Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 20

Let $G=x^2/a^2+y^2/a^2+z^2/b^2-1$. Its gradient cannot vanish on the ellipsoid, since the origin is not on the surface. Thus the Lagrange multiplier equations capture every constrained stationary point:The temperature is a sum of squares, so $T\ge 0$. Equality requires $x=0$ and $yz=0$. The surface constraint then gives the four coldest points:These are all such points, because a sum of nonnegative terms can vanish only when each term vanishes.

Solve the Lagrange multiplier equations by separating cases according to which coordinates vanish, so that division by a coordinate loses no stationary points.

If $x\ne 0$ and $y=z=0$, the surface constraint givesIf $x\ne 0$ and exactly one of $y,z$ is nonzero, the equation for that nonzero coordinate forces $\lambda =0$, while the $x$ equation gives $\lambda =2a^2{x}^2>0$. Hence there are no stationary points in that case.

If $x=0$ and $y,z\ne 0$, the two remaining equations give $\lambda =a^2{z}^2=b^2{y}^2$. The surface constraint therefore has equal contributions from $y$ and $z$, giving four stationary points:If all three coordinates are nonzero, the equations implyso $z^2=2x^2$ and $y^2=2a^2{x}^2/b^2$. Substitution in the surface constraint gives, with $D=4a^2+b^2$,Thus the remaining eight stationary points and their temperatures areFor example, their common temperature follows from $x^4+y^2{z}^2=a^4{b}^4/D^2+4a^6{b}^2/D^2=a^4{b}^2/D$. The cases with $x=0$ and $yz=0$ are exactly the coldest points already listed. This exhausts all possibilities, giving fourteen other stationary points in addition to those four.

For $a=b=1$, the stationary points found in parts (a) and (b) have temperatures $0$, $1$, $1/4$ and $1/5$. The unit sphere is compact and $T$ is a continuous function, so the extreme value theorem ensures that a hottest point exists; it must be one of those stationary points. ThusFor a direct bound, put $s=x^2\in [0,1]$. The identity $(y^2-z^2{)}^2\ge 0$ gives $y^2{z}^2\le (y^2+z^2{)}^2/4=(1-s{)}^2/4$, andsince $1-s^2-(1-s{)}^2/4=(1-s)(5s+3)/4\ge 0$. Equality forces $s=1$, giving exactly the two points above.