 Solution

Use $g=x^2+y^2+z^2$. The Lagrange multiplier equations areTaking the dot product with $(x,y,z{)}^T$ gives $2T=2\lambda$, so $\lambda =T$.

If $\lambda \ne 0$, the first and third equations give $x=z$. The middle equation then gives $x=\lambda y$, so $2{\lambda }^2=1$. The norm constraint gives $x^2=z^2=1/4$ and $y^2=1/2$. Thus the hottest points and their temperature areThe coldest points areFor $\lambda =0$, the equations give $y=0$, $z=-x$, and hence the other two constrained stationary points areThey are saddle points of the restriction to the sphere: with $u=(x+z)/\sqrt{2}$ and $v=y$ as local tangent coordinates there, $T=\sqrt{2}uv$ changes sign in every neighbourhood. The surface is compact and smooth, so the listed extreme temperatures are global. Here “zero gradient on the surface” means zero surface gradient; the ambient gradient at the hottest and coldest points is normal to the sphere, rather than zero.