Engineering Optimization: Theory and Practice, Fourth Edition

88 Classical Optimization Techniques

Proof: The proof is similar to that of Theorem 2.4.

Notes:

1.If

Q=

∑n

i= 1

∑n

j= 1

∂^2 L

∂xi∂xj

(X∗,λ∗)dxidxj

is negative for all choices of the admissible variationsdxi,X∗will be a con-

strained maximum off (X).

2.It has been shown by Hancock [2.1] that a necessary condition for the quadratic

formQ, defined by Eq. (2.43), to be positive (negative) definite for all admissi-

ble variationsdXis that each root of the polynomialzi, defined by the following

determinantal equation, be positive (negative):

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

L 11 − z L 12 L 13... L 1 n g 11 g 21... gm 1

L 21 L 22 − zL 23... L 2 n g 12 g 22... gm 2

..

.

Ln 1 Ln 2 Ln 3... Lnn− zg 1 n g 2 n... gmn

g 11 g 12 g 13... g 1 n 0 0 ... 0

g 21 g 22 g 23... g 2 n 0 0 ... 0

..

.

gm 1 gm 2 gm 3... gmn 0 0 ... 0

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

= 0 (2.44)

where

Lij=

∂^2 L

∂xi∂xj

(X∗,λ∗) (2.45)

gij=

∂gi

∂xj

(X∗) (2.46)

3.Equation (2.44), on expansion, leads to an (n−m)th-order polynomial inz. If

some of the roots of this polynomial are positive while the others are negative,

the pointX∗is not an extreme point.

The application of the necessary and sufficient conditions in the Lagrange multiplier

method is illustrated with the help of the following example.

Example 2.10 Find the dimensions of a cylindrical tin (with top and bottom) made

up of sheet metal to maximize its volume such that the total surface area is equal to

A 0 = 42 π.

SOLUTION Ifx 1 andx 2 denote the radius of the base and length of the tin, respec-

tively, the problem can be stated as

Maximizef (x 1 , x 2 ) =πx 12 x 2