Multivariate Calculus - Bruce E. Shapiro

LECTURE 15. CONSTRAINED OPTIMIZATION: LAGRANGE
MULTIPLIERS 123
Using equation 15.8 in 15.4 gives
Using equation 15.8 in 15.3 gives
Using equation 15.9 in 15.6 gives
Substituting equation 15.10 gives
Hence
From equation 15.9
b(aλ − 2c) = a(bλ − 2c)
abλ − 2bc = abλ − 2ac
2bc = 2ac
b = a (15.8)
h = V/a 2 (15.9)
4ac = a 2 λ ⇒ λ = 4c/a (15.10)
V
a 2 =
V
a 2 =
4ac
aλ − 2c
(15.11)
4ac = 2a (15.12)
4c − 2c
a = b = (V/2) 1/3 (15.13)
h = V a −2 = V V/2 −2/3 = V 1/3 2 2/3 = (4V ) 1/3 (15.14)
The solution is a = b = (V/2) 1/3 and h = (4V ) 1/3 .
Theorem 15.2 Lagrange’s Method: Two Constraints An extreme value of
the the function f(x, y, z) subject to the constraints
g(x, y, z) = 0
and
occurs when
for some numbers λ and µ
h(x, y, z) = 0
∇f = λ∇g + µ∇h
Example 15.7 Find the maximum and minimum of
f(x, y, z) = −x + 2y + 2z
on the ellipse formed by the intersection of the cylinder
x 2 + y 2 = 2
with the plane
y + 2z = 1
Math 250, Fall 2006 Revised December 6, 2006.