# Critical Point at Origin: Max, Min, or Neither

For this problem, there's actually 4 distinct problems. I have no idea what to do for these. Would you have to do something called the hessian matrix,which I am confused by. Can someone run me through one of these problems step by step so I can do the other three on my own?

Thanks!!

• You can start from: en.wikipedia.org/wiki/Hessian_matrix#Second_derivative_test and find the eigenvalues of the Hessian matrix in your cases. if you do this and write here your result we can help you. – Emilio Novati Mar 25 '18 at 20:51
• In the future, please take the time to enter the content of your question as text instead of posting a picture of it. It’s only fair that you should take some of your own time to formulate the question if you expect others to spend their time to help you. Images are neither searchable nor accessible to people who use screen readers. Use MathJax to format your mathematical expressions; you can find a quick reference here. – amd Mar 26 '18 at 2:32
• Your three questions so far are all of a kind: they involve exercising basic knowledge about critical points of multivariable functions. Your saying that you have no idea about how to approach any of them suggests that you’re trying to do these exercises without having studied the material in the first place, or that it didn’t stick when you did. If so, I highly recommend reviewing that material before coming back with another similar question. – amd Mar 26 '18 at 2:41

Sure. Take $f(x)=x^2+xy+y^2$. First, we find the partial derivatives with respect to $x$ and $y$.

$\frac{\partial }{\partial x}f(x,y)= 2x+y.$

$\frac{\partial }{\partial y}f(x,y)= 2y+x.$

From this, we can find the critical points.
Then we simply use this formula:

$D(x,y)=f_{xx}f_{yy}-(f_{xy})^2.$

When you plug your critical points into $D(x,y)$,

If $D>0$ AND $f_{xx}<0$, then this is a relative maximum.

If $D<0$ AND $f_{xx}>0$, then this is a relative minimum.

If $D<0$, then this is a saddle point.

If $D=0$, then the test is inconclusive.

I hope this helps!