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

2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 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


A. p =q B. p = 2q


C. p = 3q D. q = 3p


In the given question the constraints are


2x + y ≤ 10,


x + 3y ≤ 15,


and x, y ≥ 0


The function is to maximize Z = px + qy


The corner points are (0, 0), (5, 0), (3, 4) and (0, 5).


The value of Z at these points are



Since the maximum value of z occurs on (3, 4) and (0, 5)


Hence the value on (3, 4) = Value on (0, 5)


3p+ 4q = 5q


3p = q


Value of Z will be maximum if q =3p


Hence D is the correct answer.


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