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Minimizing the Surface Area

  1. Mar 15, 2008 #1
    [SOLVED] Minimizing the Surface Area

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

    A box has a bottom with one edge 8 times as long as the other. If the box has no top and the volume is fixed at V, what dimensions minimize the surface area?

    2. Relevant equations

    V = lwh

    SA (with no top) = lw + 2lh + 2wh

    3. The attempt at a solution

    l = x
    w = 8x
    h = V/(8x^2)

    Finding an equation for the surface area.

    SA = lw + 2lh + 2wh
    SA = x(8x) + 2x(V/(8x^2)) + 2(8x)(V/(8x^2))
    SA = 8x^2 + V/(4x) + 2V/x

    Finding the derivative of the equation in order to set it equal to zero to find the critical points, so the minimum can be found.

    (d SA)/(d x) = 16x - V/(4x^2) - 2V/(x^2)
    (d SA)/(d x) = (64x^3 - 9V) / (4x^2)

    (d SA)/(d x) = 0
    (64x^3 - 9V) / (4x^2) = 0
    (64x^3 - 9V) = 0
    64x^3 = 9V
    x^3 = (9V)/64
    x = ((9V)/64)^(1/3)

    Plugging the solution into the equations for the dimensions.

    l = x = ((9V)/64)^(1/3)
    w = 8x = 8((9V)/64)^(1/3)
    h = V/(8(((9V)/64)^(1/3))^2)



    I am unsure if I did the right steps in order to find the solution.
    Also, I am not very confident in the work I did for each step.
     
  2. jcsd
  3. Mar 15, 2008 #2
    It looks good to me, except you can simplify 64^(1/3) into 4 so that you can reduce your answers and it looks cleaner.
     
  4. Mar 15, 2008 #3
    Thanks a lot for looking over it. I just put it into the online grader and it is, in fact, correct.
     
  5. Apr 10, 2008 #4
    I appreciate it.
     
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