Multivariate Calculus - Bruce E. Shapiro

118
LECTURE 15.
CONSTRAINED OPTIMIZATION: LAGRANGE
MULTIPLIERS
maximize f which becomes a function of a single variable. Instead, we will use a
different method: We want to maximize the function f(x, y) subject to the constraint
g(x, y) = 0 where
g(x, y) = M − 2x − 2y
According to Lagrange’s method, the optimum (maximum or minimum) occurs
when
∇f(x, y) = λ∇g(x, y)
for some number λ. Thus
(y, x, 0) = −λ (2, 2, 0)
This gives us three equations in three unknowns (x, y, and λ; the third equation is
the original constraint)
y = −2λ
x = −2λ
M = 2x + 2y
We can find λ by substituting the first two equations into the third:
Hence
and the total area is
M = 2(−2λ) + 2(−2λ) = −8λ ⇒ λ = −M/8
x = y = M/4
f(M/4, M/4) = M 2 /16
To test whether this is a maximum or a minimum, we need to pick some other
solution that satisfies the constraint, say x = M/8. Then y = M/2 − x = M/2 −
M/8 = 3M/8 and the area is
f(M/8, 3M/8) = 3M 2 /64 < 4M 2 /64 = M 2 /16
Since this area is smaller, we conclude that the point (M/4, M/4) is the location of
the maximum (and not the minimum) area.
Example 15.2 Verify the previous example using techniques from Calculus I.
Solution. We want to maximize f(x, y) = xy subject to M = 2x + 2y. Solving for
y gives
y = M/2 − x
hence
Setting the derivative equal to zero,
( ) M
f(x, y) = x
2 − x = Mx
2 − x2
0 = M 2 − 2x
Revised December 6, 2006. Math 250, Fall 2006