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Homework Help: Find the right shape

  1. Jul 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Imagine having a rectangle with horizontal sides ##x## and vertical sides ##y##. The surface of that rectangle is than ##S=xy##.
    Now imagine the sides ##y## are rigid - they don't bend nor do they change their length, while the ##x## sides is some kind of cloth and is easily bendable but does not extend (it does not stretch, therefore the length is always ##x##). Now if we for some reason change the distance between the parallel ##y## sides from ##x## to ##x-\delta x## the cloth will bend outwards.

    The question is:
    What is the shape of the cloth if we want the surface to be maximum?

    2. Relevant equations

    3. The attempt at a solution
    Since I probably wasn't clear enough in the problem statement, here is a rough sketch (I have drawn only the upper membrane. There should be the one symmetrical at the bottom of course).
    So the arc length is ##x## and the question is what is the shape of the arc? Is it a circle, an ellipse, a sin function... Whatever.

    I started this way:

    The total surface is $$S=(x-\delta x)y+2S_0$$ where ##S_0## is the surface under the cloth. If i am not mistaken it can be written as $$S_0=2\int_{-\Delta}^{\Delta} f(t)dt$$ where ##\Delta =\frac{1}{2}(x-\delta x)## and ##f(t)## being the function I am looking for.

    I tried to solve the problem using Lagrangian multiplier and setting a constraint of constant perimeter ##2x+2y=const.##

    But I didn't come very far... Is there another way? :/
  2. jcsd
  3. Jul 9, 2015 #2


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    Maximum what? Surface area? Central deflection? What quantity are you trying to maximize?
  4. Jul 9, 2015 #3
    Surface area, yes!
  5. Jul 9, 2015 #4


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    If the x-sides bend but do not stretch, then you will get some cylindrical surface but it will still have exactly the same area as the original.
  6. Jul 9, 2015 #5
    They only bend, yes. And by bending, they reduce the distance between the vertical sides for ##\delta x##.

    I have to disagree.
    Isn't that close to claiming that the circle with perimeter ##a## will have the same surface as rectangle with perimeter ##a##? The perimeter is constant, I agree with that, but I am not so sure about the surface area.
  7. Jul 9, 2015 #6


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    We can reduce it to maximising the area under a curve of given length L from origin to a point B on the x axis.
    If you have been taught calculus of variations that is probably the method you are expected to use. But you need to use the generic expression for length of an arc over dx, ##\sqrt{1+p^2}.dx##, where p=dy/dx. You will also need a Lagrange multiplier, of course, and the Euler-Lagrange equation.
    However, if we can assume a circle maximises the area of a closed loop of fixed length, there may be an easier way.
    Suppose an arc of a circle of the given length would have radius R, and mark the centre of curvature, C.
    So OC=BC=R. Suppose the angle OCB happens to be a rational multiple of pi. Can you see how to relate the problem to the area of a circle?
  8. Jul 10, 2015 #7
    Are you sure about that? Can we really? Because I was thinking about that too but what confuses me is that the size of the rectangle - therefore it's surface area - also changes from ##S=xy## to ##S=(x-\delta x )y##. So to me it looks like ##\delta x## is also a parameter that needs to optimized in order to ensure the maximum surface area. Or maybe my feelings are completely mistaken.
    Of course, if we can reduce to the maximising the area under a curve, than this is a famous isoperimetric problem also known as Dido's problem. And the solution to it is of course a circle.

    Absolutely not. I have to find a curvature that maximises the surface area. I am mentioning circular shape, because I know it is, but that is what I am trying to prove here. I don't think that we can go the other way around: assume a circular shape. Or you can convince me that we can. :)
  9. Jul 10, 2015 #8


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    The way I read the OP, it asks what shape of the bendy bit maximises the area for a given displacement of the sides. You now seem to be suggesting that the area is also to be maximised wrt the displacement, which is a different question.
    If we accept (or prove) the best is the arc of a circle for a given displacement, then it should be relatively straightforward to find the displacement that maximises it.
    Even with the interpretation I gave above, they're not exactly the same problem. One maximises the area of the entire closed loop, allowing the loop to be any smooth shape, while the other maximises it under the constraint of a straight portion of specified length. It is not completely trivial to derive the second from the solution to the first.
  10. Jul 10, 2015 #9
    Than I apologize. The displacement of the sides is not given.

    Ok, so now that we understand the problem. How should I start? What is the idea?
  11. Jul 10, 2015 #10


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  12. Jul 11, 2015 #11
    It is highly related, yes, but this time I want to use a completely different approach! I want to solve the problem completely mathematically, like I know nothing about the physics.
    The solution me and Chet got on a link you provided is ok. Great actually. But I have a feeling - and it is just a feeling - that I would get the same result (a circular shape) if I would use a constraint that the surface area has to be maximized. Now since Chet truly is a master, he helped me to find a solution but also warned me that solving this problem mathematically (perhaps through Dido's problem) is something he wouldn't be confident doing since he is not a mathematician. Therefore I decided to post the problem in Calculus and find somebody that could help me.

    I hope I am not complicating things too much.
  13. Jul 11, 2015 #12


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    You will get the same result for the same reason. If you consider a fixed δx, the shape of the membrane will be a section of arc (in the same way it has to be this shape if the rest of the figure would be the rest of the circle). You can calculate its area, add twice its value to y(x-δx) and maximize that sum.

    This solution also happens to minimize energy in the physics case.
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