2.4 Multivariable Optimization with Equality Constraints 89
subject to
2 π x 12 + 2 πx 1 x 2 =A 0 = 42 π
The Lagrange function is
L(x 1 , x 2 , λ) =πx^21 x 2 + λ( 2 π x 12 + 2 πx 1 x 2 −A 0 )
andthe necessary conditions for the maximum off give
∂L
∂x 1
= 2 πx 1 x 2 + 4 πλx 1 + 2 πλx 2 = 0 (E 1 )
∂L
∂x 2
= πx 12 + 2 πλx 1 = 0 (E 2 )
∂L
∂λ
= 2 π x 12 + 2 πx 1 x 2 −A 0 = 0 (E 3 )
Equations(E 1 ) nd (Ea 2 ) ead tol
λ= −
x 1 x 2
2 x 1 +x 2
= −
1
2
x 1
that is,
x 1 =^12 x 2 (E 4 )
and Eqs.(E 3 ) nda (E 4 ) ive the desired solution asg
x∗ 1 =
(
A 0
6 π
) 1 / 2
, x∗ 2 =
(
2 A 0
3 π
) 1 / 2
, nda λ∗= −
(
A 0
24 π
) 1 / 2
This gives the maximum value offas
f∗=
(
A^30
54 π
) 1 / 2
IfA 0 = 42 π, the optimum solution becomes
x 1 ∗ = 2 , x 2 ∗= 4 , λ∗= − 1 , and f∗= 61 π
To see that this solution really corresponds to the maximum off, we apply the suffi-
ciency condition of Eq. (2.44). In this case
L 11 =
∂^2 L
∂x^21
∣
∣
∣
∣
(X∗,λ∗)
= 2 πx∗ 2 + 4 πλ∗= 4 π
L 12 =
∂^2 L
∂x 1 ∂x 2
∣
∣
∣
∣
(X∗,λ∗)
=L 21 = 2 πx 1 ∗+ 2 πλ∗= 2 π