Page 208 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 199

x 2 y 2
2 2 2
(a)We must find the extrema of F ¼ x þ y þ z subject to the constraint conditions 1 ¼ 4 þ 5 þ
z 2
1 ¼ 0 and 2 ¼ x þ y z ¼ 0. In this case we use two Lagrange multipliers 1 ; 2 and consider
25
the function
!
x 2 y 2 z 2
2 2 2
G ¼ F þ 1 1 þ 2 2 ¼ x þ y þ z þ 1 þ þ 1 þ 2 ðx þ y zÞ
4 5 25
Taking the partial derivatives of G with respect to x; y; z and setting them equal to zero, we find
1 x 2 1 y 2 1 z
þ 2 ¼ 0; þ 2 ¼ 0; 2 ¼ 0
2 5 25
G x ¼ 2x þ G y ¼ 2y þ G x ¼ 2z þ ð1Þ
Solving these equations for x; y; z,we find
2 2 5 2 25 2
; ;
x ¼ y ¼ z ¼ ð2Þ
1 þ 4 2 1 þ 10 2 1 þ 50
From the second constraint condition, x þ y z ¼ 0, we obtain on division by 2 ,assumed dif-
ferent from zero (this is justified since otherwise we would have x ¼ 0; y ¼ 0; z ¼ 0, which would not
satisfy the first constraint condition), the result
2 5 25
¼ 0
1 þ 4 þ 2 1 þ 10 þ 2 1 þ 50
Multiplying both sides by 2ð 1 þ 4Þð 1 þ 5Þð 1 þ 25Þ and simplifying yields
2
17 1 þ 245 1 þ 750 ¼ 0 or ð 1 þ 10Þð17 1 þ 75Þ¼ 0
from which 1 ¼ 10 or 75=17.
Case 1: 1 ¼ 10.
1 1 5 2 2
2
6
3
From (2), x ¼ 2 ; y ¼ 2 ; z ¼ 2 . Substituting in the first constraint condition, x =4 þ y =5þ
2
2
ffiffiffiffiffiffiffiffiffiffi
z =25 ¼ 1, yields 2 ¼ 180=19 or 2 ¼ 6 5=19. This gives the two critical points
p
p ffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffi
ð2 5=19; 3 5=19; 5 5=19 Þ; ð 2 5=19; 3 5=19; 5 5=19 Þ
2
2
2
The value of x þ y þ z corresponding to these critical points is ð20 þ 45 þ 125Þ=19 ¼ 10.
Case 2: 1 ¼ 75=17:
34 17 17 2 . Substituting in the first constraint condition,
7 4 28 p ffiffiffiffiffiffiffiffi
From (2), x ¼ 2 ; y ¼ 2 ; z ¼
2
2
2
x =4 þ y =5 þ z =25 ¼ 1, yields 2 ¼ 140=ð17 646 Þ which gives the critical points
ffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi
p p p p p p
ð40= 646; 35 646; 5= 646 Þ; ð 40= 646; 35= 646; 5= 646 Þ
2
2
2
The value of x þ y þ z corresponding to these is ð1600 þ 1225 þ 25Þ=646 ¼ 75=17.
Thus, the required maximum value is 10 and the minimum value is 75/17.
2
2
2
(b)Since x þ y þ z represents the square of the distance of ðx; y; zÞ from the origin ð0; 0; 0Þ,the problem
is equivalent to determining the largest and smallest distances from the origin to the curve of intersec-
2
2
2
tion of the ellipsoid x =4 þ y =5 þ z =25 ¼ 1 and the plane z ¼ x þ y.Since this curve is an ellipse, we
p ffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffi
have the interpretation that 10 and 75=17 are the lengths of the semi-major and semi-minor axes of
this ellipse.
The fact that the maximum and minimum values happen to be given by 1 in both Case 1 and
Case 2 is more than a coincidence. It follows, in fact, on multiplying equations (1)by x, y, and z in
succession and adding, for we then obtain
1 x 2 2 1 y 2 2 1 z 2
2 2 2
2 5 25
2x þ þ 2 x þ 2y þ þ 2 y þ 2z þ 2 z ¼ 0
!
2
2
2
2
2
2
i.e., x þ y þ z þ 1 x þ y þ z þ 2 ðx þ y zÞ¼ 0
4 5 25
2
2
2
Then using the constraint conditions, we find x þ y þ z ¼ 1 .
For a generalization of this problem, see Problem 8.76.