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Thread: Finding the Maximum Radius for a Chord.

  1. #1
    May 2010

    Finding the Maximum Radius for a Chord.

    Hey guys,
    I want to find the Minimum/Maximum Radius for the following equation:
    Area of a circle segment - Math Open Reference (Central angle in radians)

    That is, assuming that Theta = 180 degrees. I think that's all the information I need to work it out, but I'm puzzled as how to figure out the final answer. It's a tiny part of a larger equation and I think my head is about to implode from the amount of times I've had to rethink how I approach the question. x.x

    The question was taken from:
    I'm not even sure whether I'm approaching this right.
    Last edited by LawfulSarcastic; Aug 22nd 2010 at 08:51 AM.
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  2. #2
    MHF Contributor

    Apr 2005
    I don't know what you mean by "Minimum/Maximum Radius". The minimum or maximum radius to do what? The formula you post is for finding the area of a circle segment of any radius. There is no "minimum or maximum radius" except for the obvious minimum: 0 radius with 0 area. But the radius can be as large as you please making the area as large as you please.

    As for the math problem website you give, Are you looking for the angle for the opening that will make that circular gutter as large as possible for a given Circumference? I think you may be overlooking the "for a given circumference" condition. Without that we can make the area as large as we wish by taking the angle closer and closer to 0.

    Start with a circle of radius R. It has circumference 2\pi R and area \pi R^2. If we cut an angle \theta, measured in radians, out of the circle, we cut R\theta out of the circumference, leaving (2\pi- \theta)R and we cut, using that formula you found, (R^2/2)(\theta- sin(\theta)) leaving a cross-section area of (R^2/2)(2\pi- \theta- sin(\theta).

    For a given circumference, C= (2\pi- \theta)R, we have 2\pi- \theta= C/R so that \theta= 2\pi- C/R.
    Since sine is periodic with period 2\pi, [tex]sin(\theta)= sin(2\pi- C/R)= sin(-C/R)= -sin(C/R) so the area is given by (R^2/2)(C/R+ sin(C/R)).

    Now, it is not a matter of differentiating that with respect to /theta to determine the "optimum" \theta. As I said, that would be \theta= 0 to maximize the cross section area. The problem is simply to compare that, for a given circumference C with the areas the same cross section lengths of the other two figures to see which gives the greatest area. There is no calculus required.
    Last edited by HallsofIvy; Aug 22nd 2010 at 09:48 AM.
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  3. #3
    May 2010
    I think you may have just nailed it. Thank you very much and sorry for pestering you across two boards.
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