Q27: Maximise $Z = x + 2y$ graphically: an unbounded region

Each constraint is bounded by a straight line: write it with "$=$" in place of the inequality sign, and plot two points of it.

For $x - y = 0$⁠: $x = 0$ gives $y = 0$⁠, and $x = 2$ gives $y = 2$⁠.

For $x - 2y = -2$⁠: $y = 0$ gives $x = -2$⁠, and $x = 0$ gives $y = 1$⁠.

A point that is not on a line tells us which side of the line to keep. At $(1, 0)$⁠: $x - y = 1$⁠, which is $\ge 0$⁠, and $x - 2y = 1$⁠, which is $\ge -2$⁠. Both are true, so keep the side containing $(1, 0)$ for both lines.

$x \ge 0$⁠, $y \ge 0$ keep only the first quadrant. The region is unbounded: every point $(n, 0)$ with $n \ge 0$ satisfies all four constraints, however large $n$ is.

The corner A is where the two lines meet, so solve their equations together. From $x - y = 0$⁠, $x = y$⁠; substitute in the other equation.

The other corner is the origin $O$⁠, where $x - y = 0$ meets the $x$-axis.

Corner point method: work out $Z$ at every corner of the feasible region. For a bounded region, the largest of these values is the maximum.

Here the region is unbounded, so $6$ is only a candidate. It must be checked before we can call it the maximum.

The rule for an unbounded region: $M = 6$ is the maximum only if the open half-plane $x + 2y > 6$ has no point in common with the feasible region.

$(8, 0)$ is feasible: $8 - 0 \ge 0$⁠, $8 - 0 \ge -2$⁠, $8 \ge 0$ and $0 \ge 0$⁠. It gives $Z = 8 + 2(0) = 8$⁠, more than $6$⁠. In fact $Z = n$ at $(n, 0)$⁠, so $Z$ grows without limit.

Watch out: do not write "maximum $= 6$" just because $6$ is the largest corner value.