As far as part 1 is concerned, you are given , do you not see that F(x,y)= 0 if and only if either or ? And those are equivalent to and . Now since F is continuous and 0onlyon those curves, F can change sign (go from positive to negative or vice-versa) only on those curves. So it is sufficient to test a single point in each of the regions bounded by those curves to see what sign F has in each.

Now which is 0 if x= 0 or if , and so . If x= 0, then 2y= 0 so y= 0. If then so that , from which x= 0. Yes, (0, 0) is the only critical point.

"Any straight line" through the origin is of the form y= ax, for some a, or x= 0 (the vertical line through the origin). In the first case, y= ax, . The derivative of that is which clearly has x= 0 as a 0. The second derivative is which is positive at x= 0 so F(x, ax) has a local minimum there. If x= 0, which clearly has a minimum at y= 0.