Engineering Optimization: Theory and Practice, Fourth Edition

90 Classical Optimization Techniques

L 22 =

∂^2 L

∂x 22

∣

∣

∣

∣

(X∗,λ∗)

= 0

g 11 =

∂g 1

∂x 1

∣

∣

∣

∣

(X∗,λ∗)

= 4 πx 1 ∗+ 2 πx 2 ∗= 61 π

g 12 =

∂g 1

∂x 2

∣

∣

∣

∣

(X∗,λ∗)

= 2 πx 1 ∗= 4 π

Thus Eq. (2.44) becomes ∣

∣ ∣ ∣ ∣ ∣ ∣

4 π−z 2 π 16 π

2 π 0 −z 4 π

16 π 4 π 0

∣ ∣ ∣ ∣ ∣ ∣ ∣

= 0

that is,

272 π^2 z + 192 π^3 = 0

This gives

z= −^1217 π

Since the value ofzis negative, the point(x 1 ∗, x 2 ∗) orresponds to the maximum ofc f.

Interpretation of the Lagrange Multipliers. To find the physical meaning of the

Lagrange multipliers, consider the following optimization problem involving only a

single equality constraint:

Minimizef (X) (2.47)

subject to

̃

g(X)=b or g(X)=b−

̃

g(X)= 0 (2.48)

wherebis a constant. The necessary conditions to be satisfied for the solution of the

problem are

∂f

∂xi

+λ

∂g

∂xi

= 0 , i= 1 , 2 ,... , n (2.49)

g= 0 (2.50)

Let the solution of Eqs. (2.49) and (2.50) be given byX∗, λ∗, andf∗= f(X∗).

Suppose that we want to find the effect of a small relaxation or tightening of the

constraint on the optimum value of the objective function (i.e., we want to find the

effect of a small change inbonf∗). For this we differentiate Eq. (2.48) to obtain

db−d

̃

g= 0