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Thread: For fun :)

  1. #1 For fun :) 
    Forum Masters Degree bit4bit's Avatar
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    I worked out this problem today, and thought it was quite neat, so I posted it for anyone else to try:

    An open-topped box, has dimensions a, b, (base) and c (height). If the surface area of the box is 100m<sup>2</sup> (outside only), find the largest volume of the box. The faces of the box are 2D (0 thickness).

    It probably won't be a hard question for some of you but I thought it was quite a nice little application of what I'm learning.


    Chance favours the prepared mind.
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  3. #2  
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    One method is to use calculus. Another method is the following:

    We have ab + 2bc + 2ca = 100.

    By the AM–GM inequality,



    which on simplification becomes



    So the maximum volume is 500*⁄*3√3*m<sup>3</sup>, attained when ab = 2bc = 2ca = 100*⁄*3, i.e. when a = b = 10*⁄*√3*m, c = 5*⁄*√3*m.

    This is a typical math-olympiad problem – and problems involving AM–GM abound in math olympiads.


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  4. #3  
    Forum Masters Degree bit4bit's Avatar
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    Nice, I didn't know about that method, but it's the right answer anyway. I just used the calculus method which I though was quite nice...

    Surface area = 100 = ab+2ac+2bc

    Then get volume as a function of only a and b, V(a,b):

    c=(100-ab) / 2(a+b)

    Volume=abc= ab(100-ab) / 2(a+b)

    Then you can find a maximum point by using ∇f=0

    ∇V(a,b) = <∂V/∂a, ∂V,∂b>

    using the quotient rule:

    V= ab(100-ab) / 2(a+b) = u/v

    ∂V/∂a = [v∂u/∂a - u∂v/∂a] / v<sup>2</sup>
    =[(2a+2b)(100-2ab<sup>2</sup>)-2(100ab-a<sup>2</sup>b<sup>2</sup>)] / (2a+2b)<sup>2</sup>

    Which simplifies to:

    ∂V/∂a = [200b<sup>2</sup>-2a<sup>2</sup>b<sup>2</sup>-4ab<sup>3</sup>] / 4(a+b)<sup>2</sup>

    Then a similar thing for ∂V/∂b gives:

    ∂V/∂b = [200a<sup>2</sup>-2a<sup>2</sup>b<sup>2</sup>-4a<sup>3</sup>b] / 4(a+b)<sup>2</sup>

    so for ∇V=0,

    [200b<sup>2</sup>-2a<sup>2</sup>b<sup>2</sup>-4ab<sup>3</sup>] / 4(a+b)<sup>2</sup>=0

    [200a<sup>2</sup>-2a<sup>2</sup>b<sup>2</sup>-4a<sup>3</sup>b] / 4(a+b)<sup>2</sup>=0

    Thats only true if the numerator is 0, so:

    200b<sup>2</sup>-2a<sup>2</sup>b<sup>2</sup>-4ab<sup>3</sup>=0 (1)
    200a<sup>2</sup>-2a<sup>2</sup>b<sup>2</sup>-4a<sup>3</sup>b=0 (2)

    a and b are non-zero, since if they were 0, then volume would be zero, so solving simultaneously:

    (1) / 2b<sup>2</sup> --> 0=100-a<sup>2</sup>-2ab
    (2) / 2a<sup>2</sup> --> 0=100-b<sup>2</sup>-2ab

    100-a<sup>2</sup>=100-b<sup>2</sup>
    a=b for positive values.

    So then 0=100-a<sup>2</sup>-2a<sup>2</sup>,
    3a<sup>2</sup>=100

    a=√(100/3) = (10√3)/3 = b

    c= 100-ab / 2(a+b) = 100- [10<sup>2</sup>*3 / 9] / [40√3 / 3]
    = (100 / [40√3 /3]) - (100/3 / [40√3 /3])
    =(300 / 40√3) - (100 / 40√3)
    =200 /40√3 = 5/√3 = 5√3/3

    Then V=abc = (10√3/3)<sup>2</sup>(5√3/3) = (100/3)*(5√3/3)

    = 500√3/9 m<sup>2</sup>

    It's kind of tedious, especially now I've seen your method.
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  5. #4  
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    Until recently, I would have tackled the problem using calculus as well. It was only recently that I learned to harness the power of the AM–GM inequality.
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  6. #5  
    Forum Masters Degree bit4bit's Avatar
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    Lol, i'm having a little read of it now, I can't understand why it's true, but i'm having a little look on the wiki article. It seems theres a few different proofs for it.
    Chance favours the prepared mind.
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  7. #6  
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    The proof by induction on the wiki article should be the easiest one of all to follow.
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