Analyzing Critical Points of f(x,y)=Ax2+E

[xokaitt]

Homework Statement

Let f(x,y)=Ax²+E where A and E are constants. What are the critical points of f(x,y)? Determine whether the critical points are local maxima, local minema, or saddle points.


2. The attempt at a solution

First I found the first partial derivatives with respect to x and y
\partialf/\partialx=2Ax
\partialf/\partialy=0
\Rightarrow 2Ax=0,
\Rightarrow x=0 for any constant A.

Therefore, all points lying on the y-axis are critical points.
(i.e. C.P.'s = (0,n), n\inR.)

Now, we have to find the second partial's with respect to x and y.
\partial²f/\partialx²=2A
\partial²f/\partialy²=0
and
\partial²f/\partialx\partialy=0

Therefore Df=(\partial²f/\partialx²)(\partial²f/\partialy²)-(\partial²f/\partialx\partialy)² at (0,n) , n\inR.
\Rightarrow Df=(2A)(0)-(0)²=0

This is where I get stuck. Now that Df=0, how do I determine whether or not the critical pts are local extrema or saddle pts?

From plotting the function on Mathematica, I know that these critical points are in fact saddle points, but I don't know how to mathematically state that.

Thanks!

 

[Mark44]

It seems to me that you are making this much harder than it needs to be by not sketching a graph of this surface. Since y doesn't appear explicitly in the formula for the function, this surface is a cylinder with parabolic cross section, and with its axis of symmetry in the direction of the y-axis. IOW, the surface looks something like a trough. If A > 0, the trough opens upward, and all critical points are global minima. If A < 0, the trough opens downward, and all critical points are global maxima. All critical points lie on a line that is parallel to the y-axis.