Can anyone show me whether my answer below is correct and complete? Specifically, I am not sure whether or not I defined the extrema in explicit-enough terms. Also, the graph of the function using http://web.monroecc.edu/manila/webfiles/calcNSF/JavaCode/CalcPlot3D.htm seems to present a somewhat different answer than I am getting using equations below. Perhaps someone might be able to point out other things I could improve to make this answer more correct. This is question 14.7.12 in the seventh edition of Stewart Calculus.
"Find the local maximum and minimum values, and saddle points, of $f(x,y)=xy + \frac{1}{x} +\frac{1}{y}$."
Here is my work:
$f_x(x,y) = y+\frac{-1}{x^2}=0$, so $yx^2 =1$
$f_y(x,y)=x+\frac{-1}{y^2}=0$, so $xy^2=1$
Thus, $yx^2=xy^2=1$ and $\frac{xy^2}{yx^2}=1=\frac{y}{x}=\frac{x}{y}$, so the critical points occur when $y=x$
$f_{xx}=\frac{2}{x^3}$
$f_{yy}=\frac{2}{y^3}$
$f_{xy}=1$
Use Second Derivatives Test: $D=D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2$
For $x=y$, we have $D=\frac{4}{x^6}-1>0$, thus, there are no saddle points.
$f_{xx} = \frac{2}{x^3}=\begin{cases}<0 ,& x < 0 \\udf ,& x=0\\>0,& x>0\end{cases}$
Thus, $x=y$ defines local maxima when $x<0$ and defines local minima when $x>0$
Highest maxima occur as $x\to\infty$ and as $x\to 0^+$
Lowest minima occur as $x\to -\infty$ and as $x\to 0^-$