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: Concept Overview:
In Linear Programming Problems (LPPs), if the optimal value (maximum or minimum) of the objective function is achieved at two different corner points of the feasible region, then this optimal value is also attained at every point on the line segment connecting these two points. This condition arises when the slope of the objective function matches the slope of the constraint line forming the edge between these corner points.
Step 2: Method:
If the objective function Z yields the same optimal value at two points, \((x_1, y_1)\) and \((x_2, y_2)\), then the following holds:
\[ p x_1 + q y_1 = p x_2 + q y_2 \] Step 3: Calculation:
The objective function is given by Z = px + qy.
The problem states that the optimal value occurs at corner points (3, 4) and (0, 5).
This implies that the value of Z is identical at these two points.
Calculating Z at (3, 4):
\[ Z_1 = p(3) + q(4) = 3p + 4q \] Calculating Z at (0, 5):
\[ Z_2 = p(0) + q(5) = 5q \] Equating the two values:
\[ Z_1 = Z_2 \] \[ 3p + 4q = 5q \] Solving for the relationship between p and q:
\[ 3p = 5q - 4q \] \[ 3p = q \] Step 4: Conclusion:
The relationship between p and q is q = 3p.