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Question

The corner points of the feasible region determined by the following system linear inequalities:
2x+y10,x+3y15,xy0 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

A
p=q
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B
p=2q
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C
p=3q
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D
q=3p
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Solution

The correct option is D q=3p
The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points, (3,4) and (0,5)
Value of Z at (3,4)= Value of Z at (0,5)
p(3)+q(4)=p(0)+q(5)
3p+4q=5q q=3p

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