Boxed in

$V$ is maximum when the rectangular box is a cube. This may be explained by the AM-GM inequality. Consider the side lengths of the rectangular boxes be $a$ , $b$ and $c$ , then we have $\dfrac{a+b+c}{3} \ge \sqrt[3]{abc} = \sqrt[3]{V}\quad \Rightarrow V$ is maximum when $a=b=c$ .

Let the sides of the cube be $x$ then $L^2+ x^2+x^2+x^2 = 3x^2 \quad \Rightarrow x = \dfrac {L}{\sqrt{3}}$

$\Rightarrow V = x^3 = \left( \dfrac {L}{\sqrt{3}}\right)^3 \quad \Rightarrow V^2 = \dfrac {L^6}{27} \quad \Rightarrow a + b + c = 1+0+27 = \boxed{28}$