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| bit4bit |
 Posted: Mon May 12, 2008 1:58 pm Post subject: For fun :) |
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Forum Ph.D.
Joined: 14 Jul 2007
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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 100m2 (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. 
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| JaneBennet |
| Posted: Mon May 12, 2008 4:26 pm Post subject: |
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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 m3, 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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| bit4bit |
| Posted: Tue May 13, 2008 5:54 am Post subject: |
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Forum Ph.D.
Joined: 14 Jul 2007
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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] / v2
=[(2a+2b)(100-2ab2)-2(100ab-a2b2)] / (2a+2b)2
Which simplifies to:
∂V/∂a = [200b2-2a2b2-4ab3] / 4(a+b)2
Then a similar thing for ∂V/∂b gives:
∂V/∂b = [200a2-2a2b2-4a3b] / 4(a+b)2
so for ∇V=0,
[200b2-2a2b2-4ab3] / 4(a+b)2=0
[200a2-2a2b2-4a3b] / 4(a+b)2=0
Thats only true if the numerator is 0, so:
200b2-2a2b2-4ab3=0 (1)
200a2-2a2b2-4a3b=0 (2)
a and b are non-zero, since if they were 0, then volume would be zero, so solving simultaneously:
(1) / 2b2 --> 0=100-a2-2ab
(2) / 2a2 --> 0=100-b2-2ab
100-a2=100-b2
a=b for positive values.
So then 0=100-a2-2a2,
3a2=100
a=√(100/3) = (10√3)/3 = b
c= 100-ab / 2(a+b) = 100- [102*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)2(5√3/3) = (100/3)*(5√3/3)
= 500√3/9 m2
It's kind of tedious, especially now I've seen your method.
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| JaneBennet |
| Posted: Tue May 13, 2008 6:02 am Post subject: |
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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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| bit4bit |
| Posted: Tue May 13, 2008 6:09 am Post subject: |
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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.
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| JaneBennet |
| Posted: Tue May 13, 2008 6:19 am Post subject: |
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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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