3 14 3 4 are and 2 1 2 1 x x x x The maximum value of the objective function is

3 14 3 4 are and 2 1 2 1 x x x x the maximum value of

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3 14 , 3 4 are and 2 1 2 1 = = x x x x The maximum value of the objective function is 10 3 14 2 3 4 5 . 0 = × + × = z , which was by [5]. Now consider a non-linear programming problem, which differs from the linear programming problem only in that the objective function: ( ) ( ) . 4 20 5 . 3 10 2 2 2 1 + = x x z (5) Imagine that it is desired to minimize the objective function. Observe that here we have a separable objective function. The graphical solution of this problem is given in Fig.2 Fig. 2 Optimal solution by graphical method The region representing the feasible solution is, of course, precisely the same as that for
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