Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / F (Multivariable Calculus)

The full-rank constraint qualification gives a multiplier vector $\lambda \in {\mathbb{R}}^2$ such thatEquivalently the gradient of the objective lies in the span of the two constraint gradients. The point also satisfies $f(x_0)=0$.

The constraint intersection is nonempty, closed and bounded, hence compact, so extrema exist. The two constraint gradients are $(x/2,2y,2z)$ and $(1,1,-1)$. If the first were $t$ times the second, the plane condition would give $2t+t/2+t/2=0$, so $t=0$ and all coordinates would vanish, contradicting the ellipsoid equation. Their rank is therefore two at every feasible point, justifying the Lagrange multiplier equations.

These equations areEliminating $\mu$ gives $1=\lambda (5x/2+2y)$ and $2=\lambda (2x+4y)$. The first equation makes $\lambda \ne 0$; subtracting twice the first from the second forces $x=0$. Hence $z=y$, and the ellipsoid gives $y=z=\pm 1/\sqrt{2}$. ThusAll candidates were found, and compactness ensures they include the extrema.