Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 6E (Methods)

For a differentiable equality constraint, the Lagrange multiplier method finds candidate extrema by solvingThe condition requires $\nabla g\ne 0$ at the point: every tangent direction to the constraint is perpendicular to $\nabla g$, and at a extremum the directional derivative of $f$ vanishes in every such direction. Thus $\nabla f$ is parallel to $\nabla g$. One must then classify candidates and check any singular or boundary cases.

For the ellipsoid, its constraint gradient never vanishes on the surface, and its compactness guarantees a maximum and a minimum. The Lagrange multiplier equations becomeIf $\lambda =0$, then $x=y=0$ and $z=\pm c$, giving value $0$. If $\lambda \ne 0$, then $z=0$; neither $x$ nor $y$ can vanish on this part of the surface. Combining the first two equations gives $4{\lambda }^2=a^2{b}^2$, and then $y=\pm (b/a)x$. The constraint yields $x^2=a^2/2$ and $y^2=b^2/2$. ThereforeThe maximum occurs at $(a/\sqrt{2},b/\sqrt{2},0)$ and its negative; the minimum at $(a/\sqrt{2},-b/\sqrt{2},0)$ and its negative. To verify these are the global extrema, useEquality is attained at exactly the points just found. The remaining two candidates have value $0$ between the extremal values.