We can use the second derivative test to find the extreem values.

and

The 2nd equation gives rise to two different cases x=0 and y=0.

Case I: if x=0 then

Case II: If y=0 then,

Now we can make the Hessian matrix

Now we can check the critical points found above

Case I:

So both of these are Saddle points.

Case I:

Since both eigen values of the Hessian are positive the point in a local minimum.

For Part b) all we need to do is calculate the maximum and minimum on the boundary.

Since the boundary is a circle with radius , it can be parameterized by

If we put these into the original function we get

This is a function of one variable and its extreem values can be found using methods from first semester calculus.

Work not shown, but you will find that is a critical point. If you put this back into the parameteric equations you get

.