# Thread: Did I do this right? (Local max/min/partial derivatives)

1. ## Did I do this right? (Local max/min/partial derivatives)

Let $f(x,y) = e^{-xy}(x^2 + y^2)$

1) Find polynomial equations for the critical points of f.

I just took both partials and found:

$f_{x}=-ye^{-xy}(x^2 + y^2) + 2x(e^{-xy}) = e^{-xy}(-x^2y-y^3+2x)$

$f_{x}=-xe^{-xy}(x^2 + y^2) + 2y(e^{-xy}) = e^{-xy}(-x^3-xy^2+2y)$

It's asking for polynomial equations, so do I just factor the exponent e out and write the equations left?

2) Find all the second order partial derivatives of f.

$f_{xx}=e^{-xy}(y^4+x^2y^2-4xy+2)$

$f_{yy}=e^{-xy}(x^4+x^2y^2-4xy+2)$

$f_{xy}=e^{-xy}(x^3y+xy^3-3x^2-3y^2)$

As it should be, I found $f_{xy} = f_{yx}$.

3) It can be shown that the critical points of f are (0,0), (1,1) and (-1,-1). Classify these points.

By plugging the points into $f_{xx}f_{yy} - (f_{xy})^2$, I got that (0,0) was a minimum, (1,1) was a maximum, and (-1,-1) was a saddle point.

You probably are wondering what I need help with; I really just wanted reassurance on the first part and wanted to make sure I was getting the concept right. I have a test coming up soon, and I wanted to make sure these review problems were getting done correctly so that I was prepared.

2. Originally Posted by wilcofan3
Let $f(x,y) = e^{-xy}(x^2 + y^2)$

1) Find polynomial equations for the critical points of f.

I just took both partials and found:

$f_{x}=-ye^{-xy}(x^2 + y^2) + 2x(e^{-xy}) = e^{-xy}(-x^2y-y^3+2x)$

$f_{y}=-xe^{-xy}(x^2 + y^2) + 2y(e^{-xy}) = e^{-xy}(-x^3-xy^2+2y)$

It's asking for polynomial equations, so do I just factor the exponent e out and write the equations left?
It requires you to observe that $e^u$ is never zero for real $u$, so:

$f_{x}=-ye^{-xy}(x^2 + y^2) + 2x(e^{-xy}) = e^{-xy}(-x^2y-y^3+2x)=0$

implies that:

$-x^2y-y^3+2x=0$

and similarly:

$f_{y}=-xe^{-xy}(x^2 + y^2) + 2y(e^{-xy}) = e^{-xy}(-x^3-xy^2+2y)=0$

implies that:

$-x^3-xy^2+2y=0$

Which is a pair of simultaneous polynomial equations for the critical points.

CB