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Exercise 12.1 · Q10

Q.Find the value of the following: Maximise Z=x+yZ = x + y, subject to x−y≤−1x - y \le -1, −x+y≤0-x + y \le 0, x,y≥0x, y \ge 0.

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The constraints x−y≤−1x - y \le -1 and −x+y≤0-x + y \le 0 together with non-negativity create an infeasible region — no point satisfies all conditions simultaneously. Hence, the maximum does not exist; the problem has no feasible solution.


Why This Happens — The Core Idea

In Linear Programming, the first thing we always check is whether the constraints actually define a region where all conditions hold at once. If they don't, there's nothing to maximise. This problem is a classic trap: the constraints look simple, but they contradict each other when you combine them with x,y≥0x, y \ge 0.

Let’s see why.


Step-by-Step Reasoning

1. Rewrite the constraints in a clearer form.

We have:

  • x−y≤−1x - y \le -1 → y≥x+1y \ge x + 1
  • −x+y≤0-x + y \le 0 → y≤xy \le x
  • x≥0x \ge 0, y≥0y \ge 0

So the first constraint says yy must be at least x+1x + 1. The second says yy must be at most xx.

2. Can both hold at the same time?

If y≥x+1y \ge x + 1 and y≤xy \le x, then we need:

x+1≤y≤xx + 1 \le y \le x

which implies x+1≤xx + 1 \le x, i.e. 1≤01 \le 0. That’s impossible.

No matter what xx and yy are, these two inequalities cannot be satisfied together.

Watch out

A common mistake is to graph each inequality separately and look for an overlapping region — but here the overlap is empty. Don’t assume a solution exists just because each inequality individually has solutions.

3. What about the non-negativity constraints? …

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