\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

$$ f(x,y)= \left ( 1 + \frac{1}{x} \right ) \left ( 1 +
\frac{1}{y} \right ) \left ( \frac{1}{x} + \frac{1}{y} \right ) $$


\textbf{Answer}

\begin{eqnarray*}
f(x,y) & = & \left ( 1 + \frac{1}{x} \right ) \left ( 1 +
\frac{1}{y} \right ) \left ( \frac{1}{x} + \frac{1}{y} \right )\\
& = & \frac{(x+1)(y+1)(x+y)}{x^2y^2}\\
f_1(x,y) & = & -\frac{(y+1)(xy+x+2y)}{x^3y^2}\\
f_2(x,y) & = & -\frac{(x+1)(xy+y+2x)}{x^2y^3}\\
A = f_{11}(x,y) & = & \frac{2(y+1)(xy+x+3y)}{x^4y^2}\\
B = f_{12}(x,y) & = & \frac{2(xy+x+y)}{x^3y^3}\\
C = f_{22}(x,y) & = & \frac{2(x+1)(xy+y+3x)}{x^2y^4}
\end{eqnarray*}
For critical points

$\begin{array}{lrlcrl}
& y = & -1 & \ \rm{or} \ & xy+x+2y & =0\\
\rm{and} \ \ & x = & -1 & \ \rm{or} \ & xy+y+2x & =0
\end{array}$

If $y=-1$, then $x=-1$ or $x-1=0$.

If $x=-1$, then $y=-1$ or $y-1=0$.

If $x \ne -1$ and $y \ne -1$, then $x-y=0$, so $x^2+3x=0$.

Thus $x=0$ or $x=-3$. However, the definition of $f$ excludes
$x=0$. Thus, the only critical points are: $$(1,-1), \ (-1,1), \
(-1,-1), \ \textrm{and} \ (-3,-3).$$

At $(1,-1), \ (-1,1)$ and $(-1,-1)$ we have $AC=0$ and $B \ne
0$. Therefore these three points are saddle points of $f$.

At $(-3,-3)$, $A=C=4/243$ and $B=2/243$, so $AC>B^2$. Therefore $f$
has a local minimum value at $(-3,-3)$.


\end{document}