Q30: Maximise $z = 500x + 300y$ graphically

Each constraint is drawn first as a straight line, by replacing the inequality sign with $=$⁠. Two points fix a line, and the easiest two are where it meets the axes: put $y = 0$ to get the x-intercept, then $x = 0$ to get the y-intercept.

For example, for $L_1$⁠: $y = 0$ gives $x = 12$⁠, and $x = 0$ gives $2y = 12$⁠, so $y = 6$⁠.

The same for $L_2$ and $L_3$⁠:

Each line divides the plane into two halves. A test point not on the line tells us which half satisfies the inequality: if the point satisfies it, shade its side; if not, shade the other side. The origin is the easiest test point when the line does not pass through it.

$x, y \ge 0$ keeps us in the first quadrant. The feasible region is the part that satisfies every constraint at once. It is bounded because it fits inside a finite box.

C is where $L_1$ and $L_2$ cross, so it satisfies both equations. Doubling the second equation makes the $y$ terms equal, $2y$ in each, and subtracting removes $y$⁠.

Put $x = 4$ into $x + 2y = 12$⁠:

The other corners lie on the axes, so we read them from the intercepts: A(0, 4) and B(0, 6) on the y-axis, E(5, 0) and D(6, 0) on the x-axis.

Put each corner into $z = 500x + 300y$⁠. For example, at C(4, 4):

The other four corners:

Corner point theorem: when the feasible region is bounded, $z$ has a maximum and a minimum, and each occurs at a corner point. So comparing the values at the corners is enough.