Question

Corner points of the feasible region for an LPP are $(0, 2), (3, 0), (6, 0), (6, 8)$ and $(0, 5)$. Let $z = 4x + 6y$ be the objective function. The minimum value of $z$ occurs at:

• A. Only $(0, 2)$
• B. Only $(3, 0)$
• C. The mid-point of the line segment joining the points $(0, 2)$ and $(3, 0)$
• D. Any point on the line segment joining the points $(0, 2)$ and $(3, 0)$

Answer 1

The Correct Option is D

1. List the corner points:

Given points: \( (0, 2) \), \( (3, 0) \), \( (6, 0) \), \( (6, 8) \), \( (0, 5) \).

2. Evaluate the objective function \( z = 4x + 6y \) at each point:

\[ \begin{align*} (0, 2) & : z = 4(0) + 6(2) = 12 \\ (3, 0) & : z = 4(3) + 6(0) = 12 \\ (6, 0) & : z = 4(6) + 6(0) = 24 \\ (6, 8) & : z = 4(6) + 6(8) = 72 \\ (0, 5) & : z = 4(0) + 6(5) = 30 \\ \end{align*} \]

3. Determine the minimum value:

The minimum value of \( z \) is 12, achieved at both \( (0, 2) \) and \( (3, 0) \).

Since the feasible region is convex, the minimum also occurs at all points on the line segment joining \( (0, 2) \) and \( (3, 0) \).

Correct Answer: (D) Any point on the line segment joining the points (0, 2) and (3, 0)

Answer 2

Let's calculate the value of $ z $ at each corner point:

• (0, 2): $ z = 4(0) + 6(2) = 12 $
• (3, 0): $ z = 4(3) + 6(0) = 12 $
• (6, 0): $ z = 4(6) + 6(0) = 24 $
• (6, 8): $ z = 4(6) + 6(8) = 72 $
• (0, 5): $ z = 4(0) + 6(5) = 30 $

The minimum value of $ z $ is 12, and it occurs at both $ (0, 2) $ and $ (3, 0) $.

The line segment connecting $ (0, 2) $ and $ (3, 0) $ can be represented by the equation:

$$ y = -\frac{2}{3}x + 2 $$

Substitute this into the objective function:

$$ z = 4x + 6\left(-\frac{2}{3}x + 2\right) = 4x - 4x + 12 = 12 $$

This shows that the value of $ z $ remains constant ($ 12 $) at every point along the line segment between $ (0, 2) $ and $ (3, 0) $.

Therefore, the minimum value of $ z $ occurs at any point on the line segment joining $ (0, 2) $ and $ (3, 0) $.