Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 3D / Solution

Because the constraint determinesthe constrained function isIts stationary points satisfyWhen $yz=0$, these equations giveWhen $yz\ne 0$, subtracting the two bracketed equations gives $z^2=a^2{y}^2$, and substitution then giveswith the two signs chosen independently. These are exactly the constrained stationary points; equivalently they follow from the Lagrange multiplier equations.

Under the additional restriction $x\ge 0$, the constraint gives the compact elliptical disc $a^2{y}^2+z^2\le b^2$. On its boundary $x=0$, the objective is zero. At the four nonzero interior stationary points,with the sign determined by $yz$. Therefore