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Maximum volume

  1. Nov 21, 2007 #1
    1. The problem statement, all variables and given/known data
    The volume of a right circular cone is V = [(pie)(r^2)(h)]/3 and it ssurface area is S = (pie)(r)(r^2+h^2)^(1/2), where r is the base radius and h is the height of the cone. Find the dimensions of the cone with surface area 1 and maximum volume.

    3. The attempt at a solution
    I think the only difficult part of this question is the math, because its quite difficult. I'm finding V' to be
    [tex]\pi r[r+(4/\pi^2r^2)-4r^2]/6[(1/\pi^2r^2)-r^2][/tex]
    Which i can't find any zero's for, can someone double check this?

    Steps to finding the derivative:,
    1) Set S equal to 1 and solve for h,
    2) stuff h into volume and take derivate, unless you know of a better way?
     
  2. jcsd
  3. Nov 21, 2007 #2

    Avodyne

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    Your method is correct, but I get something simpler. After solving for h and plugging into V, simplify as much as possible before taking the derivative.
     
  4. Nov 21, 2007 #3
    is this what you simplified it down to:
    [tex]\sqrt{(1/9)r^2[1-\pi^2r^4]}[/tex].

    If so, i got as a derivative:
    [tex](1/2)[(1/9)r^2(1-\pi^2r^4)]^(-1/2) [(1/9)r^2(-4\pi r^3) + (1-\pi^2r^4)(2/9)\pi][/tex]
    which doesn't simplify down much nicer...
     
  5. Nov 21, 2007 #4

    Avodyne

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    You're missing an r in the last term, but that's it. You want to set this to zero, so you can cancel out everything not in square brackets, and then pull out common factors and cancel those too ...
     
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