Thomas Calculus 13th [Solutions]

1048 Chapter 14 Partial Derivatives
36. M (6 z) i 2yj xk and g 2xi 2yj 2zk so that M g (6 z) i 2yj xk
(2xi 2yj 2 zk ) 6 z 2 x , 2y 2 y , x 2z
1 or y 0.
CASE 1: 1 6 z 2x and x 2z 6 z 2( 2 z) z 2 and x 4. Then
2 2 2
( 4) y 2 36 0 y 4.
CASE 2: y 0, 6 z 2 x , and
36 z 6 or z 3. Now
x 2 2 2 2 2
2 z x 6 2 x 6 6 0
2z
z x 2z
z z x z z z
2 2
z 6 x 0 x 0; z 3 x 27 x 3 3.
Therefore we have the points 3 3, 0, 3 , (0, 0, 6), and ( 4, 4, 2). Then M 3 3, 0, 3 27 3 60
106.8, M 3 3, 0, 3 60 27 3 13.2, M (0, 0, 6) 60, and M ( 4, 4, 2) 12 M ( 4, 4, 2).
Therefore, the weakest field is at ( 4, 4, 2).
37. Let g 1 ( x, y, z) 2x y 0 and g2 ( x, y, z) y z 0 g1 2 i j, g2
j k , and f 2xi 2j 2zk
so that f g1 g2 2xi 2j 2 zk (2 i j) ( j k) 2xi 2j 2zk 2 i ( ) j k
2x 2 , 2 , and 2 z x . Then 2 2z x x 2z 2 so that 2x
y 0
2( 2z 2) y 0 4z 4 y 0. This equation coupled with y z 0 implies z 4 and 4
3
y 3 .
2 2
Then x 2 so that 2 , 4 , 4 is the point that gives the maximum value f 2 , 4 , 4 2 2 4 4
3
3 3 3
3 3 3 3 3 3
4
3 .
38. Let g 1 ( x, y, z) x 2y 3z 6 0 and g2 ( x, y, z) x 3y 9z 9 0 g1
i 2j 3 k,
g2 i 3j 9 k , and f 2xi 2yj 2zk so that f g1 g2 2xi 2yj 2zk
( i 2j 3 k) ( i 3j 9 k ) 2 x , 2y
2 3 , and 2z 3 9 . Then 0 x 2y 3z
6
1 ( ) (2 3 ) 9 27 6 7 17 6; 0 x 3y 9z
9
2 2 2
1 ( ) (3 9 ) 27 81 9 34 91 18. Solving these two equations for and gives
2 2 2 2
240
59 and 78 81 2 3 123
3 9
x , y , and z 9
59 2 59 2 59
2 59 . The minimum value is
21,771
f 81 , 123 , 9 369 . (Note that there is no maximum value of f subject to the constraints because at
59 59 59 2
59 59
least one of the variables x, y, or z can be made arbitrary and assume a value as large as we please.)
2 2 2
39. Let f ( x, y, z)
x y z be the square of the distance from the origin. We want to minimize f ( x, y, z)
subject to the constraints g 1 ( x, y, z) y 2z 12 0 and g 2 ( x, y, z) x y 6 0. Thus
f 2xi 2yj 2 zk, g1
j 2 k , and g2
i j so that f g1 g2 2 x , 2 y , and
2z 2 . Then 0 y 2z
12 2 12 5 1 12 5 24;
2 2 2 2
0 x y 6 6 1 6 2 12. Solving these two equations for and gives
2 2 2 2
4 and 4 x 2, y 4, and z 4. The point (2, 4, 4) on the line of intersection is
2
2
closest to the origin. (There is no maximum distance from the origin since points on the line can be arbitrarily
far away.)
40. The maximum value is f 2 , 4 , 4 4 from Exercise 33 above.
3 3 3 3
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