Take partial derivatives with respect to x and y.




Set them = 0,
Solve that
Get (x,y) = (0,0), (5,10)
Critical points (x,y,z) = (0,0,1), (5,10,-65/6)
Second partial test. Get the four second partial derivatives:




at (0,0) is -1, at (5,10) is 9


at (0,0) is -2, at (5,10) is -2




at (0,0) is -2, at (5,10) is -2


at (0,0) is 1, at (5,10) is 1


We test (x,y) = (0,0)






So the point (0,0,10) is a saddle point.
We test (x,y) = (5,10)






So the point (5,10,65/6) is determined by the sign of fxx,
which is 9, which is positive, so concave upward there and is a
relative minimum.
Edwin