Local minimums and maximums of function of three variables I got such function:
$f(x_1, x_2, x_3) = x_1 x_2 x_3(4-x_1-x_2-x_3)$
I need to find all local minimums and maximums of this function.
I calculated partial derivatives and I got that the only points where extremums of this function can be located are:
$(1,1,1)$, $(0,0,a)$, $(0, a, 0)$, $(a, 0, 0)$, where $a$ is real.
I don't know how to deal with this $a$, when calculating this matrix for determining if an extremum is a minimum/maximum.
How to deal with it? Can someone show please? 
 A: So the function is
$$
f(x,y,z) = 4xyz - x^2yz - xy^2z - xyz^2
$$
so
$$
\nabla f
 = \begin{pmatrix}
4yz - 2xyz - y^2z - yz^2\\
4xz - 2xyz - x^2z - xz^2\\
4xy - 2xyz - xy^2 - x^2y
\end{pmatrix}
 = \begin{pmatrix}
yz(4 - 2x - y - z)\\
xz(4 - 2y - x - z)\\
xy(4 - 2z - y - x)
\end{pmatrix}
$$
and we must solve $\nabla f = \vec{0}$. The first equation yields 3 cases:
Case I $x \ne 0, y \ne 0, z \ne 0$
Then,
$$
\begin{pmatrix}
2x + y + z = 4\\
2y + x + z = 4\\
2z + y + x = 4
\end{pmatrix}
$$
so $x=y=z=1$.
Case II $y=0$
The only meaningful equation is then the second one, which reduces to $xz(4-x-z)=0$. So we get $(0,0,a), (a,0,0), (4-a,0,a)$ for any real $a$.
All other symmetric cases to the ones we dealt with (which have zeros for one or more of the variables), so you get (1,1,1) and 6 families of solutions: $(a,0,0), (0,a,0), (0,0,a), (4-a,0,a), (0,4-a,a), (a,4-a,0)$.
Plug these back into $f$ to get 0 for all of them except $f(1,1,1)=1$. To classify these, as real minima or saddle points, you can use the 2nd derivative tests. So you possibly have an uncountable infinity of minima which all result in 0 value, and one maximum with value of 1.
Note that the absolute minimum of this surface occurs as $-\infty$, so the minima candidates are only local, not absolute.
