Corner points of the feasible region determined by the system of linear constraints are (0, 4), (3, 3), (4, 0). Let $Z= px+ qy (p> 0, q > 0).$ Then the condition on p and q so that minimum of Z occurs at (4, 0) and (3, 3) is :

$p=3q$

The correct answer is Option (1) → $p=3q$

To have the minimum occur at (4, 0) and (3, 3), the values of Z at these two points must be equal, and both must be less than Z at (0, 4).

Equate Z at (4, 0) and (3, 3):

$4p=3p+3q$

$p=3q$