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Maximum Volume

  1. Oct 22, 2007 #1
    I am a little stuck on this problem :frown:

    If the total surface area (including the area of the top and bottom ends) of a cylinder is to be kept fixed (=A), what is its maximum possible volume?

    For such cylinders of fixed total area, plot Volume(V) v/s Radius(R) clearly indicating the values of R for which the volume is maximum and zero.

    The total surface area will be [itex]A = 2\pi R(L + R)[/itex] where L is the length of the cylinder. Here, A = constant and I have to determine the maximum possible volume. I don't know how to proceed, should I express the volume in terms of the area and do something?

    Kindly guide me on this...
     
  2. jcsd
  3. Oct 22, 2007 #2

    Gokul43201

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    Fixing the area, A, sets up a relationship between the radius R and length L. This then allows you to write down an equation for the volume that involves only one variable (either L or R). From here, it's just the matter of taking a derivative, etc.
     
  4. Oct 22, 2007 #3
    Thank you so much, I will try it and post my solution soon. :smile:
     
  5. Oct 26, 2007 #4
    Surface area A = constant.
    [tex] A = 2\pi R (L + R)[/tex]

    [tex]L = {{A - 2\pi R^2}\over 2\pi R}[/tex]

    [tex]V = \pi R^2 L[/tex]

    Putting the value of L:
    [tex]V = {AR\over 2} - \pi R^3[/tex]

    Solving for dV/dR = 0 for maximum:
    [tex]R = \sqrt{{A\over 6\pi}}[/tex]

    [tex]V_{max} = {A\over 2}\sqrt{{A\over 6\pi}} - \pi \left({A \over 6\pi}\right)^{3/2}[/tex]


    V = 0 for [tex]R = \sqrt{{A\over 2\pi}}[/tex]

    Should I assume some value for A when I plot V v/s R?
     
    Last edited: Oct 26, 2007
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