Minimise Z = 4x + 3y subject to 2x + y ≥ 8, x + 3y ≥ 9, x ≥ 0 and y ≥ 0…

Step 1. The values. At $(0, 8)$, $Z = 24$. At $(3, 2)$, $Z = 12 + 6 = 18$. At $(9, 0)$, $Z = 36$. The smallest is $M = 18$.

Step 2. The slopes. The edge $2x + y = 8$ has slope $-2$. The edge $x + 3y = 9$ has slope $-\dfrac{1}{3}$. The line $4x + 3y = 18$ has slope $-\dfrac{4}{3}$. Since $-2 < -\dfrac{4}{3} < -\dfrac{1}{3}$, this line passes through $(3, 2)$ and stays outside the region on both sides of it.

Step 3. Conclude. The region lies on the side with $4x + 3y \ge 18$. The half-plane $Z < 18$ has no point in common with the region.