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 Definition:
In Linear Programming Problems (LPP), if the objective function's optimal value (maximum or minimum) is attained at two different corner points of the feasible region, then this optimal value also applies to all points on the line segment connecting these two points. This condition arises when the objective function's slope matches the slope of the constraint line that forms the boundary between these corner points.

Step 2: Core Principle:
If the objective function Z yields the identical optimal value at two distinct points, denoted as \((x_1, y_1)\) and \((x_2, y_2)\), then the following equality holds:
\[ p x_1 + q y_1 = p x_2 + q y_2 \]

Step 3: Derivation:
Given the objective function Z = px + qy.
The problem specifies that the optimal value is achieved 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 established between p and q is q = 3p.