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:
I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.
edit 1:(add more details)
$f(x,y)=(x-2)^2+y^2+5 \implies (x-2)^2+y^2=f(x,y)-5=R^2$
you want to find max and min of $f(x,y)$ is same to find max and min of $R$.
note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$
if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.
Best Answer
Note that $$f(x,y)=(x-2y)^2+2\ .$$ So at any point where $x=2y$ (and there are infinitely many such points) the value of $f(x,y)$ is $2$, and for any other point the value is greater than $2$. So these points are local and absolute minimum points.