Engineering Optimization: Theory and Practice, Fourth Edition

430 Nonlinear Programming III: Constrained Optimization Techniques

subject to

x 1 +x 2 +x 3 ≤ 06 (E 2 )

x 1 ≤ 63 (E 3 )

xi≥ 0 , i= 1 , 2 , 3 (E 4 )

By introducing new variables as

y 1 =x 1 , y 2 =x 2 , y 3 =x 1 +x 2 +x 3 (E 5 )

or

x 1 =y 1 , x 2 =y 2 , x 3 =y 3 −y 1 −y 2 (E 6 )

the constraints of Eqs.(E 2 ) ot (E 4 ) an be restated asc

0 ≤y 1 ≤ 63 , 0 ≤y 2 ≤ 06 , 0 ≤y 3 ≤ 06 (E 7 )

where the upper bound, for example, ony 2 is obtained by settingx 1 =x 3 = 0 in

Eq. (E 2 ). The constraints of Eq. (E 7 ) will be satisfied automatically if we define new

variableszi, i= 1 , 2 , 3 , as

y 1 6 sin= 3 2 z 1 , y 2 = 0 sin 6 2 z 2 , y 3 = 0 sin 6 2 z 3 (E 8 )

Thus the problem can be stated as an unconstrained problem as follows:

Maximizef (z 1 , z 2 , z 3 )

=y 1 y 2 (y 3 −y 1 −y 2 ) (E 9 )

= 2 160 sin^2 z 1 sin^2 z 2 ( 0 sin 6 2 z 3 − 6 sin 3 2 z 1 − 0 sin 6 2 z 2 )

The necessary conditions of optimality yield the relations

∂f

∂z 1

= 592 ,200 sinz 1 cosz 1 sin^2 z 2 ( ins^2 z 3 −^65 sin^2 z 1 − ins^2 z 2 )= 0 (E 10 )

∂f

∂z 2

= 185 ,400 sin^2 z 1 sinz 2 cosz 2 (^12 sin^2 z 3 − 103 sin^2 z 1 − ins^2 z 2 ) = 0 (E 11 )

∂f

∂z 3

= 592 ,200 sin^2 z 1 sin^2 z 2 sinz 3 cosz 3 = 0 (E 12 )

Equation (E 12 ) gives the nontrivial solution as cosz 3 = 0 or sin^2 z 3 =. Hence 1

Eqs.(E 10 ) nda (E 11 ) ield siny^2 z 1 =^59 and sin^2 z 2 =^13. Thus the optimum solution is

given byx∗ 1 0 in.,= 2 x 2 ∗= 0 in., 2 x 3 ∗= 0 in., and the maximum volume 2 =8000 in^3.

7.12 Basic Approach of the Penalty Function Method

Penalty function methods transform the basic optimization problem into alternative

formulations such that numerical solutions are sought by solving a sequence of