I am trying to find points on $z= (2x^2+y^2)e^(-x^2-y^2)$ where the tangent planes are horizontal. Taking the partial derivatives with respect to x and y and setting both to zero. After simplifying the two equations I get:
$$2=2x^2+3y^2$$ $$3=2x^2+3y^2$$
I know one solution is $(0,0)$, but can anyone help me find the others just by using algebra.
The partial derivatives are zero when
$$2x(2x^2+y^2-2)=0$$ and
$$2y(2x^2+y^2-1)=0.$$
Go through each case:
You already found $(0,0)$.
Now suppose $x=0$ and $y\neq 0$. Then from the second equation $y^2=1$. This gives two solutions $(0,-1)$ and $(0,1)$.
Next, suppose $x\neq 0$ and $y=0$. Then from the first equation $x^2=1$. This gives two more solutions $(-1,0)$ and $(1,0)$.
Finally, suppose $x\neq 0$ and $y\neq 0$. Then from the two equations:
$$2x^2+y^2=2$$
and
$$2x^2+y^2=1$$
which have no solution (they are contradictory).