Page 207 - Schaum's Outline of Theory and Problems of Advanced Calculus

198 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8

@S 64 2 @S 64 2
x y ¼ 64; xy ¼ 64
@x ¼ y x 2 ¼ 0 when ð3Þ @y ¼ x y 2 ¼ 0 when ð4Þ
3
Dividing equations (3) and (4), we find y ¼ x so that x ¼ 64 or x ¼ y ¼ 4 and z ¼ 2.

2 128 128 128 > 0. Hence, it follows that
For x ¼ y ¼ 4, ¼ S xx S yy S xy ¼ 3 3 1 > 0 and s xx ¼ 3
x y x
the dimensions 4 ft 4ft 2ft give the minimum surface.

LAGRANGE MULTIPLIERS FOR MAXIMA AND MINIMA

8.24. Consider Fðx; y; zÞ subject to the constraint condition Gðx; y; zÞ¼ 0. Prove that a necessary
condition that Fðx; y; zÞ have an extreme value is that F x G y F y G x ¼ 0.
Since Gðx; y; zÞ¼ 0, we can consider z as a function of x and y,say z ¼ f ðx; yÞ.A necessary condition
that F½x; y; f ðx; yފ have an extreme value is that the partial derivatives with respect to x and y be zero. This
gives
F x þ F z z x ¼ 0 F y þ F z Z y ¼ 0
ð1Þ ð2Þ
Since Gðx; y; zÞ¼ 0, we also have
G x þ G x z x ¼ 0 ð4Þ G y þ G z z y ¼ 0
ð3Þ
From (1) and (3)we have (5) F x G x F x G x ¼ 0, and from (2) and (4)we have (6) F y G z F z G y ¼ 0. Then
from (5) and (6)we find F x G y F y G x ¼ 0:
The above results hold only if F z 6¼ 0; G z 6¼ 0.

8.25. Referring to the preceding problem, show that the stated condition is equivalent to the conditions
x ¼ 0; y ¼ 0 where ¼ F þ G and is a constant.
If x ¼ 0; F x þ G x ¼ 0. If y ¼ 0; F y þ G y ¼ 0. Elimination of between these equations yields
F x G y F y G x ¼ 0.
The multiplier is the Lagrange multiplier.Ifdesired we can consider equivalently ¼ F þ G where
x ¼ 0; y ¼ 0.

2
2
8.26. Find the shortest distance from the origin to the hyperbola x þ 8xy þ 7y ¼ 225, z ¼ 0.
2
2
We must find the minimum value of x þ y (the square of the distance from the origin to any point in
2
2
the xy plane) subject to the constraint x þ 8xy þ 7y ¼ 225.
2
2
2
2
According to the method of Lagrange multipliers, we consider ¼ x þ 8xy þ 7y 225 þ ðx þ y Þ.
Then
x ¼ 2x þ 8y þ 2 x ¼ 0 or ð1Þ ð þ 1Þx þ 4y ¼ 0
y ¼ 8x þ 14y þ 2 y ¼ 0 or ð2Þ 4x þð þ 7Þy ¼ 0
From (1) and (2), since ðx; yÞ 6¼ð0; 0Þ,we must have

þ 1 4 2
¼ 0; i:e:; þ 8 9 ¼ 0or ¼ 1; 9
4 þ 7
2
2
2
Case 1: ¼ 1. From (1)or(2), x ¼ 2y and substitution in x þ 8xy þ 7y ¼ 225 yields 5y ¼ 225, for
which no real solution exists.
2
2
2
Case 2: ¼ 9. From (1)or(2), y ¼ 2x and substitution in x þ 8xy þ 7y ¼ 225 yields 45x ¼ 225.
2
2
2
2
2
25 ¼ 5.
Then x ¼ 5; y ¼ 4x ¼ 20 and so x þ y ¼ 25. Thus the required shortest distance is p ffiffiffiffiffi
2
2
2
8.27 (a) Find the maximum and minimum values of x þ y þ z subject to the constraint conditions
2 2 2
x =4 þ y =5 þ z =25 ¼ 1 and z ¼ x þ y.(b) Give a geometric interpretation of the result in (a).