Engineering Optimization: Theory and Practice, Fourth Edition

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