= Solution
Solve the <Lagrange multiplier> equations by separating cases according to which coordinates vanish, so that division by a coordinate loses no <stationary points>.
If $x\ne0$ and $y=z=0$, the surface constraint gives
$$
\boxed{(x,y,z)=(\pm a,0,0),\qquad T=a^4.}
$$
If $x\ne0$ and exactly one of $y,z$ is nonzero, the equation for that nonzero coordinate forces $\lambda=0$, while the $x$ equation gives $\lambda=2a^2x^2>0$. Hence there are no <stationary points> in that case.
If $x=0$ and $y,z\ne0$, the two remaining equations give $\lambda=a^2z^2=b^2y^2$. The surface constraint therefore has equal contributions from $y$ and $z$, giving four <stationary points>:
$$
\boxed{(x,y,z)=\left(0,\varepsilon\frac a{\sqrt2},\eta\frac b{\sqrt2}\right),\qquad T=\frac{a^2b^2}{4},\quad\varepsilon,\eta\in\{-1,1\}.}
$$
If all three coordinates are nonzero, the equations imply
$$
\lambda=2a^2x^2=a^2z^2=b^2y^2,
$$
so $z^2=2x^2$ and $y^2=2a^2x^2/b^2$. Substitution in the surface constraint gives, with $D=4a^2+b^2$,
$$
x^2=\frac{a^2b^2}{D},\qquad y^2=\frac{2a^4}{D},\qquad z^2=\frac{2a^2b^2}{D}.
$$
Thus the remaining eight <stationary points> and their <temperatures> are
$$
\boxed{(x,y,z)=\left(\varepsilon_1\frac{ab}{\sqrt D},\varepsilon_2\frac{\sqrt2a^2}{\sqrt D},\varepsilon_3\frac{\sqrt2ab}{\sqrt D}\right),\qquad T=\frac{a^4b^2}{4a^2+b^2},\quad\varepsilon_1,\varepsilon_2,\varepsilon_3\in\{-1,1\}.}
$$
For example, their common <temperature> follows from $x^4+y^2z^2=a^4b^4/D^2+4a^6b^2/D^2=a^4b^2/D$. 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.
Back to article page
