Given:
$$\tag 1 f(x,y) = x^3 - 12xy + 8y^3$$
Find and classify all critical points.
The critical points are: $(0,0)$ and $(2,1)$.
The partial derivatives are:
- $f_x = 3x^2 -12 y$
- $f_y = -12x + 24y^2$
- $f_{xx} = 6x$
- $f_{yy} = 48y$
- $f_{xy} = f_{yx} = -12$
The Hessian determinant is given by:
$$\det(H) = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{vmatrix} \ $$
If you are using the Hessian, there are four conditions you need to test:
- $(01)$ $f_{xx} \gt 0$ and $\det H \gt 0 \rightarrow$ local minimum
- $(02)$ $f_{xx} \lt 0$ and $\det H \gt 0 \rightarrow$ local maximum
- $(03)$ $\det(H) \lt 0 \rightarrow$ saddle point
- $(04)$ $\det(H) = 0 \rightarrow$ no statement can be made using this approach
For $(0,0)$, condition $(4)$ tells us that $\det H = 0$, so nothing can be said about this critical point, neither a min or max. Of course, we could have also looked at $$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2.$$
Since $g(x,y) \lt 0$, this is not an extremum.
For $(2,1)$, we have: $f_{2,1} = 12 > 0$ and $\det H = 432 >0 \rightarrow$ a local minimum. The value of $f(x,y)$ at this minimum is $f(2,1) = -8$.
Plots are given by:
If there are no restrictions on x,
from $f_x=0$ you get that $y=(-1\pm \sqrt{65})/2$ or $x=(2n+1)\frac{\pi}{2}$, where n is any integer, and
from $f_y=0$ you get that $y=-1/2$ or $x=n\pi$, where n is any integer.
Therefore the stationary points are of the form $(n\pi, \frac{-1\pm\sqrt{65}}{2})$ and $((2n+1)\frac{\pi}{2}, -\frac{1}{2}).$
Now you need to test each of these points using the Second Partials Test.
Best Answer
Let $f(x,y)=x^{3}+x^{2}-xy+y^{2}+5$ and notice that $f\in \mathcal{C}^{+\infty}(\Omega\subseteq \mathbb{R}^{2}, \mathbb{R})$ then since $f$ is differentiable the critical points is the same as stationary points. A point $(x,y)\in \Omega$ is critical point if $\nabla f(x,y)=0$. Here we have $\nabla f(x,y)=(3x^{2}+2x-y,-x+2y)=0$ iff $(x,y)$ is $(-1/2,-1/4)$ or $(0,0)$.
The determinant of Hessian matrix for $f$ is given by $$\det Hf(x,y):=\det \begin{bmatrix} f_{xx}(x,y) & f_{xy}(x,y)\\ f_{yx}(x,y) & f_{yy}(x,y)\end{bmatrix}=\det \begin{bmatrix} 6x+2 & -1 \\ -1 & 2\end{bmatrix} $$
Remark:
Hence