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Math Help - maximizing doughnut

  1. #1
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    maximizing doughnut

    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.
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  2. #2
    Grand Panjandrum
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    Re: maximizing doughnut

    Quote Originally Posted by quantoembryo View Post
    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.
    f(x,y,z) is a function defined on all of \mathbb{R}^3 , now for points constrained to be on the surface of the doughnut how many critical points are there.

    Parameterise the surface of the doughnut so that the surface is {\bf{x}}=(x(u,v), y(u,v), z(u,v)) then f({\bf{x}})=z(u,v) and the question is asking about the critical points of z(u,v).

    (By the way this can be done by inspection you do not actually have to do the parameterisation etc if you have good visio-spatial visulasation skills)

    CB
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