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Math Help - Find the minimum & maximum area of a box!

  1. #1
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    Find the minimum & maximum area of a box!

    Here's another question Ive come across during revision



    So

    Volume is
    w*d*h = 32000
    w^2*h = 32000

     h = \frac{32000}{w^2}


    Total area is base of box + 4 sides -
    w^2 + 4wh
    w^2 + \frac{128000}{w}

    And then I got the derivative of f(x) = w^2 + \frac{128000}{w}

    And where this derivative = 0, there was a local minimum. This was where w = 40, so that makes d = 40 and h = 20 for the dimensions of the box that will use the minimum amount of material.

    I take it this is all correct?

    If it is, what I'm wondering is how I would be able to find the dimensions that would use the maximum amount of material? I only got one critical point when differentiating and that turned out to be a local minmum so how is it possible to find the maximum amount of material used?
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  2. #2
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    The math doesn't lie: if there is no maximum, then it is theoretically possible to use an infinite amount of material to build a box of that volume, unless there are any bounding values for the individual dimensions.
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  3. #3
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    For example, if you were to make the length of a side of the base, w, equal to 0.0001, then the height would be \frac{32000}{0.000001}= 32000000000 and so the total surface area would be 0.000001+ 4(32000000000)(0.001)= 128000000.000001. By making the base smaller and smaller, you can make the height, and so the surface area, arbitrarily large.

    More generally, the derivative for f(w) you got was f'= 2w- \frac{128000}{w^2}= \frac{2w^3- 128000}{w^2}= 2\frac{w^3- 64000}{w^2}= 2\frac{(w- 40)(w^2+ 40w+ 1600)}{w^2}. Yes, f'(40)= 80- 80= 0. If w is larger than 50 then w- 40>0 and the other numbers are clearly positive so f'> 0. That is, f is increasing for all w> 40 and so there is no maximum surface area.
    Last edited by HallsofIvy; April 13th 2011 at 11:09 AM.
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  4. #4
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    That's pretty crazy hehe, thanks for clearing it up for me.
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