Thomas Calculus 13th [Solutions]

1044 Chapter 14 Partial Derivatives
CASE 2:
2 2
x 0 y z 1, which has no solution.
Therefore the points on the unit circle
origin. The minimum distance is 1.
2 2
x y 1, are the points on the surface
2 2 2
x y z 1 closest to the
20. Let
2 2 2
f ( x, y, z)
x y z be the square of the distance to the origin. Then f 2x 2y 2z
i j k and
g yi xj k so that f g 2xi 2yj 2 zk ( yi xj k ) 2 x y, 2 y x,
and 2z
y
y
x 2y y 0 or 2.
2 2
CASE 1: y 0 x 0 z 1 0 z 1.
CASE 2: 2 x y and
2 2
z 1 x ( 1) 1 0 x 2 0, so no solution.
CASE 3: 2 x y and z 1 ( y) y 1 1 0 y 0, again.
Therefore (0, 0,1) is the point on the surface closest to the origin since this point gives the only extreme value
and there is no maximum distance from the surface to the origin.
2 2 2
21. Let f ( x, y, z)
x y z be the square of the distance to the origin. Then f 2xi 2yj 2zk and
g yi xj 2zk so that f g 2xi 2yj 2 zk ( yi xj 2 zk ) 2 x y , 2 y x , and
2z 2z 1 or z 0.
2
CASE 1: 1 2x y and 2y x y 0 and x 0 z 4 0 z 2 and x y 0.
CASE 2: z 0 xy 4 0 y 4 . Then 2 x 4
x , and 8 x 8 x x
x
x
2 x
x 2
4
x 16 x 2. Thus, x 2 and y 2, or x 2 and y 2.
Therefore we get four points: (2, 2, 0), ( 2, 2, 0), (0, 0, 2). and (0, 0, 2) But the points (0, 0, 2) and
(0, 0, 2) are closest to the origin since they are 2 units away and the others are 2 2 units away.
2
2
2 2 2
22. Let f ( x, y, z)
x y z be the square of the distance to the origin. Then f 2xi 2yj 2zk and
2
2
g yzi xzj xyk so that f g 2 x yz, 2 y xz , and 2z xy 2x xyz and 2y yxz
2 2
x y y x z x x( x)( x) 1 x 1 the points are (1, 1, 1), (1, 1, 1), ( 1, 1, 1),
and ( 1, 1, 1).
23. f i 2j 5k and g 2xi 2yj 2zk so that f g i 2j 5 k (2xi 2yj 2 zk) 1 2 x ,
2 2 y , and 1 1
2 2 2
5 2 z x , y 2 x , and z 5 5 x x ( 2 x) (5 x) 30 x 1.
2
2
Thus, x 1, y 2, z 5 or x 1, y 2, z 5. Therefore f (1, 2, 5) 30 is the maximum value and
f ( 1, 2, 5) 30 is the minimum value.
24. f i 2j 3k and g 2xi 2yj 2zk so that f g i 2j 3 k (2xi 2yj 2 zk) 1 2 x ,
2 2 y , and 1 1
2 2 2
3 2 z x , y 2 x , and z 3 3 x x (2 x) (3 x) 25 x 5 . Thus,
2
2 14
x 5 , y 10 , z 15 or x 5 , y 10 , z 15 . Therefore f 5 , 10 , 15 5 14 is the
14 14 14
14 14 14
14 14 14
maximum value and f 5 , 10 , 15 5 14 is the minimum value.
14 14 14
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