For x ≥ 0 and y ≥ 0 satisfying x + 3y ≤ 15 and 2x + y ≤ 10, what is the greatest possible value of 3x + 5y?

The maximum of a linear expression over a feasible region occurs at a vertex. The constraints $x \geq 0$, $y \geq 0$, $x + 3y \leq 15$, and $2x + y \leq 10$ give vertices $(0,0)$, $(5,0)$, $(0,5)$, and the intersection of the two lines. Solving $x + 3y = 15$ and $2x + y = 10$: from the second, $y = 10 - 2x$, so $x + 30 - 6x = 15$, giving $x = 3$, $y = 4$. Evaluate $3x + 5y$: 0, 15, 25, and $9 + 20 = 29$. The maximum is 29, choice B. Choice A (25) ignores the interior vertex, choice C (15) uses only the $x$-intercept, and choice D (35) is not attainable.