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Homework Help: Cylinder Optimization for a Minimum?

  1. Mar 26, 2009 #1
    Question: Cost of producing cylindrical can determined by materials used for wall and end pieces. If end pieces are 3 times as expensive per cm2 as the wall, find dimensions (to nearest millimeter) to make a can at minimal cost with volume of 600 cm3.

    Relevant equations: a) V=600cm3=[pi]r2h

    b) surface area = 2[pi]r2 + [pi]rh

    The miraculous attempt:

    a) solve h in terms of [pi]r2: 600=[pi]r2h ; 600/[pi]r2=h

    b) find surface area equation using h: SA=2[pi]r2 + 2[pi]r(600/[pi]r2 ; SA=2[pi]r2 + 2(600/r) ; SA=2[pi]r2 + 1200/r

    c) first derivative: SA'= 4[pi]r - (600/r2) = 0

    I am regrettably hitting a brick wall on what should happen next. Additional information is that I think the cost looks something like: 2[pi]r2 = (2[pi]rh)(3), with the multiplication by 3 being the fact that the round end pieces are three times as expensive.

    Any and all further help would be greatly appreciated.

    Last edited: Mar 26, 2009
  2. jcsd
  3. Mar 26, 2009 #2
    The constraint equation is as you have it
    V = pi*r^2*h = 600

    The cost function is the surface area, weighted by the production cost factors:
    C = 2*pi*r*h + 3*2*pi*r^2
    where the first term is the cost of producing the side wall and the second term is the cost of producing the two ends, with the factor 3 included to reflect the higher manufacturing cost for the ends.

    Your previous approach was to solve the constraint for h, so let us do this again:
    h = V/(pi*r^2)
    and then substitute into the cost function
    C = 2*pi*r*V/(pi*r^2) + 3*2*pi*r^2
    = 2*V/r + 6*pi*r^2

    Now differentiate wrt r ...
    dC/dr = -2*V/r^2+12*pi*r
    set the derivative equal to zero...
    0 = -2*V/r^2 + 12*pi*r
    =V - 6*pi*r^3
    r^3 = V/(6*pi)
    and so forth...
    (all assuming that I did not make any mistakes!)
  4. Mar 26, 2009 #3
    Thanks a million! The math seems to check out on paper as well, with a radius of 7.3 cm and a height of 3.6 cm. (Volume would end up being 602.7 cm3.)

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