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52.
For 0
,
x
, }
5
3
p
}
, the area of the rectangle is given by
A
(
x
)
5
(2
x
)(8 cos 0.3
x
)
5
16
x
cos 0.3
x
.
Then
A
9
(
x
)
5
16
x
(
2
0.3 sin 0.3
x
)
1
16(cos 0.3
x
)(1)
5
16(cos 0.3
x
2
0.3
x
sin 0.3
x
)
Solving
A
9
(
x
)
5
0 graphically, we find that the critical
point occurs at
x
<
2.868 and the corresponding area is
approximately 29.925 square units.
53.
The cost (in thousands of dollars) is given by
C
(
x
)
5
40
x
1
30(20
2
y
)
5
40
x
1
600
2
30
ˇ
x
2
w
2
w
1
w
4
w
4
w
.
Then
C
9
(
x
)
5
40
2
}
2
ˇ
x
2
w
30
2
w
1
w
4
w
4
w
}
(2
x
)
5
40
2
}
ˇ
x
2
w
3
2
w
0
x
1
w
4
w
4
w
}
.
Solving
C
9
(
x
)
5
0, we have:
}
ˇ
x
2
w
3
2
w
0
x
1
w
4
w
4
w
}5
40
3
x
5
4
ˇ
x
2
w
2
w
1
w
4
w
4
w
9
x
2
5
16
x
2
2
2304
2304
5
7
x
2
Choose the positive solution:
x
51}
ˇ
48
7
w
}
<
18.142 mi
y
5
ˇ
x
2
w
2
w
1
w
2
w
2
w
5 }
ˇ
36
7
w
}
<
13.607 mi
54.
The length of the track is given by 2
x
1
2
p
r
, so we have
2
x
1
2
p
r
5
400 and therefore
x
5
200
2
p
r
. Then the area
of the rectangle is
A
(
r
)
5
2
rx
5
2
r
(200
2
p
r
)
5
400
r
2
2
p
r
2
,for0
,
r
, }
2
p
00
}
.
Therefore,
A
9
(
r
)
5
400
2
4
p
r
and
A
0
(
r
)
52
4
p
, so the
critical point occurs at
r
5 }
1
p
00
}
m
and this point corresponds to the maximum rectangle area
because
A
0
(
r
)
,
0 for all
r
.
The corresponding value of
x
is
x
5
200
2
p
1
}
1
p
00
}
2
5
100 m.
The rectangle will have the largest possible area when
x
5
100 m and
r
5 }
1
p
00
}
m.
55.
Assume the profit is
k
dollars per hundred grade B tires and
2
k
dollars per hundred grade A tires.
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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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