The corner points of the feasible region associated with the LPP: Maximise Z = px + qy, p, q > 0 subject to 2x + y ≤ 10, x + 3y ≤ 15, x,y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). If optimum value occurs at both (3, 4) and (0, 5), then

The correct option is q = 3p.

Step-by-step Explanation:

In a Linear Programming Problem (LPP), the objective function is given as:

$Z=px+qy$

where $p,q>0$.

We are given that the optimum (maximum) value of the objective function $Z$ occurs at two corner points, namely $(3,4)$ and $(0,5)$. According to the corner point method, if the optimal value occurs at two corner points, the value of the objective function must be equal at both of these points.

Let us calculate the value of $Z$ at the corner point $(3,4)$ by substituting $x=3$ and $y=4$:

$Z_{(3,4)}=p\left(3\right)+q\left(4\right)=3p+4q$

Next, let us calculate the value of $Z$ at the corner point $(0,5)$ by substituting $x=0$ and $y=5$:

$Z_{(0,5)}=p\left(0\right)+q\left(5\right)=5q$

Since the optimum value is the same at both corner points, we equate the two expressions:

$3p+4q=5q$

Subtracting $4q$ from both sides of the equation, we get:

$3p=5q-4q$

Simplifying the right side gives:

$3p=q$

Rearranging the equation yields:

$q=3p$

Hence, the relation between $p$ and $q$ for the optimum value to occur at both points is $q=3p$.