Well, you can probably see by inspection that (x,y)=(0,0) is a minimum, but let's look at the critical points - where .

so the only critical point is (0,0). But we have to look at the boundary, that is, where . Solving for and substituting gives:

, which has a maximum at x=0, and minima at x=1 and -1. The minima are greater than the minimum at (0,0), so we can ignore them. But the maximum at x=0 gives us maxima at (x,y)=(0,1) and (0,-1).

- Hollywood