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Q. 11.     The corner points of the feasible region determined by the following system of linear inequalities:

               2x+y \leq 10,x+3y \leq 15,x,y\geq 0 are (0,0),(5,0),(3,4) and (0,5). Let Z=px+qy,                where p,q > 0. Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is

               (B)\; p=2q

Answers (1)

At (0, 0) Z = 0 

At (5, 0) Z = 5p 

At (3, 4) Z = 3p +4q 

At (0, 5) Z = 5q 

∴ maximum of Z  3p + 4q = 5q  ∴ 3p = q

Posted by

Satyajeet Kumar

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