<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\usepackage{epsfig}
\begin{document}
\parindent=0pt

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

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

\textbf{Question}

Find the maximum and minimum values of

$$f(x,y)=xy-y^2$$

On the disk $x^2+y^2\le 1$.


\textbf{Answer}

For critical points
\begin{eqnarray*}
0 &amp; = &amp; f_1(x,y) = y\\
0 &amp; = &amp; f_2(x,y) = x-2y
\end{eqnarray*}
So the only critical point is $(0,0)$. This lies inside the disk, with
$f(0,0)=0$.

The boundary of the disk is given by the equations for a circle
\begin{eqnarray*}
x &amp; = &amp; \cos t\\
y &amp; = &amp; \sin t\\
\textrm{with }-\pi \le t \le \pi &amp; &amp; 
\end{eqnarray*}
On this circle
\begin{eqnarray*}
g(t) &amp; = &amp; f(\cos t, \sin t) = \cos t \sin t -\sin^2t\\
&amp; = &amp; \frac{1}{2} \left [ \sin 2t + \cos 2t -1 \right ]\\
g(0) &amp; = &amp; g(2\pi) =0\\
g'(t) &amp; = &amp; \cos 2t - \sin 2t
\end{eqnarray*}

The critical points of $g$ must satisfy
\begin{eqnarray*}
\cos 2t &amp; = &amp; \sin 2t\\
\textrm{i.e. }\tan 2t &amp; = &amp; 1\\
&amp; &amp; \\
\textrm{So }2t &amp; = &amp; \pm\frac{\pi}{4}, \ \ \textrm{or }
\pm\frac{5\pi}{4}\\
\Rightarrow t &amp; = &amp; \pm \frac{\pi}{8}, \ \ \textrm{or }
\pm\frac{5\pi}{8}
\end{eqnarray*}

This gives
\begin{eqnarray*}
g \left ( \frac{\pi}{8} \right ) &amp; = &amp; \frac{1}{2\sqrt{2}} - \frac{1}{2}
+ \frac{1}{2\sqrt{2}}\\
&amp; = &amp; \frac{1}{\sqrt{2}}-\frac{1}{2} &gt; 0\\
g \left ( -\frac{\pi}{8} \right ) &amp; = &amp; -\frac{1}{2\sqrt{2}}
-\frac{1}{2} + \frac{1}{2\sqrt{2}}\\
&amp; = &amp; -\frac{1}{2}\\
g \left ( \frac{5\pi}{8} \right ) &amp; = &amp; -\frac{1}{2\sqrt{2}} -
\frac{1}{2} - \frac{1}{\sqrt{2}}\\
&amp; = &amp; -\frac{1}{2\sqrt{2}} - \frac{1}{2}\\
g \left ( -\frac{5\pi}{8} \right ) &amp; = &amp; \frac{1}{2\sqrt{2}} -
\frac{1}{2} - \frac{1}{2\sqrt{2}}\\
&amp; = &amp; - \frac{1}{2}
\end{eqnarray*}

So, on the disk
\begin{eqnarray*}
\textrm{min}(f) &amp; = &amp; -\frac{1}{\sqrt{2}}-\frac{1}{2}\\
\textrm{max}(f) &amp; = &amp; \frac{1}{\sqrt{2}}-\frac{1}{2}
\end{eqnarray*}

\end{document}

</pre></body></html>