# Thread: multivarible extremum point

1. ## multivarible extremum point

Let f(x,y)=((x^(2)y^(2))/(x^(2)+y^(2))), classify the behavior of f near the critical point (0,0).

2. Originally Posted by bobby77
Let f(x,y)=((x^(2)y^(2))/(x^(2)+y^(2))), classify the behavior of f near the critical point (0,0).
I don't know how you are supposed to approach this on your course,
but the following works:

Along the $\displaystyle x \,$ and $\displaystyle y\,$ axes the function is a constant equal to zero. Along
any other ray through the origin we may put $\displaystyle y=\lambda x\,$, when:

$\displaystyle f(x,\lambda x)=\frac{\lambda^2 x^2}{1+\lambda^2}\,$

which has a quadratic like mininum at $\displaystyle x=0\,$.

A surface plot shows what this looks like (see attachment)

(this assumes that we give the function a value of $\displaystyle 0\,$ at $\displaystyle x=y=0\,$)

RonL

3. Originally Posted by CaptainBlank
I don't know how you are supposed to approach this on your course,
but the following works:
I think he wants it done through the second partials test.

4. Originally Posted by ThePerfectHacker
I think he wants it done through the second partials test.
In this case that will be inconclustve (I think, I haven't checked)

RonL

5. Originally Posted by CaptainBlack
In this case that will be inconclustve (I think, I haven't checked)

RonL
You mean, a saddle point.

I agree with that, because by different views (planes) the point is both a maximum and minimum. Hence it is neither.

6. Originally Posted by ThePerfectHacker
You mean, a saddle point.

I agree with that, because by different views (planes) the point is both a maximum and minimum. Hence it is neither.
No the determinant of the Hessian is indeterminate. The point is tippling on
the edge of being a saddle point, but is not quite there.

The point is a minima.

RonL