Q.Show that the LPP: Maximise $Z=x+y$ subject to $x-y\ge -1,\ -x+y\le 1$ (equivalently $y-x\le1$), $x\ge 0,\ y\ge 0$ has no maximum value.

Both constraints reduce to one strip, leaving an unbounded region on which $Z=x+y$ grows without limit.

The constraints. $x-y\ge-1$ and $-x+y\le1$ are the same condition written two ways: both say $y-x\le1$, i.e. $y\le x+1$. With $x\ge0,\ y\ge0$, the feasible region is the part of the first quadrant on or below the line $y=x+1$ — an unbounded region stretching infinitely up-right.

Why no maximum. Consider feasible points along the line $x=y$ (which satisfies $y\le x+1$ since $x\le x+1$ always, and $x,y\ge0$). At such a point $Z=x+y=2x$. As $x\to\infty$, $Z=2x\to\infty$. So for any number $M$ we can find a feasible point with $Z>M$ — the open half-plane $x+y>M$ always meets the region. Therefore $Z$ has no maximum on this feasible region. …

Reporting the largest corner value as "the maximum" on an unbounded region. When the open half-plane $x+y>M$ meets the region for every $M$, there is genuinely no …