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\textbf{Applications of Partial Differentiation}

\textit{\textbf{Extremes}}
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\textbf{Question}

Find and classify the critical points of the function

$$f(x,y)=x^2 y e^{-(x^2+y^2)}$$


\textbf{Answer}

\begin{eqnarray*}
f_1(x,y) & = & 2xy(1-x^2)e^{-(x^2+y^2)}\\
f_2(x,y) & = & x^2(1-2y^2)e^{-(x^2+y^2)}\\
A = f_{11}(x,y) & = & 2y(1-5x^2+2x^4)e^{-(x^2+y^2)}\\
B = f_{12}(x,y) & = & 2x(1-x^2)(1-2y^2)e^{-(x^2+y^2)}\\
C = f_{22}(x,y) & = & 2x^2y(2y^2-3)e^{-(x^2+y^2)}
\end{eqnarray*}
For critical points
\begin{eqnarray*}
xy(1-x^2) & = & 0\\
x^2(1-2y^2) & = & 0
\end{eqnarray*}
The critical points are $(0,y)$ $\forall y$, $(\pm 1, 1/\sqrt{2})$,
and $(\pm 1,-1/\sqrt{2})$.

Obviously $f(0,y)=0$.

Also $(x,y)>0$ if $y>0$ and $x \ne 0$, and $f(x,y)<0$ if $y<0$ and $x
\ne 0$.

Thus $f$ has a local minimum at $(0,y)$ if $y>0$, and a local maximum
if $y<0$. the origin is a saddle point.

At $(\pm 1, 1/\sqrt{2})$: $A=C=-2\sqrt{2}e^{-3/2}$, $B=0$, and so
$AC>B^2$. Thus $f$ has local maximum values at these two points. 

At $(\pm 1, -1/\sqrt{2})$: $A=C=2\sqrt{2}e^{-3/2}$, $B=0$, and so
$AC>B^2$. Thus $f$ has local minimum values at these two points. 

Since $f(x,y)\to 0$ as $x^2+y^2 \to \infty$, the value $$f(\pm 1,
1/\sqrt{2}) = e^{-3/2}/\sqrt{2}$$ is the absolute maximum value for
$f$, and the value $$f(\pm 1,
-1/\sqrt{2}) = -e^{-3/2}/\sqrt{2}$$ is the absolute minimum value.


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