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
Given the vertices of the feasible region are:
O (0,0)
A (5,0)
B (3,4)
C (0,5)
Also given the objective function is Z = px+qy
Now substituting O,A,B and C in Z
As per the condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5)
Then we can equate Z values at B and C, this gives
3p + 4q = 5q
3p = 5q – 4q
3p = q
Therefore the answer is option D i.e. q = 3p.