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Homework Help: Multivariable Derivates?

  1. Aug 23, 2010 #1
    if Cost(r)=0.01(2[tex]\pi[/tex]r2+v/[tex]\pi[/tex]r2(2[tex]\pi[/tex]r))+0.015(4[tex]\pi[/tex]r+v/[tex]\pi[/tex]r2)
    how does one show for the optimal radius/height for any volume i.e. where the cost will be the minimum for that volume?

    all my attempts have ended with 0=0

    help please? i'll give you cookehz ^^
    Last edited: Aug 23, 2010
  2. jcsd
  3. Aug 23, 2010 #2


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    The first thing I would do is go ahead and replace "v" by [itex]\pi r^2 h[/itex]:
    [itex]cost(r, h)= 0.01(2\pi r^2+[v/(\pi r^2)][/itex][itex](2\pi r)+0.015(4\pi r+v/(\pi r^2)[/itex]
    [itex]cost(r, h)= 0.01(2\pi r^2+[(\pi r^2h)/(\pi r^2)][/itex][itex](2\pi r)+0.015(4\pi r+(\pi r^2 h)/(\pi r^2)[/itex]
    [itex]cost(r, h)= 0.01(2\pi r^2+ h(2\pi r)+ 0.015(4\pi r+ h))[/itex]

    Now differentiate with respect to both r and h:
    Treating h as if it were a constant,
    [tex]cost_r(r, h)= 0.01(4\pi r+ 2\pi h+ 0.015(4\pi))[/tex]
    [tex]cost_h(r, h)= 0.01(2\pi r+ 0.015)[/tex]

    To find the optimum height, set those both equal to 0 and solve the two equations for r and h. (Since there is no "h" in the second equation , it is particularly easy to solve it for r. Although it looks to me like there is a sign error- that will give a negative value for r.)
  4. Aug 23, 2010 #3
    i don't think this is right or if it is how did you get it so that the second part(0.15(4pir+h) is also multiplied by 0.01 as the 0.01 and 0.015 "segments" of the equation are seperated completely by the addition sign? (i'm only first year calculus) so if it's a law or something please explain :D
  5. Aug 23, 2010 #4


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    i think its because your original brackets were a little ambiguous - why not use the method but with what you've got?
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