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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 $$f(x,y)=xye^{-x^2-y^2}.$$


\textbf{Answer}

\begin{eqnarray*}
f(x,y) & = & xye^{-x^2-y^2}\\
f_1(x,y) & = & y(1-2x^2)e^{-(x^2+y^4)}\\
f_2(x,y) & = & x(1-4y^4)e^{-(x^2+y^4)}
\end{eqnarray*}

For critical points $y(1-2x^2)=0$ and $x(1-4y^4)=0$. The critical
points are: $$(0,0), \ \ \ \left ( \pm
\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ), \ \ \ \left ( \pm
\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \right ).$$
We have
\begin{eqnarray*}
f(0,0) & = & 0\\
f\left ( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) & = & 
f\left ( -\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \right ) = \frac{1}{2}e^{-3/4}>0\\
f\left ( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) & = & 
f\left ( \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \right ) = -\frac{1}{2}e^{-3/4}<0
\end{eqnarray*}

Since $f(x,y)\to\infty$ as $x^2+y^2\to\infty$, the maximum and minimum
of $f$ are
$$\frac{1}{2}e^{-3/4} \ \ \rm{and} \ \ -\frac{1}{2}e^{-3/4}$$
respectively.


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