Notes for Math 210 - by M. Flashman

Suppose f(x,y) = A x2 + B xy + C y2 .

To begin assume that A is not 0
We reorganize this function by completing the square on the x's.
So f(x,y) = A( x2  + B/A xy  +B2 /4A2  y2 )  + (C - B2 /4A)y2  or
f(x,y) = A( x + B/(2A) y)2 )  + ((4AC - B2 )/4A2)y2  .
Now assume 4AC-B2  >0.
If  A >0  then f(x,y) >0 for all (x,y) but (0,0). So (0,0) is a minimum for f.
If  A < 0  then  f(x,y) < 0 for all (x,y) but (0,0). So (0,0) is a maximum for f.

Now assume that 4AC-B2  <0.
If y = 0 and x>0 then f(x,y) has its sign the same as A.
But when x = -B/2A y and y>0, f(x,y) has it sign opposite to that of A.
Thus no matter how close we consider (x,y) to (0,0), there will always be points where f(x,y) > 0 and where f(x,y)<0.
Thus (0,0) is neither a maximum nor a minimum for f(x,y) ... it is a saddle.

If A = 0 and B is not 0 then 4AC-B2  < 0 and

f(x,y) = y (Bx + Cy).
The line Bx+Cy = 0 passes through (0,0) and divides the plane into two regions- one where Bx+Cy is positive and one where Bx+Cy is negative. Since B is not 0, the line Bx+Cy = 0 is not horizontal, so there are points in the plane with y>0  in each of the two regions.
If y>0 and Bx+Cy is positive, then f(x,y) is positive.
If y>0 and Bx+Cy is negative then f(x,y) is negative.
Thus, no matter how close we consider (x,y) to (0,0), there will always be points where f(x,y) > 0 and where f(x,y)<0.
Thus (0,0) is neither a maximum nor a minimum for f(x,y) ... it is a saddle.

Summary: If 4AC-B2  < 0, then (0,0) is saddle point for f.
If 4AC-B2  > 0, and A>0 the (0,0) is a minimum for f.
If 4AC-B2  > 0, and A<0 the (0,0) is a maximum for f.