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Math Help - Minimum area of cylinder

  1. #1
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    Minimum area of cylinder

    I have to find the minimum surface area of a right cylinder using numerical method
    I have did it with differentation, I have looked back at some of my work and in all my books,
    I can get that the surface area = 2pi(r^2) + 2V/r

    And found that using arithmetic mean-geometric mean inequality

    SA= 2pi(r^2) + 2V/r

    SA= 2pi(r^2) + V/r + V/r

    This the part I dont quite understand, how it changes to this
    SA> 3 x cubedrt2pi{(r^2) (V/r)(V/r)}

    I can just do it and be done but I realy prefer to understand what I am doing
    any advise would be great thx
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  2. #2
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    Quote Originally Posted by pmh118 View Post
    I have to find the minimum surface area of a right cylinder using numerical method
    I have did it with differentation, I have looked back at some of my work and in all my books,
    I can get that the surface area = 2pi(r^2) + 2V/r

    And found that using arithmetic mean-geometric mean inequality

    SA= 2pi(r^2) + 2V/r

    SA= 2pi(r^2) + V/r + V/r

    This the part I dont quite understand, how it changes to this
    SA> 3 x cubedrt2pi{(r^2) (V/r)(V/r)}

    I can just do it and be done but I realy prefer to understand what I am doing
    any advise would be great thx
    1. I assume that the volume of the cylinder is a constant.

    2. The surface area of a cylinder is:

    a_s = 2 \cdot \pi r^2 + 2 \pi r \cdot h

    3. The volume of the cylinder is:

    V = \pi r^2 \cdot h

    If you calculate \dfrac{2V}r=2 \dfrac{\pi r^2 \cdot h}r = 2 \pi r \cdot h you'll get indeed the curved surface of the cylinder twice.

    4. To get the minimum surface area you have to differentiate

    a_s(r)=2 \cdot \pi r^2+\dfrac{2V}r

    a'_s(r)=4 \cdot  \pi r - \dfrac{2V}{r^2}

    5. Now solve a'_s(r)=0 for r:

    4 \cdot  \pi r - \dfrac{2V}{r^2}=0~\implies~4 \cdot  \pi r^3 =2V

    So after moving some stuff around you'll get: r = \sqrt[3]{\dfrac V{2\pi}}

    6. Now plug in this value into the equation of the surface area and you'll get your result.
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  3. #3
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    Thank you for your response maybe I need to clarify a bit better
    Yes the Volume is constant
    I have already sucssesfully found the minimum Radius value with differentiation and proved it was minimum with the second derivitive
    And also worked out the Surface area with this Radius value given me the minimum surface area.
    But I found a new formula that works for the minimum surface area with using only the volume I just dont understand how derive this new formula


    My problem is I dont understand how to get from:

    2\pi}{r^2}+\frac Vr + \frac Vr

    to

    3\sqrt[3]{2\pi}{V^2}

    I hope this LaTex works
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