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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes}}
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\textbf{Question}

Find the minimum and maximum values of $$f(x,y)=
\frac{x}{(1+x^2+y^2}$$


\textbf{Answer}

\begin{eqnarray*}
f(x,y) & = & \frac{x}{(1+x^2+y^2)}\\
f_1(x,y) & = & \frac{1+y^2-x^2}{(1+x^2+y^2)^2}\\
f_2(x,y) & = & \frac{-2xy}{(1+x^2+y^2)^2}\\
 & & \rm{i.e.} \ y=0 \ \rm{or} \ z=0 \ \rm{or} \
x=\pm\frac{1}{\sqrt{2}}\\
\end{eqnarray*}

For critical points, $x^2-y^2=1$ and $xy=0$. The critical points are
$(\pm 1, 0)$. $f(\pm 1,0)=\pm \frac{1}{2}$.

Since $f(x,y)\to 0$ as $x^2+y^2 \to \infty$, the maximum and minimum
values of $f$ are $1/2$ and $-1/2$ respectively.


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