Finding extrema of a function of multiple variables I had to find extrema of the following function: \begin{align} z=x^3y^2(2-x-y)\end{align}
I identified first order partial derivatives, which are: \begin{align} z_x^{'}=y^2(3x^2(2-x-y)-x^3) 
\\\\
z_y^{'}=x^3(2y(2-x-y)-y^2)\end{align}
I then equated the two expressions to $0$ to identifiy critical points for further analysis (points which would be checked for extrema, in my case, by the means of cunstructing and evaluating a Hessian matrix).
The solutions for the system of equations I contrived were $y=0; x=0; x=1, y=2/3$.
However, as I was solving the system and going through all the possibilities, I got some roots that were special cases of the $y=0, x\in{\Bbb{R}}$ and $x=0, y\in{\Bbb{R}}$ solutions. Here is an example: \begin{cases}
&3x^2(2-x-y)-x^3=0 \\
&x^3(2y(2-x-y)-y^2)=0 \\ \end{cases} \begin{align}x_1=0,y_1=2\\\\x_2=3/2, y_2=0\\\\x_3=1, y_3=2/3 \end{align}
Are those points ($x_1=0,y_1=2; x_2=3/2, y_2=0$) critical? If not, then what's the gist here, in what way are those points special?
Thank you!
 A: Every point of the form $(x,0)$ is a critical point of $f$, as well as every point of the form $(0,y)$. And the Hessian of $f$ at a point $(x,y)$ is$$\begin{bmatrix}-6 x y^2 (2 x+y-2) & x^2 (-8 x-9 y+12) y \\ x^2 (-8 x-9 y+12) y & -2 x^3 (x+3 y-2)\end{bmatrix},$$and therefore the Hessian is the null matrix at those points. So, you deduce nothing from the Hessian.
A: $$z=2x^3y^2-x^4y^2-x^3y^3$$
as you said you want the partial derivates to be zero, so:
$$z_x=6x^2y^2-4x^3y^2-3x^2y^3=0$$
$$z_y=4x^3y-2x^4y-3x^3y^2=0$$
and for an extrema both must be true, and so there will be an extrema:
$$\forall(x,y)\in\mathbb{R}^2\text{s.t.}\begin{Bmatrix}z_x,z_y=0\end{Bmatrix}$$
However for equations like these it can be quite difficult to find the points, but we can check your points:
$$(x,y)=(0,2)\Rightarrow z_x=0,z_y=0$$
however you can see from our expressions that if $x=0$ then $z=0$ for any value of $y$ so along the line $x=0,z_y=0$ as well. Whilst these are maxima/minima they are not points but rather lines which can be the case.
