Multivariate Calculus - Bruce E. Shapiro

LECTURE 15. CONSTRAINED OPTIMIZATION: LAGRANGE
MULTIPLIERS 119
or x = M/4. This is the same answer we found using Lagrange’s Method.
Why does Lagrange’s method work? The reason hinges on the fact that the
gradient of a function of two variables is perpendicular to the level curves. The
equation
∇f(x, y) = λ∇g(x, y)
says that the normal vector to a level curve of f is parallel to a normal vector of
the curve g(x, y) = 0. Equivalently, the curve g(x, y) = 0 is tangent to a level curve
of f(x, y). Why is this an extremum? Suppose that it is not an extremum. Then
we can move a little to the left or the right along g(x, y) = 0 and we will go to a
higher or lower level curve of f(x, y). But this is impossible because we are tangent
to a level curve, so if we move infinitesimally in either direction, we will not change
the value of f(x, y). Hence the value of f(x, y) must be either a maximum or a
minimum at the point of tangency.
Example 15.3 . Find the maximum and minimum value of the function
f(x, y) = x + y
on the ellipse
3x 2 + 4y 2 = 25
Solution. We use Lagrange’s method with the constraint
setting ∇f = λ∇g to give
Our system of equations is
g(x, y) = 3x 2 + 4y 2 − 25 = 0
(1, 1) = λ (6x, 8y)
x = 1
6λ
y = 1
8λ
3x 2 + 4y 2 = 25
Substituting the first two equations into the third equation gives
25 = 3
36λ 2 + 4
64λ 2
25λ 2 = 1 12 + 1 16 + 12
=
16 (16)(12) = 28
192 = 7
48
λ 2 7
=
(48)(25) = 7
1200
√
7
λ = ±
1200 20√ = ± 1 7
3
Math 250, Fall 2006 Revised December 6, 2006.