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Tin can with the least amount of tin

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the smallest r and h for a tin can with Volume a) 1L b) V so that the amount of tin used is minimal

    2. Relevant equations
    [itex]V=r^{2}\pi h[/itex]
    [itex]A=2\pi(r^{2}+h)[/itex]

    3. The attempt at a solution
    Since the area has to be minimal, I expressed r from V
    [itex]r^{2}=\frac{V}{\pi h}[/itex]
    and inputed it into the equation of A
    [itex]A=2\pi(\frac{V}{\pi h}+h)[/itex]

    Now I get the first derivative:
    [itex]\frac{dh}{dA}=2\pi(\frac{-V}{\pi^{2}h^{2}}+1)[/itex]

    I set this to equal 0, do the algebra and end up with
    [itex]-1=\frac{-V}{\pi^{2}h^{2}} \\
    -\pi^{2}h^{2}=-V \\
    \pi^{2}h^{2}=V \\
    h^{2}=\frac{V}{\pi^{2}}\\
    h=\frac{\sqrt{V}}{\pi}
    [/itex]

    When I input this into the equation for r I get:
    [itex]r=\frac{\sqrt[4]{V}}{\sqrt{\pi}}[/itex]

    If I set [itex]V=1[/itex] it gives me ~0.32 for h and ~0.56 for r

    The solution book tells me the solution is
    [itex]r=\sqrt[3]{V(2\pi)}[/itex]
    [itex]h=2\sqrt[3]{V(2\pi)}
    [/itex]

    Which would give me totally different values for h and r

    Where did I make a mistake? :(
     
    Last edited: Oct 30, 2013
  2. jcsd
  3. Oct 30, 2013 #2

    Mark44

    Staff: Mentor

    I cleaned up your LaTeX, which was pretty hard to read.
    Your formula for area is wrong. The total surface area should include the cylinder, which is made from a rectangular piece, plus two circular end pieces.
     
  4. Oct 30, 2013 #3
    Thank you, but I looked into my formula booklet just now and it tells me that the area is [itex]2r^{2}\pi+2r\pi h [/itex]which is [itex]2\pi(r^{2}+h)[/itex]

    I just saw that I forgot to write it into my LaTeX above, but I still calculated with it, so that can't be it Oo

    EDIT: I just saw I forgot the second r...
    Nevermind... Thank you

    >.<
     
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