
a
5
b
.
Graphical support:
[0, 20] by [
2
30, 110]
9.
Let
x
be the length in meters of each side that adjoins the
river. Then the side parallel to the river measures
800
2
2
x
meters and the area is
A
(
x
)
5
x
(800
2
2
x
)
5
800
x
2
2
x
2
for 0
,
x
,
400.
Therefore,
A
9
(
x
)
5
800
2
4
x
and the critical point occurs at
x
5
200. Since
A
9
(
x
)
.
0 for 0
,
x
,
200 and
A
9
(
x
)
,
0
for 200
,
x
,
400, the critical point corresponds to the
maximum area. The largest possible area is
A
(200)
5
80,000 m
2
and the dimensions are 200 m
(perpendicular to the river) by 400 m (parallel to the river).
Graphical support:
[0, 400] by [
2
25,000, 90,000]
10.
If the subdividing fence measures
x
meters, then the
pea patch measures
x
m by
}
21
x
6
}
m and the amount of fence
needed is
f
(
x
)
5
3
x
1
2
}
21
x
6
} 5
3
x
1
432
x
2
1
. Then
f
9
(
x
)
5
3
2
432
x
2
2
and the critical point (for
x
.
0)
occurs at
x
5
12. Since
f
9
(
x
)
,
0 for 0
,
x
,
12 and
f
9
(
x
)
.
0 for
x
.
12, the critical point corresponds to the
minimum total length of fence. The pea patch will measure
12 m by 18 m (with a 12-m divider), and the total amount
of fence needed is
f
(12)
5
72 m.
Graphical support:
[0, 40] by [0, 250]
11. (a)
Let
x
be the length in feet of each side of the square
base. Then the height is
}
5
x
0
2
0
}
ft and the surface area
(not including the open top) is
S
(
x
)
5
x
2
1
4
x
1
}
5
x
0
2
0
}
2
5
x
2
1
2000
x
2
1
. Therefore,
S
9
(
x
)
5
2
x
2
2000
x
2
2
5
}
2(
x
3
2
x
2
1000)
}
and the critical
point occurs at
x
5
10. Since
S
9
(
x
)
,
0 for 0
,
x
,
10
and
S
9
(
x
)
.
0 for
x
.
10, the critical point corresponds
to the minimum amount of steel used. The dimensions
should be 10 ft by 10 ft by 5 ft, where the height is 5 ft.
(b)
Assume that the weight is minimized when the total
area of the bottom and the four sides is minimized.
2
a
2
1
(400
2
a
2
)
}}
2
ˇ
4
w
0
w
0
w
2
w
a
w
2
w
156
Section 4.4