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\begin{center}
\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes within restricted domains}}
\end{center}

\textbf{Question}

The temperature of a point of a disk $D$ is given by

$T=(x+y)e^{-x^2-y^2}$

Where $D$ is defined as all points such that $x^2+y^2 \le 1$.

Find the maximum on minimum values of temperature on $D$.


\textbf{Answer}

For critical points
\begin{eqnarray*}
0 = \frac{\partial T}{\partial x} &amp; = &amp; \left ( 1 - 2x(x+y) \right )
e^{-x^2-y^2}\\
0 = \frac{\partial T}{\partial y} &amp; = &amp; \left ( 1 -2y(x+y) \right )
e^{-x^2-y^2}\\
&amp; &amp; \\
\Rightarrow 2x(x+y) &amp; = &amp; 1 = 2y(x+y)\\
\Rightarrow x &amp; = &amp; y, \ \ \ \textrm{and }4x^2=1\\
\Rightarrow x &amp; = &amp; y = \pm \frac{1}{2}
\end{eqnarray*}
The two critical points are then $(\frac{1}{2}, \frac{1}{2})$ and
$(\frac{1}{2}, \frac{1}{2})$. Both of these lie inside the disk, with
$T=\pm e^{-1/2}$.

Parameterising on the boundary of the disk gives
\begin{eqnarray*}
x &amp; = &amp; \cos t\\
y &amp; = &amp; \sin t
\end{eqnarray*}
with $0\le t \le 2\pi$.
$$\Rightarrow T = (\cos t +\sin t)e^{-1} = g(t), \ \ \ (0\le t\le
2\pi)$$
Now $g(0)=g(2\pi)=e^{-1}$, so for critical points of $g$
\begin{eqnarray*}
0 = g'(t) &amp; = &amp; (\cos t - \sin t)e^{-1}\\
\Rightarrow \tan t &amp; = &amp; 1\\
\Rightarrow t &amp; = &amp; \pi/t, \ \ \ \textrm{or }t=5\pi/4
\end{eqnarray*}
This gives
\begin{eqnarray*}
g(\pi/t) &amp; = &amp; \sqrt{2}e^{-1}\\
\textrm{and } g(5\pi/4) &amp; = &amp; -\sqrt{2}e^{-1}
\end{eqnarray*}

As $e^{-1/2}&gt;\sqrt{2}e^{-1}$, min$f=-e^{-1/2}$ and max$f=e^{-1/2}$.


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