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Minimum Surface Area Cylinder using Euler-Lagrange Equation?

  1. Feb 1, 2006 #1
    So, I've been reading Thornton and Marion's "Classical Dynamics of Particles and Systems" and have gotten to the chapter on the calculus of variations. In trying the end of chapter problems, I find I'm totally baffled by 6-9: given the volume of a cylinder, find the ratio of the height to the radius that minimizes the surface area.

    Now, just using elementary calculus, the solution is easy enough: take the volume, solve for the height, plug that back in the equation for area so you only have the radius, differentiate and set the result to zero, and solve. Voila, the ratio is 2.

    That's not my problem... how would you set this up to solve it using the Euler-Lagrange equation? I''m not really seeing how it fits... although it seems to me that it must. I've tried using the ratio as the independent variable, and the radius... but I somehow am not imposing the condition that the volume must be specified. (When I tried, I ended up with a cylinder of zero surface area... which I suppose *is* minimal, but...)

    Thanks.
     
  2. jcsd
  3. Feb 1, 2006 #2

    HallsofIvy

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    I don't see how that is a "calculus of variations" problem to begin with! Calculus of variations is concerned with minimizing or maximizing integrals- and I don't see any good way of representing the surface area as an integral.

    I wonder if this problem wasn't intended, rather, as an example of the "duality" property- that minimizing the surface area, for a given volume, gives the same answer as maximizing the volume for a given surface area.
     
  4. Feb 1, 2006 #3
    Well, volume and surface area are both integrals, although because we know the answer, it's convenient to forget that.

    Anyway, I see my problem here... I insisted on looking for a function with an independent variable. Bzzzt, wrong. Let R and H *both* be dependent variables... then Euler-Lagrange just becomes the method of Lagrange multipliers. (Hurm, there might be a connection here... :smile: )

    Thanks for responding!
     
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