Q30: Minimise $Z = 13x - 15y$ graphically

Each inequality is bounded by a straight line, found by writing "=" in place of the inequality sign. Two points fix a line, and the easiest are where it meets the axes.

For $x + y = 7$⁠: $y = 0$ gives $x = 7$⁠, and $x = 0$ gives $y = 7$⁠.

For $2x - 3y + 6 = 0$⁠: $y = 0$ gives $2x = -6$⁠, so $x = -3$⁠. And $x = 0$ gives $-3y = -6$⁠, so $y = 2$⁠.

A line splits the plane into two sides, and an inequality holds on one whole side. Testing one point that is not on the line tells us which side. The origin is the easiest test point.

At $(0, 0)$⁠: $x + y = 0$⁠, which is $\le 7$⁠. And $2x - 3y + 6 = 6$⁠, which is $\ge 0$⁠.

$x \ge 0$⁠, $y \ge 0$ keep us in the first quadrant. The feasible region is where all four conditions hold at once.

B is where the two lines cross, so its coordinates satisfy both equations. Solve them together.

From $x + y = 7$⁠, $x = 7 - y$⁠. Then $2(7 - y) - 3y = 14 - 5y$⁠, so $-5y = -20$ and $y = 4$⁠.

The second line $2x - 3y + 6 = 0$ is the same as $2x - 3y = -6$⁠. Then:

Check: $3 + 4 = 7$ and $2(3) - 3(4) + 6 = 0$⁠.

Corner point method: if $Z$ has a minimum on the feasible region, it occurs at a corner. So we only need $Z$ at the corners:

At O both terms are 0.

The smallest of $0$⁠, $91$⁠, $-21$⁠, $-30$ is $-30$⁠, at C.

On a bounded feasible region, $Z$ always has both a maximum and a minimum, so the smallest corner value is the answer.

Watch out: if the region were unbounded, we would also need to check that the half-plane $13x - 15y < -30$ has no point in common with the region.