Solving critical points and classifying the maxima, minima and saddle points

Question

I need help to find critical points of the function: $$f(x,y,z)={(x^2+2y^2+3z^2) e ^{-(x^2+y^2+z^2)}}$$ Then I have to classify these critical points as local maxima/minima or saddle points.

What I would do is to take the partial derivatives $\frac{\partial f}{\partial x}(x, y, z)$, $\frac{\partial f}{\partial y}(x, y, z)$ and $\frac{\partial f}{\partial z}(x, y, z)$.

But now I will get

$\frac{\partial f}{\partial x}(x, y, z)$ = $ {2x-6e^{-x^2-y^2-z^2}xz^2} $

$\frac{\partial f}{\partial y}(x, y, z)$ = $ {4y-6e^{-x^2-y^2-z^2}yz^2} $

$\frac{\partial f}{\partial z}(x, y, z)$ = $ 3(-2e^ {-x^2-y^2-z^2}z^3+2e^{-x^2-y^2-z^2}z) $

Normally I would put them in the matrix form and solve the system of equation for x, y and z.

But I don't know how to solve the system of equation with these partial derivatives. What I am doing wrong or what should be done in order to solve this problem?

Answer 1

Computing the partial derivatives, we then have $$\frac{\partial f(x, y, z)}{\partial x}=-2 x \left(x^2+2 y^2+3 z^2-1\right)e^{-(x^2+y^2+z^2)}$$ $$\frac{\partial f(x, y, z)}{\partial y}=-2 y \left(x^2+2 y^2+3 z^2-2\right)e^{-(x^2+y^2+z^2)}$$ $$\frac{\partial f(x, y, z)}{\partial z}=-2 z \left(x^2+2 y^2+3 z^2-3\right)e^{-(x^2+y^2+z^2)}$$ which means that you need to solve $$x \left(x^2+2 y^2+3 z^2-1\right)=0$$ $$y \left(x^2+2 y^2+3 z^2-2\right)=0$$ $$z \left(x^2+2 y^2+3 z^2-3\right)=0$$ for which the solutions are quite simple $$\{\{ 0, 0, -1\}, \{0, 0, 1\}, \{0, -1, 0\}, \{0, 1, 0\}, \{-1, 0, 0\}, \{1, 0, 0\}, \{0, 0, 0\}\}$$

Now, your turn.

Answer 2

The derivative with respect to $x$ should be $$\frac{\partial f(x,y,z)}{\partial x}=2xe^{-(x^2+y^2+z^2)}+(x^2+2y^2+3z^2)e^{-(x^2+y^2+z^2)}\cdot 2x$$ by the product and chain rule. And the other derivatives analogously. This is $$\frac{\partial f(x,y,z)}{\partial x}=e^{-(x^2+y^2+z^2)}(2x-(2x(x^2+2y^2+3z^2))$$ You have to solve the system $$2x-2x(x^2+2y^2+3z^2)=0$$ $$4y-2y(x^2+2y^2+3z^2)=0$$ $$6z-2z(x^2+2y^2+3z^2)=0$$ for $x,y,z$