Question

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

• A. p = q
• B. p = 2q
• C. p = 3q
• D. q = 3p

Hint:

For an objective function Z = ax + by, the slope of the iso-profit/iso-cost line is -a/b. When multiple optimal solutions exist, this slope is equal to the slope of one of the boundary lines of the feasible region. In this case, the line connecting (3,4) and (0,5) has a slope of \((5-4)/(0-3) = -1/3\). The slope of Z = px + qy is -p/q. So, \(-p/q = -1/3\), which gives \(q=3p\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In a Linear Programming Problem (LPP), if the optimal (maximum or minimum) value of the objective function occurs at two distinct corner points of the feasible region, then the optimal value also occurs at every point on the line segment joining these two points. This happens when the slope of the objective function is the same as the slope of the constraint line that forms the edge between those two points.
Step 2: Key Formula or Approach:
If the objective function Z has the same optimal value at two points, say \((x_1, y_1)\) and \((x_2, y_2)\), then:
\[ p x_1 + q y_1 = p x_2 + q y_2 \] Step 3: Detailed Explanation:
The objective function is Z = px + qy.
The problem states that the optimum value occurs at both corner points (3, 4) and (0, 5).
This means the value of Z is the same at these two points.
Value of Z at (3, 4):
\[ Z_1 = p(3) + q(4) = 3p + 4q \] Value of Z at (0, 5):
\[ Z_2 = p(0) + q(5) = 5q \] Set the two values equal to each other:
\[ Z_1 = Z_2 \] \[ 3p + 4q = 5q \] Now, solve for the relationship between p and q:
\[ 3p = 5q - 4q \] \[ 3p = q \] Step 4: Final Answer:
The relationship between p and q is q = 3p.