Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / F (Multivariable Calculus)

A two-dimensional embedded submanifold is a subset which near every point can be straightened by a smooth local coordinate change in the ambient space to an open piece of the coordinate plane ${\mathbb{R}}^2\times \{0\}$. Equivalently locally it has a regular smooth parametrization of rank two, with the induced local topology.

A tangent vector at $x$ is the derivative ${\gamma }^{\prime }(0)$ of a smooth curve in the submanifold with $\gamma (0)=x$. These derivatives form the two-dimensional tangent space $T_{x}M$. A normal vector is a vector orthogonal to every tangent vector. Its space is $(T_{x}M{)}^{\perp }$, which has $\boxed{\dim (T_{x}M{)}^{\perp }=1}$ in the three-dimensional ambient Euclidean space.

Define $F(x)=x_1^2/a^2+x_2^2/b^2+x_3^2/c^2-1$. Its gradient is $2(x_1/a^2,x_2/b^2,x_3/c^2)$. It cannot vanish on $F^{-1}(0)$, because the only possible zero is the origin and that point has $F=-1$. The regular level-set theorem therefore makes the ellipsoid a two-dimensional embedded submanifold. Its normal space isand the tangent space is the orthogonal complement of that line.

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 $ar,bs,ct$, with $r^2+s^2+t^2=1$. Its volume is $8abcrst$. 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 $r=s=t=1/\sqrt{3}$. Alternatively the arithmetic–geometric mean inequality applied to $r^2,s^2,t^2$ proves global maximality. Thus