Past exam of the options II course of the University of Oxford / 2008 printed variant / F / F2

A two-dimensional embedded submanifold is locally flattened by a continuously differentiable coordinate diffeomorphism of an ambient open neighbourhood onto a coordinate plane, or equivalently locally admits a regular two-parameter chart. A tangent vector at $x$ is the velocity ${\gamma }^{\prime }(0)$ of a differentiable curve in $M$ through $x$. All such vectors form the two-dimensional tangent space $T_{x}M$, the image of the differential of a regular parametrization. Normal vectors form its orthogonal complement $(T_{x}M{)}^{\perp }$ in ${\mathbb{R}}^3$, which has dimension one.

The defining function $h(x)=x_1^2/a^2+x_2^2/b^2+x_3^2/c^2-1$ is smooth and has gradient $2(x_1/a^2,x_2/b^2,x_3/c^2)$. This cannot vanish on its zero level, since the origin does not lie there. The regular level-set theorem therefore makes $M$ a two-dimensional submanifold, with tangent space $\ker Dh(x)$ and normal space

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