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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes within restricted domains}}
\end{center}

\textbf{Question}

$f(x,y)=xye^{-xy}$ has domain $x&gt;0$, $y&gt;0$.

Show that $\ds \lim_{x\to\infty} f(x,kx)=0$.

Is there a limit for $f$ as $(x,y)$ moves arbitrarily far from the
origin yet still remaining in the first quadrant?

Is there a maximum value for $f$ in the given domain?


\textbf{Answer}

\begin{eqnarray*}
f(x,kx) &amp; = &amp; kx^2e^{-kx^2} \to 0\\
\textrm{as }x\to\infty &amp; &amp; \textrm{if }k&gt;0\\
\textrm{and }f(x,0) &amp; = &amp; f(0,y) = 0\\
\end{eqnarray*}
$\Rightarrow f(x,y) \to  0$ as $(x,y) \to \infty$ along any straight
line from the origin in $Q$.

But $\ds f \left ( x, \frac{1}{x} \right )$ and $f(x,0)=0$, $\forall
x&gt;0$, even though $\left ( x, \frac{1}{x} \right ) $ and $(x,0)$ move
arbitrarily close as $x$ increases. And so $f$ has no limit as
$x^2+y^2\to \infty$.

See that $f(x,y)=re^{-r}=g(r)$ on the hyperbola $xy=r&gt;0$.

\begin{eqnarray*}
g(r) &amp; \to &amp; 0\\
\textrm{as } r &amp; \to &amp; 0, \ \ \textrm{or }r \to \infty\\
\textrm{and }g'(r) &amp; = &amp; (1-r)e^{-r} = 0\\
\Rightarrow r &amp; = &amp; 1
\end{eqnarray*}
And so $f(x,y)$ is less than $g(1)=1/e$ everywhere on $Q$.

$\Rightarrow$ $f$ has a maximum value on $Q$. 


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