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Math Help - Surface area and volume?

  1. #1
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    Surface area and volume?

    How do you find the dimensions for cylinder that, given an area, has the least volume but the most surface area? Is there a plugin formula? Could you give me, say, an example such as the volume = 1L? (The 1 litre is not the actual question but it would help me to understand)

    Thank you in advance - any help is much appreciated.

    Mike
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  2. #2
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    The only "plug-in" formulas I can think of are the formulas for volume and surface area of a cylinder: V= \pi r^2h and A= 2\pi(r^2+ rh).

    No, I cannot give you "an example such as the volume = 1L" because in your question, it is the surface area that is given, not the volume. I can show how to get the minimum volume given that the surface area is 1 m^2.

    With A= 2\pi (r^2+ rh)= 1, we can solve for h: h= \frac{1- 2\pi r^2}{2\pi r}. Putting that into the formula for volume, V= \frac{1}{2}(r- 2\pi r^3). The extreme (maximum or minimum) values occur where the dervative is 0: \frac{1}{2}(1- 6\pi r^2)= 0 or r= \sqrt{\frac{1}{6\pi}}
    Last edited by CaptainBlack; July 1st 2009 at 03:26 AM. Reason: fix latex
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  3. #3
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    Surface area and volume of a cylinder

    Hello everyone -

    I'm not sure whether we're understanding the question correctly here. For a given surface area the minimum volume of a cylinder is (essentially) zero. Think of a cylinder with (almost) zero height, whose circular ends are each one-half of the given surface area, and you'll see what I mean.

    Similarly, for a given volume, the surface area can be made as large as you like, by taking a small enough height.

    Grandad
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  4. #4
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    I'm so sorry everyone, I made a mistake in the question. It was meant to be:

    How do you find the dimensions for cylinder that, given an volume, has the optimum surface area? Is there a plugin formula? Could you give me, say, an example such as the volume = 1L? (The 1 litre is not the actual question but it would help me to understand)

    Thank you in advance - any help is much appreciated.

    Mike
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  5. #5
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    Dimensions of the cylinder :  \pi R^2 h

    Total surface area =  2 \pi R h + 2 \pi R^2

    We can let  V =   \pi R^2 h ~~ and change the subject

     h = \frac{V}{ \pi R^2 }

    Then combine the equations ,

    Total surface area =  2 \pi R (\frac{V}{ \pi R^2 }) + 2 \pi R^2

     = 2  ( \frac{V} R + \pi R^2 )

    You can see that it is a function of only R , you can find the extrema by taking its derivative .
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  6. #6
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    Here is a non calculus solution:

    <br />
V = \pi r^2 h <br />

    <br />
SA = 2 \pi r^2 + 2 \pi r h<br />

    let <br />
\alpha = \frac{h}{r}<br />

    <br />
SA = \frac{2 V}{h} + \frac{V}{2r}<br />

    <br />
SA = 2 \pi \left(\frac{V}{\pi \alpha}\right)^\frac{2}{3} + 2 V \left(\frac{V}{\pi \alpha}\right)^\frac{-1}{3}<br />

    <br />
SA = 2 \pi^\frac{1}{3} V^\frac{2}{3}\left(\frac{1}{\alpha^\frac{2}{3}} + \frac{\alpha^\frac{1}{3}}{2} + \frac{\alpha^\frac{1}{3}}{2} \right)<br />

    using AM-GM inequality

    <br />
SA \geq 2 \pi^\frac{1}{3} V^\frac{2}{3}   3\sqrt[3]{\frac{1}{4}}<br />

    with equality when

    <br />
\alpha = 2<br />

    so the minimum surface area occurs when the height is twice the radius. You can also find the minimum using calculus if you like.
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  7. #7
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    Hello Tesla23

    Welcome to Math Help Forum!
    Quote Originally Posted by Tesla23 View Post
    Here is a non calculus solution:

    <br />
V = \pi r^2 h <br />

    <br />
SA = 2 \pi r^2 + 2 \pi r h<br />

    let <br />
\alpha = \frac{h}{r}<br />

    <br />
SA = \frac{2 V}{h} + \frac{V}{2r}<br />

    \color{red}<br />
SA = 2 \pi \left(\frac{V}{\pi \alpha}\right)^\frac{2}{3} + 2 V \left(\frac{V}{\pi \alpha}\right)^\frac{-1}{3}<br />

    <br />
SA = 2 \pi^\frac{1}{3} V^\frac{2}{3}\left(\frac{1}{\alpha^\frac{2}{3}} + \frac{\alpha^\frac{1}{3}}{2} + \frac{\alpha^\frac{1}{3}}{2} \right)<br />

    using AM-GM inequality

    <br />
SA \geq 2 \pi^\frac{1}{3} V^\frac{2}{3}   3\sqrt[3]{\frac{1}{4}}<br />

    with equality when

    <br />
\alpha = 2<br />

    so the minimum surface area occurs when the height is twice the radius. You can also find the minimum using calculus if you like.
    An ingenious solution!

    I didn't find it easy to see how you arrived at the line I've coloured red. May I suggest:

    <br />
V = \pi r^2 h <br />

    <br />
SA = 2 \pi r^2 + 2 \pi r h<br />

    let <br />
h =\alpha r<br />

    Then V = \pi r^3\alpha \Rightarrow r = \left(\frac{V}{\pi\alpha}\right)^{\tfrac13} and therefore h = \alpha\left(\frac{V}{\pi\alpha}\right)^{\tfrac13}

    \Rightarrow SA = 2\pi\left(\frac{V}{\pi \alpha}\right)^{\tfrac23} + 2\pi\left(\frac{V}{\pi\alpha}\right)^{\tfrac13}\al  pha\left(\frac{V}{\pi\alpha}\right)^{\tfrac13}

     = 2 \pi \left(\frac{V}{\pi \alpha}\right)^{\tfrac23} + 2 V \left(\frac{V}{\pi \alpha}\right)^{-\tfrac13}

    ...etc

    Grandad
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  8. #8
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    Can I just resurrect this delightful thread?

    I imagine this is a standard question, but google is so glutted with hyper-physics and stuff these days, its hard to find a standard solution.

    I have a liter of water and i want to minimize the surface area of a closed cylinder.

    I assume I want to minimize: S=2\pi rh+2\pi r^2

    with the constraint V=\pi r^2h=1

    Can somebody start me off if I wanted to use the Lagrange method?

    Can I transform S and V so that

    S=f(x,y,z)= ...

    and

    V=g(x,y,z)= ...

    Or do I do something else?
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  9. #9
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    Quote Originally Posted by billym View Post
    Can I just resurrect this delightful thread?

    I imagine this is a standard question, but google is so glutted with hyper-physics and stuff these days, its hard to find a standard solution.

    I have a liter of water and i want to minimize the surface area of a closed cylinder.

    I assume I want to minimize: S=2\pi rh+2\pi r^2

    with the constraint V=\pi r^2h=1

    Can somebody start me off if I wanted to use the Lagrange method?

    Can I transform S and V so that

    S=f(x,y,z)= ...

    and

    V=g(x,y,z)= ...

    Or do I do something else?
    Set f(r, h) = 2pi*rh + 2pi*r^2 and g(r, h) = pi*r^2*h. Then you want to solve grad f = lambda * grad g, g(r, h) = 1.
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