Stationary points of a multivariable function

Question

This question might just be a quick one but I'm slightly confused by the answer I've been provided for this question.

I have the function: $f(x,y) = (x^2+2y^2)e^{-y^2 - x^2}$

I found the partial derivatives:

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

$f_y = (4y-2y(x^2+2y^2))e^{-y^2-x^2}$

I know that the stationary points can be found where $f_x=0$ and/or where $f_y=0$, so that's what I did, I set both $f_x$ and $f_y$ to equal $0$:

\begin{align} \tag{1} (2x-2x(x^2+2y^2))e^{-y^2-x^2}&=0\\ \tag{2} (4y-2y(x^2+2y^2))e^{-y^2-x^2}&=0 \end{align}

I then divided both $(1)$ and $(2)$ by $e^{-y^2-x^2}$ and simplified both equations, this is what I was left with: \begin{align} \tag{1'} x(1-x^2-2y^2)=0\\ \tag{2'} y(2-x^2-2y^2)=0 \end{align} This is where I was slightly confused. On the answer sheet I'm provided, the only stationary points they find are $(0,0)$, $(1,0)$ and $(-1,0)$

I understand how $(0,0)$ was found but I'm not sure how $(\pm1,0)$ was found. Did they make the two equations equal to each other then solve?

Answer 1

Let us solve your last system systematically. In $(1')$ there are two possibilities, $x = 0$ or $x^2 = 1-2y^2$.

In the former case, substituting $x = 0$ in $(2')$ yields $y(2-2y^2) = 0$, so you get $y = 0$, $y = 1$ and $y = -1$.

In the latter case, substituting $x^2 = 1-2y^2$ in $(2')$ yields $y = 0$. Now substituting $y = 0$ in $(1')$ yields $x(1-x^2) = 0$, so you get $x = 0$, $x = 1$ and $x = -1$.

All in all, you have five critical points: $(0,0)$, $(0,\pm 1$), $(\pm 1,0)$.

Wolfram Alpha gives you a nice diagram of the critical points:

Answer 2

If you have $$AB=0\\CD=0$$ then you have the following solution $A=C=0$, $A=D=0$, $B=C=0$, $B=D=0$. So in your case, $A=x$, $B=1-x^2-2y^2$, $C=y$, $D=2-x^2-2y^2$. Note that you can't have $B=D=0$ since they differ by $1$. So you need to check $$x=y=0\\x=2-x^2-2y^2=0\\y=1-x^2-2y^2=0$$ The solutions are $(0,0)$, $(0,\pm1)$, $(\pm1,0)$