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

There are real Lagrange multipliers ${\lambda }_1,{\lambda }_2$ such thatEquivalently $Dg(x_0)={\lambda }^{T}Df(x_0)$. The rank assumption says the two constraint gradients are linearly independent; it rules out an abnormal constraint point.

Use constraints $h_1=x^2/4+y^2+z^2-1$ and $h_2=x+y-z$. Both constraints and the objective are continuously differentiable. The feasible set is a nonempty closed subset of a bounded ellipsoid, hence compact, so the extreme value theorem guarantees both extrema.

The gradients $(x/2,2y,2z)$ and $(1,1,-1)$ are independent everywhere feasible. If the first were $t$ times the second, then $x=2t$, $y=t/2$, $z=-t/2$. The plane constraint would give $3t=0$, forcing the origin, which is not on the ellipsoid. Thus the Lagrange multiplier theorem applies at every feasible point.

Its equations areHere $\lambda \ne 0$, since otherwise the first two equations require $\mu =1=2$. Subtract twice the first equation from the second, and use $\mu =2\lambda z$ and $z=x+y$; the result is $0=-3\lambda x$, so $x=0$. Then $z=y$ and the ellipsoid gives $y=\pm 1/\sqrt{2}$. The only candidates are therefore $(0,\pm 1/\sqrt{2},\pm 1/\sqrt{2})$, with matching signs. Consequently