Thomas Calculus 13th [Solutions]

Section 4.6 Applied Optimization 303
65 . (a) If y cot x 2 csc x where 0 x , then y (csc x) 2 cot x csc x . Solving y 0 cos x 1
(b)
x 4 . For 0 x we have y 0 and y 0 when
4
4
x . Therefore, at
value of y 1.
2
x there is a maximum
4
The graph confirms the findings in (a).
66. (a) If y tan x 3 cot x where 0 x
2 2
x , then y sec x 3csc x . Solving y 0 tan x 3 x ,
2
3
but
3 is not in the domain. Also, 2 2
y 2sec x tan x 6csc x cot x 0 for all 0 x . Therefore at
2
x there is a minimum value of y 2 3.
3
(b)
The graph confirms the findings in (a).
2
2
67. (a) The square of the distance is D( x)
x 3
2
x 0 x 2 x 9 , so D ( x) 2x 2 and the critical
2
4
point occurs at x 1. Since D ( x ) 0 for x 1 and D ( x ) 0 for x 1, the critical point corresponds to the
5
minimum distance. The minimum distance is D (1) .
2
(b)
The minimum distance is from the point 3 , 0 to the point (1, 1) on the graph of ,
2 y x and this occurs at
the value x 1 where D( x ), the distance squared, has its minimum value.
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