Question

For the LPP
Maximise z=x+y
subject to x-y≤-1, x+y≤2, x, y≥0, z has:

• A. Max. value=\(\frac{2}{3}\)
• B. Max. value=5
• C. Max. value=11
• D. No Max. value

Answer 1

The Correct Option is D

To solve the given Linear Programming Problem (LPP), we have the objective function to maximize: \( z = x + y \) subject to the following constraints:

• \( x - y \leq -1 \)
• \( x + y \leq 2 \)
• \( x \geq 0 \)
• \( y \geq 0 \)

First, let's analyze the constraints graphically:

1. The inequality \( x - y \leq -1 \) can be rewritten as \( x \leq y - 1 \).
2. The inequality \( x + y \leq 2 \) describes a line where all points below and including the line satisfy the inequality.

To find the feasible region, determine the intersection of these constraints:

• For \( x = 0 \), the constraints become:
  • \( 0 \leq y - 1 \rightarrow y \geq 1 \)
  • \( 0 + y \leq 2 \rightarrow y \leq 2 \)
  Combining gives \( 1 \leq y \leq 2 \).
• For \( y = 0 \), the constraints are not applicable as they contradict \( y \geq 1 \).

The feasible region is the area where \( x \) is below \( y - 1 \) and above the \( x \)-axis, and \( x \) is below \( 2 - y \).

The area of feasibility extends indefinitely in the positive direction, and for any \( x \leq y - 1 \), we can always find a higher value of \( z = x + y \) by traveling towards the line \( x + y = 2 \).

Thus, the objective function \( z = x + y \) does not have an upper bound in the feasible region. Therefore, the correct conclusion for the objective is:

No Max. value