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Maximum/Minimum Problem

  1. Dec 11, 2006 #1
    Sorry to post another problem, but hey, the more the marrier, right? :smile:

    1. The problem statement, all variables and given/known data

    maxima/minima problem
    2)a cylinder, including top and bottom, is made from material which costs "x" dollars per square inch. Suppose there is an additional cost of fabrication given by "n" dollars per inch of the circumfrence of the top and bottom of the can. find an algerbraic equation whose solution would be he radius ,r, of the can of given volume, V, whose cost is a minimum.
    * Do not have to solve equation, just find a algerbraic solution

    2. Relevant equations

    3. The attempt at a solution

    I tried this one, and I think I may be overthinking it. Would the equation just consists of the volume of the cone (equalling the money for the material) plus the additional amount of the circumfrence of top and bottom (so two times the circumfrence) for the entire cost of the can?

    Once again, thanks again!
  2. jcsd
  3. Dec 11, 2006 #2
    By the way it's worded it looks like your cost will depend on the total surface area of the closed cylinder - that is twice the area of the circles and the area of the rest. That part about the cost being n times the circumfrence also seems ambiguous - I can't really tell if they want the total circumfrence of the top and bottom or just the circumfrence of either.
  4. Dec 11, 2006 #3
    I get what you mean, and agree about the surface area. It seems I was underestimating the problem before, haha. The circumference aspect asks for both, the top and bottom, (so 2 times that). I've been messing around with it and I know that once you differntiate the problem, you should just be left with pi,r,n and x in the answer since V is a constant and the problem doesn't want anything else in the answer.
  5. Dec 11, 2006 #4
    Here, the cost of material: [tex]((2\ pi)r^2+ (2 \pi)rl)x[/tex]
    cost of fabrication: [tex](2 \pi r)n[/tex]
    Hence, the total cost is the sum of the above;
    It is also given that the volume of the cylinder is constant, so
    [tex]V=\pi r^2 l[/tex]
    from which you get [tex]l= \frac{V}{\pi r^2}[/tex]
    substituting the value of l in the initial equation gives you your equation.

    You can differentiate this equation w.r.t r and get your solution.
    Last edited: Dec 11, 2006
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