<s>[INST] Find the extrema of $f(x,y) = \left( \frac{1}{2} - x^2 + y^2 \right) \cdot e^{1-x^2-y^2}$ I am asked to find the extrema of $$f(x,y) = \left( \frac{1}{2} - x^2 + y^2 \right) \cdot e^{1-x^2-y^2}$$
The partial derivatives could be easier, but that's not the issue:
\begin{align*}
f_x &= \left( e^{1-x^2-y^2} \right) \left( 2x^3-2xy^2 -3x \right) = 0\\
f_y &= \left( e^{1-x^2-y^2} \right) \left( -2yx^2+2y^3-y\right) = 0
\end{align*}
Exponentials are never null so
\begin{align*}
x \left( 2x^2-2y^2 -3 \right) &= 0\\
y \left( -2x^2+2y^2-1\right) &= 0
\end{align*}
The extrema $(0,0)$ (saddle point) is pretty obvious but what about the other extrema: how could I find those given this system of equations?
Any help is appreciated! Thank you.
Link for the complete answer (Wolfram)
 [/INST] For $x=0$ but $y\ne 0$, we have
$$2y^2-1=0$$
$$y=\pm\sqrt{\frac{1}{2}}$$
For $y=0$ but $x\ne 0$, we have
$$2x^2-3=0$$
$$x=\pm\sqrt{\frac{3}{2}}$$
For $x,y\ne 0$, we have
$$2x^2=2y^2+3$$
$$-2y^2 -3+2y^2-1=0$$
$$-4=0$$
So no solution
</s>