A perfectly smooth two-dimensional surface that has the shape of a doughnut is suspended in the (x, y, z)-Cartesian space. Let the doughnut be positioned so that an observer facing the plane y=0 will see through the hole. For example, the doughnut is standing up vertically, rather than lying flat. Next, define f(x, y, z)=z.
1. Consider the restrictions of f to the doughnut surface. How many critical points are there?
2. How many of those critical points are of the minimum or maximum type?
(It is suffice to sketch the surface and indicate which of the points are max/min.)
Essentially in 2 space, the shape will merely be two circles with different radii, and the domain will include all points on and between the boundaries of these circles. Also, since an observer sees the hole facing y=0, the circles will be in the xz-plane.
What I don't really understand is the statement "f(x, y, z)=z". Are they restricting the domain to points on the z-axis? If so, there will be 4 critical points, 2 relative max, and two relative min.