The corner points of the bounded feasible region associated with the LPP: Maximize $Z=px+qy, p,q>0$ are (0, 0), (3.5, 0), $\left(\frac{112}{59},\frac{135}{59}\right)$ and (0, 3). If the optimum value of Z occurs at both $\left(\frac{112}{59},\frac{135}{59}\right)$ and (0, 3), then

$8p=3q$

The correct answer is Option (2) → $8p=3q$

Given: A linear programming problem with objective function Z = px + qy, where p, q > 0, and the corner points of the feasible region are:

(0, 0), (3.5, 0), (112/59, 135/59), and (0, 3)

Also, it is given that the optimum value of Z occurs at both points:

(112/59, 135/59) and (0, 3)

Then: The optimum value occurring at two distinct corner points implies that the line Z = px + qy is parallel to the line joining these two points.

Step 1: Find the slope of the line joining the points:

Slope = $\frac{135/59 - 3}{112/59 - 0} = \frac{\frac{135 - 177}{59}}{112/59} = \frac{-42/59}{112/59} = -\frac{42}{112} = -\frac{3}{8}$

Step 2: For Z = px + qy to be constant along this line, the slope of this line must be same as that of the level curve px + qy = constant.

The slope of the line px + qy = k is given by: $-\frac{p}{q}$

So, $-\frac{p}{q} = -\frac{3}{8} \Rightarrow \frac{p}{q} = \frac{3}{8}$