Find the stationary points of $$f(x) = 2x_1^3 -3x_1^2
-6x_1x_2(x_1-x_2-1)$$ Which os these points are minimizers, maximizers, local or global?
$$f(x) = 2x_1^3-3x_1^2-6x_1x_2(x_1-x_2-1)$$
$$\frac{\partial f}{\partial x_1} = 6x_1^2 -6x_1-12x_1x_2+6x_2^2+6x_2$$
$$\frac{\partial^2 f}{\partial x_1^2} = 12x_1-12x_2-6$$
$$\frac{\partial^2 f}{\partial x_2\partial x_1} = -12x_1+12x_2+6$$
$$\frac{\partial f}{\partial x_2} = -6x_1^2+12x_1x_2+6x_1$$
$$\frac{\partial^2 f}{\partial x_2^2} = 12x_1$$
You can verify the hessian matrix here
I must first see which points make the first derivatives $0$. I know that $(0,0)$ and $(-1,-1)$ do it, so let's find the others:
$$\frac{\partial f}{\partial x_2} =0 \implies -6x_1^2+12x_1x_2+6x_1 = 0\implies x_2 = \frac{x_1}{2}-\frac{1}{2}, x_1\neq 0$$
I don't even need to do $\frac{\partial f}{\partial x_1}$ because if $x_2 = \frac{x_1}{2}-\frac{1}{2}$ is not satisfied then I don't need to find the points that give $0$ in the first coordinate of $\nabla f$, because the second already doesn't give $0$. Ok, so for $x_1\neq 0, x_2 = \frac{x_1}{2}-\frac{1}{2}$.
If $x_1=0$ we have $\frac{\partial f}{\partial x_2} = 0$ and $\frac{\partial f}{\partial x_1} = 6x_2^2 +6x_2$ whch is $0$ at $x_2=0$ only, so the points where $x_1\neq 0$ are not our concern.
I think now we must solve find which poins of the form $(x_1,\frac{x_1}{2}-\frac{1}{2})$ satisfy $\frac{\partial f}{\partial x_1} = 0$:
$$\frac{\partial f}{\partial x_1}((x_1,\frac{x_1}{2}-\frac{1}{2})) = 6x_1^2 -6x_1-12x_1(\frac{x_1}{2}-\frac{1}{2})+6(\frac{x_1}{2}-\frac{1}{2})^2+6(\frac{x_1}{2}-\frac{1}{2})\implies x_1 = \pm 1$$
When $x=-1, x_2 = x_2=-1$. When $x=1, x_2=0$
So all the points are:
$$(0,0), (-1,-1),(1,0)$$
$$H(0,0) = 0\\ [x \ y]H(-1,-1)[x \ y]^T = -6x^2+12xy-12y^2\mbox{ not always positive or 0}\\ [x \ y]H(1,0)[x \ y]^T = 12y^2 -12xy+6x^2 \mbox{ not always positive or 0}$$
The $3$ critical points I found doesn't satisfy the necessary condition to be minima or maxima. So I think I should analyze the neighborhood of these points analytically. However the function is too ugly for me to do that. I think I should have found conclusive answers on my hessian. Can you see anything wrong?
Best Answer
First of all, you haven't quite well looked at the case $x_1=0$. In that case, $\frac{\partial f}{\partial x_2}=0$ for every $x_2$, but $\frac{\partial f}{\partial x_1}$ becomes $$\frac{\partial f}{\partial x_1}(0,x_2)=6x_2^2+6x_2,$$ which is zero both for $x_2=0$ and $x_2=-1$. So you missed the critical point $(0,-1)$.
As for the Hessian matrix I got $$H(0,0)=\left(\begin{matrix}-6&6\\6&0\\\end{matrix}\right),$$ $$H(-1,-1)=\left(\begin{matrix}-6&6\\6&-12\\\end{matrix}\right),$$ $$H(1,0)=\left(\begin{matrix}6&-6\\-6&12\\\end{matrix}\right),$$ $$H(0,-1)=\left(\begin{matrix}6&-6\\-6&0\\\end{matrix}\right).$$
The first one and the last one have negative determinant, and that means they represent undefined quadratic forms. This assures that the corresponding critical points are not extrema, but saddle points indeed.
The second and the third one have positive determinant, which implies they are extrema; the second one represents a negative definite quadratic form and the third one a positive definite quadratic form. This implies that at $(-1,-1)$ and $(1,0)$ there are a relative maximum and a relative minimum, respectively.