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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes within restricted domains}}
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\textbf{Question}

Find the maximum and minimum values of

$$f(x,y)=x+2y$$

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


\textbf{Answer}

As $f_1=1$ and $f_2=2$, $f$ will have no critical points. So the
minimum and maximum points must occur at the boundary. $f$ must have
minimum and maximum points as $f$ is a continuous on a closed, bounded
set in the plane.

The boundary is the circle $x^2+y^2=1$ and can be parameterised as
\begin{eqnarray*}
x &amp; = &amp; \cos t\\
y &amp; = &amp; \sin t\\
&amp; &amp; \\
\Rightarrow f(x,y)=f (\cos t, \sin t) &amp; = &amp; \cos t + 2 \sin t =g(t)
\end{eqnarray*}

For critical points of $g$
\begin{eqnarray*}
0 &amp; = &amp; g'(t) = -\sin t + 2 \cos t\\
\Rightarrow \tan t &amp; = &amp; 2\\
x &amp; = &amp; \pm 1/\sqrt{5}\\
y &amp; = &amp; \pm 2/\sqrt{5}
\end{eqnarray*}
So the critical points of $g$ are
\begin{eqnarray*}
(-1/\sqrt{5},-2/\sqrt{5}) &amp; &amp; f=-\sqrt{5}=\textrm{min}(f)\\\
(1/\sqrt{5},2/sqrt{5}) &amp; &amp; f=\sqrt{5}\textrm{max}(f)
\end{eqnarray*}

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