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Question
The corner points of the feasible region determined by the following system of linear inequalities: 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 (B) p = 2 q (C) p = 3 q (D) q = 3 p
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Solution
All constraints are given as,
2 x + y ≤ 10 x + 3 y ≤ 15 x ≥ 0 y ≥ 0
It is given that Maximum Z = p x + q y .
Corner points Value of z ( 0 , 0 ) 0 ( 5 , 0 ) 5 p ( 3 , 4 ) 3 p + 4 q ( 0 , 5 ) 5 q
It is given that maximum value of Z occurs on ( 3 , 4 ) and ( 0 , 5 ) . So it can be written as,
3 p + 4 q = 5 q 3 p = q
Then, value of Z is maximum if q = 3 p .
Thus, the correct option is (D).
All constraints are given as,
It is given that
| Corner points | Value of |
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| | |
| | |
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It is given that maximum value of
Then, value of
Thus, the correct option is (D).
Suggest Corrections
0
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