1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook