[Math] Find points at which coldest and hottest temperature on plate occur

Question

Circular hot plate given by the relationship $x^2+y^2\le4,$ heated according to the spatial temperature function $T(x,y) = 10 – x^2+2x-4y^2$.

Answer 1

Note that $$T(x,y)=11-(x-1)^2-4y^2\ .$$ This shows that $T(x,y)\leq T(1,0)=11$ for all $(x,y)\in D$, and the maximum value $11$ is taken only at the point $(1,0)$.

It is obvious that $(1,0)$ is the only stationary point of $T$ in the interior of $D$. Therefore the minimum of $T$ has to occur on the boundary $\partial D$. In order to find it we parametrize $\partial D$ by means of $$\partial D:\quad\phi\mapsto(2\cos\phi,2\sin\phi)\qquad(\phi\in{\mathbb R})$$ and consider the pullback $$f(\phi):=T(2\cos\phi,2\sin\phi)=\ldots=12\cos^2\phi+4\cos\phi-6\ .$$ The derivative $$f'(\phi)=-(24\cos\phi+4)\sin\phi$$ vanishes when $\cos\phi=-{1\over6}$ or $\sin\phi=0$. This leads to $4$ conditionally stationary points of $T$ on $\partial D$, namely $$\left(-{1\over3},{1\over3}\sqrt{35}\right),\quad \left(-{1\over3},-{1\over3}\sqrt{35}\right),\quad (2,0),\quad(-2,0)\ .\tag{1}$$ When $\cos\phi=-{1\over6}$ then $f(\phi)=-{19\over3}$, and when $\sin\phi=0$ then $f(\phi)=6\pm 4>0$. It follows that the minimum of $T$ on $D$ is $-{19\over3}$, and it is taken at the first two points listed in $(1)$.