# Classification of extrema

1. Nov 4, 2007

### Fermat

If we have a function f(x,y) for which we wish to find the stationary points, then there is one set of rules for classifying those points.

Let there be an extremum of f(x,y) at (x,y) = (a,b), and D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}², then

1) D > 0 and fxx > 0 => min
2) D > 0 and fxx < 0 => max
3) D < 0 => saddle point
4) D = 0 - indeterminate - needs deeper analysis.

Now I can never remember this lot of rules so I made up my own little set of rules - as follows.

In 2-D maths, fxx > 0 gives a min, and fxx < 0 gives a max, so I extended this to 3-D maths, as below.**

My rules
1) if fxx and fyy > 0 then a min
2) if fxx and fyy < 0 then a max
3) if fxx and fyy of opposite sign then a saddle

Now, this has always worked well for me; I find it easy to transfer the 2-D rules to a 3-D situation, and it saves me having to work out fxy.

The only thing it doesn't cover is; if fxx and fyy are the same sign (which by my rules should be either a max or a min) is it possible for D to be < 0, which would mean the TP was a saddle?

So, my questions are:

1) can anyone find an error in my rules?
2) does anyone know of a situation where fxx and fyy were both of the same sign, but the TP was a saddle?

**: here, fxx should really be d²y/dx²

2. Nov 4, 2007

### Fermat

My rules don't (always) work.

The function f(x,y)=x^2+y^2+4xy has a TP at (x,y) = (0,0) with fxx = fyy = 2, but fxy = 4 and D = -12 < 0. So f(x,y) has a saddle at the origin despite both fxx and fyy having the same sign.

Oh well, now I know.

3. Nov 4, 2007

### HallsofIvy

Surely you can make up an example where D= fxxfyy- fxy2 is negative. How about if fxx= fyy= 2 and fxy= 3? If fxx= 2 then fx= 2x+ g(y). In that case, fxy= g'(y)= 3 so g= 3y. Now that we know that fx= 2x+ 3y f(x,y) must equal x^2+ 3xy+ h(y). Then we would have fy= 2x+ h' and fyy= h"= 2 so h, since it is function of y only, is y^2+ cy+ d. Taking those constants of integration to be 0 for simplicity, f(x,y)= x^2+ 3xy+ y^2.

That has fxx= fyy= 2> 0 but fxy= 3 so D= fxxfyy- (fxy)^2= 4- 9= -5. The only critical point is at (0,0) and, although fxx and fyy are both positive, that critical point is a saddle point.