# Thread: Local extremum (2 variables)

1. ## Local extremum (2 variables)

I have to find the critical points of the function f and to tell whether they are local minimum, local maximum or saddle point.
The function is $f(x,y)=x^2y^2$.
By intuition the point (0,0) is a local minimum.
Formally, I found that $f_x(x,y)=0 \Leftrightarrow 2xy^2=0$, so $x=0$ and $y=$ any value. (I'm not convinced since y is treated as a constant and it might be 0, while in this case x could be any other number different from 0. Hence my "so" is not an implication in fact! I don't understand what I'm doing wrong)

$f_y(x,y)=0 \Rightarrow y=0$ and $x=$ any value. (Once again I don't trust my implication)
Hence the critical point is $(0,0)$.

From my class notes, if $f_{xx}(0,0)f_{yy}(0,0)-f_{xxy}(0,0)=0$, then I cannot conclude about the nature of the critical point. Unfortunately it happens here, and there's no theorem that can help me.
If it was the case of a transcendantal function, I would have approximate it by a Taylor's polynomial and checking if the polynomial reaches a minimum or a maximum or a saddle point, but the problem is that the function is a polynomial. (by the way I'm not even sure that approximating a function via a Taylor's polynomial assegurate that the approximated function reaches the same critical points. Maybe there's a theorem about it but I never read it yet).

2. All the critical points are positive-semidefinite since $Q(h,k)=2y^2h^2,\ y\in R$ or $Q(h,k)=2x^2k^2,\ x\in R$ (depending on the critical point)

Because of that, you can't use the quadratic form to see if the point is an extremum or not.

$x^2\geq 0,\ y^2\geq 0 \implies x^2y^2\geq 0$

$f(x,y)=0$ for all points $(x,y)=(0,C),\ C\in R$ or $(x,y)=(C,0),\ C\in R$, so these points are local minima ( $f'_x=0,\ f'_y=0$ here as well).

3. Originally Posted by arbolis

From my class notes, if $?!""f_{xx}(0,0)f_{yy}(0,0)-f_{xxy}(0,0)""!?=0$, then I cannot conclude about the nature of the critical point. Unfortunately it happens here, and there's no theorem that can help me.
.
The formula not like that but like this

$D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f^2_{xy}(x_0,y_0)$

4. Originally Posted by Amer
The formula not like that but like this

$D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f^2_{xy}(x_0,y_0)$
Thanks a lot! My professor wrote the $^2$ but seemed right after the $_x$. She didn't prove anything so I took the formula I saw, without understand why it is like it is. I miss a good book on the subject. (I can't borrow Stewart's one on vector calculus since too many people borrow it)