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

$2x + y ≤ 10, x + 3y ≤ 15, x ≥ 0, y ≥ 0$ are $(0, 0), (5, 0), (3, 4)$ and $(0, 5)$.

Let $Z = px + qy$, when $p, q> 0$, then find the relation between $p$ and $q$ so that the maximum of Z occurs at both points $(3, 4)$ and $(0, 5)$.

$3p=q$

The correct answer is Option (2) → $3p=q$

The values of $Z = px + qy$ at the points $(3, 4)$ and $(0,5)$ are $3p + 4q$ and $5q$ respectively. As $Z$ has maximum value at both points $(3, 4)$ and $(0,5)$, we get $3p+4q = 5q⇒3p=q$, which is the required relation between $p$ and $q$.