Multivariate Calculus - Bruce E. Shapiro

120
LECTURE 15.
CONSTRAINED OPTIMIZATION: LAGRANGE
MULTIPLIERS
Thus
x = 1
6λ = ±20 6
√
3
7 = ±10 3
√
3
7 = ± 10 √
21
and
Hence
y = 1
√ √
3 3 3 · 3
8λ = ±20 8 7 2√ = ±5 7 = ±5 2 3 · 7 = ± 15
2 √ 21
( 10
(x, y) = ± √ , 21
15
2 √ 21
)
= ± √ 5 (
2, 3 )
21 2
To determine which point gives the minimum and which point gives the maximum
we must evaluate f(x, y) at each point.
whereas
f
f
( 10
√
21
,
15
2 √ 21
)
= 35
2 √ 21
(
−√ 10 , − 15 )
21 2 √ = − 35
21 2 √ 21
so the positive solution gives the location of the maximum and the negative solution
gives the location of the minimum.
Example 15.4 Find the minimum of
f(x, y, z) = x 2 + y 2 + z 2
on the plane
x + 3y − 2z = 12
Solution. We write the constraint as
g(x, y) = x + 3y − 2z − 12 = 0
Setting ∇f = λ∇g gives
or
(2x, 2y2z) = λ (1, 3, −2)
2x = λ
2y = 3λ
2z = −2λ
x + 3y − 2z = 12
Substituting each of the first three equations into the fourth,
λ
2 + 9λ 2
+ 2λ = 12 ⇒ λ =
12
7
Revised December 6, 2006. Math 250, Fall 2006