# Thread: For fun :)

1. 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.   2.

3. 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.   4. 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.   5. 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.   6. 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.  7. The proof by induction on the wiki article should be the easiest one of all to follow.  Bookmarks
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