Find stationary points of $f(x,y) = (6 - x - y)x^2y^3$

Question

I am stuck in the middle. I need to find stationary points of function from title. I managed to differentiate and create system of equations but I have no idea how to proceed with this.

$f_x(x,y) = xy^3(12 - 3x - 2y)$

$f_y(x,y) = x^2y^2(18 -3x - 4y)$

Part below made me stuck:

$\Bigg\{\begin{array}{c}xy^3(12 - 3x - 2y) = 0\\x^2y^2(18 -3x - 4y) = 0\end{array}$

I am looking for your help.

Answer 1

Well, If $x=0$ or $y=0$, you have a solution, so every point on the coordinate axes is a stationary point of $f$. If $x\ne 0$ and $y \ne 0$ the only solution is found solving the linear system $12-3x-2y=0, \quad 18-3x-4y=0$. so, the solutions are

1. $(0,t), \quad t \in \mathbb{R}$
2. $(t,0), \quad t \in \mathbb{R}$
3. $(2,3)$