Thomas Calculus 13th [Solutions]

Section 14.8 Lagrange Multipliers 1049
41. Let g 1 ( x, y, z) z 1 0 and 2 2 2
g2 ( x, y, z) x y z 10 0 g1 k, g2
2xi 2yj 2 zk , and
2 2
2 2
f 2xyzi x zj x yk so that f g1 g2 2 xyzi x zj x yk ( k) (2xi 2yj 2 zk)
2
2
2xyz 2 x , x z 2 y , and x y 2z xyz x x 0 or yz y since z 1.
CASE 1: x 0 and
CASE 2:
2
z 1 y 9 0 (from g 2 ) y 3 yielding the points (0, 3,1).
2 2 2 2
y x z 2y x 2y (since z 1)
2 2
2y y 1 10 0 (from g 2 )
2
2
y 3 x 2 3 x 6 yielding the points 6, 3,1 .
2
3y
9 0
Now f (0, 3,1) 1 and f 6, 3,1 6 3 1 1 6 3. Therefore the maximum of f is 1 6 3 at
6, 3, 1 , and the minimum of f is 1 6 3 at 6, 3, 1 .
42. (a) Let g 1 ( x, y, z) x y z 40 0 and g2 ( x, y, z) x y z 0 g1 i j k, g2
i j k , and
w yzi xzj xyk so that w g1 g2 yzi xzj xyk ( i j k) ( i j k)
yz , xz , and xy yz xz z 0 or y x.
CASE 1: z 0 x y 40 and x y 0 no solution.
CASE 2: x y 2x z 40 0 and 2x z 0 z 20 x 10 and y 10 w (10)(10)(20)
2000
i j k
(b) n 1 1 1 2i 2j is parallel to the line of intersection the line is x 2t 10, y 2t
10,
1 1 1
z 20. Since z 20, we see that
maximum when t 0 x 10, y 10, and z 20.
2
w xyz ( 2t 10)(2t 10)(20) 4t 100 (20) which has its
43. Let g 1 ( x, y, z) y x 0 and g 2 2 2
2 ( x, y, z) x y z 4 0. Then f yi xj 2 zk, g1
i j , and
g2 2xi 2yj 2zk so that f g1 g2 yi xj 2 zk ( i j) (2xi 2yj 2 zk)
y 2 x , x 2 y , and 2z 2z z 0 or 1.
2 2 2
CASE 1: z 0 x y 4 0 2x 4 0 (since x y ) x 2 and y 2 yielding the points
2, 2, 0 .
CASE 2: 1 y 2x and x 2y x y 2( x y) 2x 2(2 x ) since x y x 0 y 0
2
z 4 0 z 2 yielding the points (0, 0, 2).
Now, f (0, 0, 2) 4 and f 2, 2, 0 2. Therefore the maximum value of f is 4 at (0, 0, 2) and the
minimum value of f is 2 at f 2, 2, 0 .
44. Let f ( x, y, z)
2
x
2
y
2
z be the square of the distance from the origin. We want to minimize f ( x, y, z)
subject to the constraints 1 ( 2y 4z 5 0 and 2 ( 4x 2 4y 2 z 2
0. f 2xi 2yj 2 zk, g1
2j 4 k , and g2 8xi 8yj 2zk so that f g1 g2
2xi 2yj 2 zk (2j 4 k) (8xi 8yj 2 zk ) 2x 8 x , 2y 2 8 y , and 2z 4 2z x 0
or 1
4 .
2 2 2
CASE 1: x 0 4(0) 4y z 0 z 2y 2y 4(2 y) 5 0 y 1 , or 2y
2
4( 2 y)
5 0
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