For x ≥ 0 and y ≥ 0 satisfying x + 2y ≤ 12 and 3x + y ≤ 16, what is the greatest possible value of 2x + 3y?

The maximum of a linear expression over a feasible region occurs at a vertex. The vertices of the region defined by $x, y \geq 0$, $x + 2y \leq 12$, and $3x + y \leq 16$ are $(0, 0)$, $(\frac{16}{3}, 0)$, $(0, 6)$, and the intersection of the two lines. From $3x + y = 16$, $y = 16 - 3x$; substituting into $x + 2y = 12$ gives $x + 32 - 6x = 12$, so $x = 4$ and $y = 4$. Evaluate $2x + 3y$: 0, $\frac{32}{3}$, 18, and $8 + 12 = 20$. The maximum is 20. A common wrong answer is 18, from only checking the axis intercepts.