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Volume Question

  1. Mar 18, 2004 #1
    Ok my teacher was going over this today, but i have no idea how to solve this.

    A Closed box with a square base is to have a volume of 16000cm^3. The top and bottme cost 3 persq/cm. while the top is 1.50 pr sq/cm Find the demison of the box that will lead to the total minium total cost. WHat is the total cost.
  2. jcsd
  3. Mar 18, 2004 #2
    So does the top cost 3/cm^2 or 1.5/cm^2?

  4. Mar 18, 2004 #3
    Whoops. my bad. THe top and bottem are $3per sq/cm and the sides are $1.5 per sq/cm
  5. Mar 18, 2004 #4
    Okay. The box is a rectangular prism. It has a base of dimensions length (l) and width (w). It also has height (h). Its volume is defined by

    V = l*w*h

    which is required to be 16000cm^3, so

    lwh = 16000.

    Additionally, the cost of the box is given by

    C = $1.5*2(lh + wh) + $3*2(lw)

    Using the equation for volume you can eliminate one of the variables, giving you the cost in terms of only 2 variables. You'll have to minimize this function using techniques you learned in class.

    Why don't you give it a shot and post what you get?

  6. Mar 19, 2004 #5
    I thought the box was a square prism ("A Closed box with a square base")...
  7. Mar 19, 2004 #6
    So it is. 3 mistakes in 1 day. Go me.

    It's easy to fix. Just let l = w and minimize the remaining variable.

  8. Mar 19, 2004 #7
    Is there actually an answer to the question if the box was not a square prism, and no other data was supplied? Can't see how myself. :smile:
  9. Mar 19, 2004 #8
    I don't think it matters either way. The top and bottom contribute most to the cost, so minimizing those areas will likely minimize the cost as well. It just makes the math a bit more difficult.

    Then again, I haven't run the numbers, so I could be (and considering the day, probably am) wrong.

  10. Mar 19, 2004 #9

    matt grime

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    Science Advisor
    Homework Helper

    You have to at least state what kind of aperture you want on the 'box'. The (smooth) surface with maximal volume per unit of surface area is the sphere; but where do you put stuff in? These problems are (often) solvable with calculus of variations (probably) subject to the smoothness constraints etc.
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