\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}
\parindent=0pt

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

\textit{\textbf{Extremes}}
\end{center}

\textbf{Question}

Find and classify the critical points of the function

$$ f(x,y)=\frac{xy}{2+x^4+y^4}$$


\textbf{Answer}

\begin{eqnarray*}
f_1 & = & \frac{(2+x^4+y^4)y-xy4x^3}{(2+x^4+y^4)^2}\\
& = & \frac{y(2+y^4-3x^4)}{(2+x^4+y^4)^2}\\
f_2 & = & \frac{x(2+x^4-3y^4)}{(2+x^4+y^4)^2}.
\end{eqnarray*}
For critical points, $y(2+y^4-3x^4) =0$ and $x(2+x^4-3y^4)=0$.

One critical point is $(0,0)$. Since $f(0,0)=0$ but $f(x,y)>0$ in the
first quadrant and $f(x,y)<0$ in the second quadrant, $(0,0)$ must be
a saddle point of $f$.

Any other critical points must satisfy $2+y^4-3x^4=0$ and
$2+x^4-3y^4=0$, that is $y^4=x^4$, or $y=\pm x$. Thus $2-2x^4=0$ and
$x=\pm 1$.

Therefore there are four other critical points: 
   $(1,1)$, $(-1,-1)$, $(1,-1)$ and $(-1,1)$.

$f$ is positive at the first two of these, and negative at the other
two.

Since $f(x,y)\to 0$ as $x^2 +y^2\to\infty$, $f$ must have maximum
values at $(1,1)$ and $(-1,-1)$, and minimum values at $(1,-1)$ and
$(-1, 1)$>

\end{document}