Local minimums and maximums of function of three variables

Question

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?

Answer 1

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.