= 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:
$$
1/x=2\lambda x/a^2,\quad1/y=2\lambda y/b^2,\quad1/z=2\lambda z/c^2.
$$
Thus the three squared scaled coordinates are equal and sum to one. The unique positive solution gives
$$
\boxed{\text{side lengths }2a/\sqrt3,\ 2b/\sqrt3,\ 2c/\sqrt3;\qquad V_{\max}=\frac{8abc}{3\sqrt3}}.
$$
Back to article page
