Question

Consider the linear programming problem:
Maximize: \(z = \alpha x + 6y\)
Subject to the constraints
\(3x + 2y \leq 60\)
\(x + 2y \leq 40\)
\(x, y \geq 0\)
If every point in the line segment joining (20, 0) and (10, 15) is optimal solution of the L.P.P, then the value of \(\alpha\) is equal to

• A. 3
• B. 4
• C. 6
• D. 8
• E. 9

Hint:

When multiple optimal solutions exist, the objective function is parallel to the corresponding constraint.

Answer 1

Answer: (E)

Step 1: Concept:
• If every point on a line segment joining two vertices is optimal, the objective function is constant along that edge.
• Hence, values of the objective function at both endpoints must be equal.

Step 2: Detailed Explanation:
• Given points: \((20,0)\) and \((10,15)\)
• Slope of the line: \[ \frac{15 - 0}{10 - 20} = \frac{15}{-10} = -\frac{3}{2} \]
• Equation of the line: \[ y = -\frac{3}{2}(x - 20) = -\frac{3}{2}x + 30 \]
• Value of objective function \(z\):
• At \((20,0)\): \[ z = 20\alpha \]
• At \((10,15)\): \[ z = 10\alpha + 90 \]
• For same optimal value: \[ 20\alpha = 10\alpha + 90 \Rightarrow 10\alpha = 90 \Rightarrow \alpha = 9 \]

Step 3: Final Answer:
• \[ \alpha = 9 \]