For x ≥ 0 and y ≥ 0 satisfying x + 3y ≤ 17 and 2x + y ≤ 14, what is the greatest possible value of 4x + 3y?

The maximum of $4x + 3y$ over the region occurs at a vertex. The constraints $x, y \geq 0$, $x + 3y \leq 17$, and $2x + y \leq 14$ give vertices $(0, 0)$, $(7, 0)$, $(0, \frac{17}{3})$, and the intersection of the two lines. From $y = 14 - 2x$ in $x + 3y = 17$: $x + 42 - 6x = 17$, so $x = 5$ and $y = 4$. Evaluate $4x + 3y$: 0, 28, 17, and $20 + 12 = 32$. The maximum is 32, choice C. Choice A (28) is the value at $(7, 0)$. Choice B (17) is the value at $(0, \frac{17}{3})$. Choice D (30) is not achieved at any vertex.