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Cylindrical Container Minimum Cost

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    An industry is to construct a cylindrical container of fixed volume
    V, which is open at the top of. Find the quotient of the height by
    radius of the base of the container so that manufacturing cost can be minimal, given that the unit area of the base costs (b) twice the unit area
    the lateral surface (a) .


    2. Relevant equations
    V = π r^2 h
    A = π r^2
    S = 2πrh (lateral surface)


    3. The attempt at a solution

    The cost of manufacturing the container is C= C_s + C_a = Α * a + S * b, where b=2a so
    C= 2a A + a S = 2πr^2 a + 2πrh a . I understand that we need to form the ratio h/r and regard that as the variable for the cost equation, but I see no way to do so... We are also give that V=constant, but other than the equation r^2 h = Constant I have no clue.
     
  2. jcsd
  3. Nov 7, 2011 #2

    gneill

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    Staff: Mentor

    I think the problem statement had the unit area cost of the base as b, and the unit area cost of the sides as a, with b = 2a. That would make your cost function
    [tex] C = A b + S a [/tex]
    and since b = 2a,
    [tex]C = 2 a A + a S = a (2 A + S) [/tex]
    [tex]C = a (2 \pi r^2 + 2 \pi r h) [/tex]
    [tex]C = 2 \pi a (r^2 + r h) [/tex]
    Now, the volume V is a constant so you can solve the volume expression for h and replace h in the above with it. You then have a cost function that involves only one variable: r. Minimize the cost function w.r.t. r.
     
  4. Nov 7, 2011 #3
    So if V is constant, then h acts as a parameter for r, and we can transform C into a function of only r by solving for r in the V equation? That feels simple however I hesitated to do so because I kept thinking of C as a multivariable equation (not that we have studied them so far) of r and h.

    So by substituting h we get [tex]C = 2 \pi a (r^2 + \frac{V}{\pi r})[/tex]. Is this correct? I proceed to study the derivative which is [tex]C'= 2 \pi a (2r + -\frac{V}{\pi r^2})[/tex] and we get that minimal cost C comes where C' = 0 and [tex]r^3 = \frac{V}{2\pi}[/tex] so we substitute V and find the ratio?
     
  5. Nov 7, 2011 #4

    gneill

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    Staff: Mentor

    There's no need to substitute for V, since it's a constant. This value of r should be the required solution for the radius (you should check to make sure that it represents a minimum for the cost function and not a maximum!).

    Use this value of r in the volume function to find the corresponding value of h.
     
  6. Nov 7, 2011 #5
    Yes that's what I am saying! Thanks a lot gneill!
     
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