= Solution
If a <differentiable> function has a constrained local extremum at a point of a smooth surface, its <derivative> vanishes in every tangent direction; equivalently its <gradient> is normal to the surface. For a regular equation $F=0$ this means $\nabla V=\lambda\nabla F$ for some real multiplier.
First the box is centred at the origin: holding the other coordinates of a <vertex of a polyhedron> fixed and comparing the two possible coordinates $u,v$ in the first direction gives $u^2=v^2$, so the distinct extremes are opposites; the other two directions give the same conclusion. Write its positive half-lengths as $a r,b s,c t$, with $r^2+s^2+t^2=1$. Its volume is $8abc\,rst$. A degenerate box has zero volume, so a positive <maximum> is interior in $r,s,t>0$. <Lagrange multipliers> applied to $\log(rst)$ give $1/r=2\lambda r$, and the analogous equations imply $r=s=t=1/\sqrt3$. Alternatively the arithmetic–geometric mean inequality applied to $r^2,s^2,t^2$ proves global maximality. Thus
$$
\boxed{\text{side lengths }\frac{2a}{\sqrt3},\frac{2b}{\sqrt3},\frac{2c}{\sqrt3},\qquad V_{\max}=\frac{8abc}{3\sqrt3}.}
$$
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