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
We have the function
$$f(x, y) = 8x^3-3x^4+48xy-12y^2$$
We have
$$f_x = -12 x^3+24 x^2+48 y, f_y = 48 x-24 y$$
If we set both equations equal to zero and simultaneously solve them, from the second equation, we get $y = 2x$. Substituting this into the first equation, we get $$ -12 x^3+24 x^2+96 x = -12 x(x-4) (x+2) =0 \implies x = -2, 0, 4$$
Using these three $x-$values, we substitute them back in $y = 2x$ and this produces three critical points:
$$(x, y) = (-2, -4), (0,0), (4, 8)$$