Find the coordinates and nature of the stationary point of $z=x^3+y^3-6xy$

Question

Find the coordinates and nature of the stationary point of $$z=x^3+y^3-6xy$$

So I have found all the partial derivatives but I'm not sure how to then find the stationary point. All I know is $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=0$ at the stationary point. $$\frac{\partial z}{\partial x}=3x^2-6y$$ $$\frac{\partial z}{\partial x}=3y^2-6x$$ $$\frac{\partial^2 z}{\partial x^2}=6x$$ $$\frac{\partial^2 z}{\partial y^2}=6y$$ $$\frac{\partial^2 z}{\partial y \partial x}=\frac{\partial^2 z}{\partial x \partial y}=0$$

Answer 1

The derivatives are:

${\partial z \over \partial x} = 3 x^2 - 6 y$

${\partial z \over \partial y} = 3 y^2 - 6 x$

Set the pair to $0$ to find the two (real) solutions: $(2,2)$ and $(0,0)$.

Take the second derivatives evaluated at those points to find a minimum at $(2,2)$ and saddle at $(0,0)$, as seen in the figure.

${\partial^2 z \over \partial x^2} = 6 x$

${\partial^2 z \over \partial y^2} = 6 y$

Answer 2

FOC: solve $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=0$ to find the critical points $(0,0)$ and $(2,2)$.

SOC: $z_{xx}=6x, z_{yy}=6y,\Delta=z_{xx}z_{yy}-z_{xy}^2=36xy-36$: $$\text{at} \ (0,0): \ \ z_{xx}=0, z_{yy}=0,\Delta=-36<0 \Rightarrow \text{saddle};\\ \text{at} \ (2,2): \ \ z_{xx}=12>0, z_{yy}=12>0,\Delta=108>0 \Rightarrow \text{min}.$$