 Solution

The Lagrange multiplier condition says that at a constrained local extremum of a continuously differentiable objective on a regular surface $h=0$, its gradient is normal to the tangent plane: $\nabla f=\lambda \nabla h$ for some scalar multiplier.

First the box must be centred at zero. If the two face coordinates in one direction are $u_i,v_i$, subtracting the ellipsoid equations for two vertices differing only in that coordinate gives $u_i^2=v_i^2$. A nondegenerate box has $u_i\ne v_i$, hence $u_i=-v_i$. Its positive half-sides $x,y,z$ therefore satisfy $x^2/a^2+y^2/b^2+z^2/c^2=1$ and its volume is $8xyz$.

The feasible closed octant is compact, the boundary gives zero volume, and some interior box has positive volume, so the maximum is interior. Apply multipliers to $\log (xyz)$ there:Thus the three squared scaled coordinates are equal and sum to one. The unique positive solution gives