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57.
The dimensions will be
x
in. by 10
2
2
x
in. by 16
2
2
x
in.,
so
V
(
x
)
5
x
(10
2
2
x
)(16
2
2
x
)
5
4
x
3
2
52
x
2
1
160
x
for 0
,
x
,
5.
Then
V
9
(
x
)
5
12
x
2
2
104
x
1
160
5
4(
x
2
2)(3
x
2
20), so
the critical point in the correct domain is
x
5
2.
This critical point corresponds to the maximum possible
volume because
V
9
(
x
)
.
0 for 0
,
x
,
2
and
V
9
(
x
)
,
0 for 2
,
x
,
5. The box of largest volume
has a height of 2 in. and a base measuring
6 in. by 12 in., and its volume is 144 in
3
.
Graphical support:
[0, 5] by [
2
40, 160]
58.
Step 1:
r
5
radius of circle
A
5
area of circle
Step 2:
At the instant in question,
}
d
d
r
t
}52}
p
2
}
m/sec and
r
5
10 m.
Step 3:
We want to find
}
d
d
A
t
}
.
Step 4:
A
5
p
r
2
Step 5:
}
d
d
A
t
} 5
2
p
r
}
d
d
r
t
}
Step 6:
}
d
d
A
t
} 5
2
p
(10)
1
2}
p
2
}
2
52
40
The area is changing at the rate of
2
40 m
2
/sec.
59.
Step 1:
x
5
x
coordinate of particle
y
5
y
coordinate of particle
D
5
distance from origin to particle
Step 2:
At the instant in question,
x
5
5 m,
y
5
12 m,
}
d
d
x
t
}52
1 m/sec, and
}
d
d
y
t
5 m/sec.
Step 3:
We want to find
}
d
d
D
t
}
.
Step 4:
D
5
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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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