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To verify that this critical point represents the minimum
distance, note that
D
0
(
x
)
52
5
}
(
x
2
1
a
2
a
2
)
3/2
}1
,
which is always positive.
We now know that
D
(
x
) is minimized when Equation 1 is
true, or, equivalently,
}
P
A
R
R
} 5 }
Q
BR
R
}
. This means that the two
right triangles
APR
and
BQR
are similar, which in turn
implies that the two angles must be equal.
47.
}
d
d
v
x
} 5
ka
2
2
kx
The critical point occurs at
x
5 }
k
2
a
k
} 5 }
a
2
}
, which represents a
maximum value
because
}
d
dx
2
v
2
}52
2
k
, which is negative for all
x
. The
maximum value of
v
is
kax
2
kx
2
5
ka
1
}
a
2
}
2
2
k
1
}
a
2
}
2
2
5 }
k
4
a
2
}
.
48. (a)
v
5
cr
0
r
2
2
cr
3
}
d
d
v
r
} 5
2
cr
0
r
2
3
cr
2
5
cr
(2
r
0
2
3
r
)
The critical point occurs at
r
5 }
2
3
r
0
}
. (Note that
r
5
0 is
not in the domain of
v
.) The critical point represents a
maximum because
}
d
dr
2
v
2
} 5
2
cr
0
2
6
cr
5
2
c
(
r
0
2
3
r
),
which is negative in the domain
}
r
2
0
}
#
r
#
r
0
.
(b)
We graph
v
5
(0.5
2
r
)
r
2
, and observe that the
maximum indeed occurs a
t v
5
1
}
2
3
}
2
0.5
5 }
1
3
}
.
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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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