572 Partial derivatives
Lagrange multiplier X should be used in the following way. When we want to
find numbers x, y, z for which f (x,y,z) takes extreme values when x, y, z satisfy the
supplementary condition u(x,y,z) = c, we seek numbers x, y, z, X for which the
function w defined by (2) takes extreme values. We illustrate use of Lagrange
multipliers by using them to obtain information about a thoroughly important
problem. Letting
u = auxx + a12xy + a13xz
- a21yx + a22Yy + a23yz
- aalzx + a32zy + a33zz,
where all = a12, a31 = a13, and a32 = a23, we seek points on the quadric surface
having the equation u = c which lie at least and (if they exist) greatest distances
from the origin. In this and similar problems, we systematically use the idea
that distances have extreme values when their squares do. To make answers
come out in the forms that are familiar when this problem is attacked by other
methods, we modify (2) by writing -A-1 in place of X and define w by the formulaw = x2 + y2 + z2 - A-l[u(x,y,z) - c].
Equating the first-order partial derivatives to 0 gives the equationsT = 2x- 2A71(aux + a12y + a13z) = 0
8x
aw_
8v - 2y -2A-1(a21x + ally + a23Z) = 0
8w
az = 2z - 2X-1(a31x + a32y + a33z) = 0
(all - X)x + a12 y + a13 z = 0
421 x + (a22 - X)y + a23 z = 0
a31 x + a32 y + (a33 - A)z = 0
which are necessary for extrema. When c 54 0, these equations and the equation
u = c cannot be simultaneously satisfied by (x,y,z) unlessall - A a12 a13
a21 a22 - A a23
a31 a32 aa3 - A=0.
In many important cases, values of A satisfying this equation can be found
(usually only approximately) and the problem can be finished.
19 To become acquainted with Lagrange multipliers by solving easy prob-
lems, find extrema of the first function when the second equation is required to
be satisfied:
(a) x2 + Y2 + z2 ax+by+cz+d=0
(b) x2+y2 y =x2-4(^2) Z2
(c)x+y+z a2+b2+2=1
Solution of part (b): Let