Thomas Calculus 13th [Solutions]

296 Chapter 4 Applications of Derivatives
2
(b) In this case V ( x) x (18 x ). Solving V ( x) 3 x(12 x) 0 x 0 or 12; but x 0 would result in
no cylinder. Then V ( x) 6 (6 x) V (12) 0 there is a maximum at x 12. The values of
x 12 cm and y 6 cm give the largest volume.
2 2
2
2 2
27. Note that h r 3 and so r 3 h . Then the volume is given by V 3 r h 3 (3 h ) h 3
h h for
3
2 2
0 h 3, and so dV r (1 r ). The critical point (for h 0 ) occurs at h 1. Since dV 0 for
dh
dh
0 h 1, and dV 0 for 1 h 3, the critical point corresponds to the maximum volume. The cone of
dh
3
greatest volume has radius 2 m, height 1 m, and volume 2
3 m .
2 2 2 2
28. Let d ( x 0) ( y 0) x y and x y
1 y b x b . We can minimize d by minimizing
a b a
2 D x 2 y 2 x 2 b
2
a
x b 2 2 b b
2 2
D x x b 2 2 2
a a
x b x b . D
2
a a
0
x
a
2 2 2
2 2
.
2 2 2 2
2 b b 0 ab
2 a
y b ab b a b
2 2
a a b
a a b a b
D
2 2
2 2
ab 2 2b
0 the critical point is a local minimum ab , a b
2 2 2
2 2 2 2
a b a
a b a b
y
b
1 that is closest to the origin.
2
D
2
2 2b
2
a
is the point on the line
29. Let S( x) x 1 , x 0 S ( x) 1 1 x 1. S ( x) 0 x 1 2
0 x 1 0 x 1. Since x 0, we
x
2 2
2
x x
x
only consider x 1. S ( x) 2 S (1) 2 0 local minimum when x 1
3 3
x
1
2
2
30. Let S( x) 1 4 x , x 0 S ( x) 1 8 x 8x
1.
S ( x) 0 8x
1 0
x
2 2
2
x
x
x
S ( x) 2 8 S 1 2 8 0 local minimum when
3 2 x 1
3
x
(1/2)
2 .
3
3
3
8x 1 0 x 1
2 .
31. The length of the wire b perimeter of the triangle circumference of the circle. Let x length of a side of the
equilateral triangle P 3 x , and let r radius of the circle C 2 r . Thus b 3x 2 r r b 3x
.
2
2
3 3 2
The area of the circle is r and the area of an equilateral triangle whose sides are x is 1 ( x) x x .
2 2 4
Thus, the total area is given by
3 3 3
A x ( b 3 x) x 3b
9 x .
2 2 2 2 2
3 2 2 3 2 3
2
3 2 b 3x
A x r x b x x
4 4 2 4 4
3
A 0 x 3b
9 x 0 x 3b
.
2 2 2 3 9
3
A 9 0 local minimum at the critical point. P 3 3b 9b
m is the length of the
2 2
3 9 3 9
triangular segment and C 2 b 3x
b 3x 9b
3
b
b m is the length of the circular segment.
2
3 9 3 9
32. The length of the wire b perimeter of the triangle circumference of the circle. Let x length of a side of the
square P 4 x , and let r radius of the circle C 2 r . Thus b 4x 2 r r b 4x
. The area of the
2
2
2
circle is r and the area of a square whose sides are x is x . Thus, the total area is given by
2 2
A x r
2
2 4
2
2 b 4x
x b x x A 2 x 4 ( b 4 x) 2 x 2b
8 x, A 0 2x 2b
8 x 0
2 4
2
x b . A 2 8 0 local minimum at the critical point. P 4 b 4b
m is the length of the
4
4 4
square segment and C 2 b 4x
b 4x b 4b
b m is the length of the circular segment.
2
4 4
2
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