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

The key idea is that the feasible region is defined by the intersection of the given linear inequalities.

Step 1: Rewrite the constraints:

$x - y \le -1$ means $y \ge x + 1$.

$-x + y \le 0$ means $y \le x$.

Also $x, y \ge 0$.

Step 2: The conditions $y \ge x+1$ and $y \le x$ cannot both be satisfied for any real $x, y$, because $x+1 > x$ for all $x$. …

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 $x, y \ge 0$.

Let’s see why.

Step-by-Step Reasoning

1. Rewrite the constraints in a clearer form.

We have:

• $x - y \le -1$ → $y \ge x + 1$
• $-x + y \le 0$ → $y \le x$
• $x \ge 0$, $y \ge 0$

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

2. Can both hold at the same time?

If $y \ge x + 1$ and $y \le x$, then we need:

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

No matter what $x$ and $y$ are, these two inequalities cannot be satisfied together.

3. What about the non-negativity constraints? …

Method: Graphical Method — Test Feasibility Before Optimising

Use this whenever you are asked to maximise or minimise a linear objective $Z=ax+by$ subject to linear inequality constraints: the very first job is to check that a feasible region even exists, then optimise over its corners.

Steps

Step 1: Rewrite every constraint in a comparable form.

Solve each inequality for $y$ (or otherwise line them up) so you can compare them directly. A constraint like $px+qy\le r$ becomes a bound on $y$, which makes contradictions between two constraints visible.

Step 2: Look for a direct contradiction before plotting.

If one constraint forces $y\ge f(x)$ and another forces $y\le g(x)$ with $f(x)>g(x)$ for every $x$ in the allowed range, no point can satisfy both — the feasible region is empty and the problem is infeasible. When that happens, stop: there is nothing to optimise, and no corner points to evaluate.

Step 3: If no contradiction, draw the lines and shade. …

Common Mistakes

Mistake 1: Assuming a maximum exists and hunting for corner points.

Why it's wrong: the constraints reduce to $y\ge x+1$ and $y\le x$, which force $x+1\le x$ — impossible. There is no feasible region and therefore no corners to evaluate. Correct approach: always test feasibility first; if two constraints contradict, report "no feasible solution" instead of producing a value.

Mistake 2: Solving $x-y=-1$ and $-x+y=0$ simultaneously and quoting that point.

Why it's wrong: those two lines are parallel (both have slope $1$), so they never meet — and even the region between them is empty. A line intersection is only meaningful if the point satisfies every inequality direction. Correct approach: check the inequality signs, not just the equations. …