Largest box under the plane

We want to maximize $xyz$ subject to $2x + 7y + z = 12$ . By AM-GM, $12 = 2x + 7y + z \ge 3 \sqrt[3]{(2x)(7y)(z)} = 3 \sqrt[3]{14xyz},$ so $xyz \le \frac{4^3}{14} = \frac{32}{7}$ . Equality occurs when $2x = 7y = z = 4$ .

The volume of the box is given by

$V = x y z = x y (a - b x - c y)$

Taking the partial derivatives of $V$ with respect to $x$ and $y$ :

$V_x = y (a - b x - c y) + x y (- b) = y (a - 2 b x - c y)$

$V_y = x (a - b x - cy ) + x y (- cy) = x (a - b x - 2 c y)$

At the critical point (the extremum), $V_x = V_y = 0$ . Since $x \ne 0$ , $y \ne 0$ , then

$2 b x + c y = a$

and

$b x + 2 cy = a$

The solution of this linear system is $x = \dfrac{a}{3b}$ , $y = \dfrac{a }{3 c }$

Clearly this critical point corresponds to a maximum of $V$ , since at $x = 0$ or $y = 0$ the volume is zero. However, one can check that by taking the second partial derivatives:

$V_{xx} = y (- 2 b)$ , $V_{xy} = (a - 2 b x - cy) - c y = a - 2 b x - c y$ , $V_{yy} = x (- 2 c)$ , hence the Hessian matrix is

$\mathbf{H} = \displaystyle \begin{bmatrix} - \dfrac{2 a b}{3 c} && -\dfrac{a}{3} \\ -\dfrac{a}{3} && - \dfrac{2 a c}{3 b} \end{bmatrix}$

Since $-H_{11} = \dfrac{2 a b}{3 c} \gt 0$ and $| \mathbf{-H} | = \dfrac{1}{3} a^2 \gt 0$ , then matrix $\mathbf{H}$ is negative definite, and hence our critical point is a local maximum.

At this point, $z = a - b x - c y = \dfrac{a}{3}$

Therefore, the maximum volume is $V_{\text{Max} } = \dfrac{a^3}{27 b c }$

Substituting $a = 12, b = 2 , c = 7$ , gives us $V_{\text{Max}} = \dfrac{12^3}{27(2)(7) } = \dfrac{32}{7}$

Thus the answer is $32 + 7 = \boxed{39}$