Crystal Math IX: cuboid

[muon2]

The thing that makes the first one easy is that given the longest and shortest edges, the intermediate edge must equal the longest edge to maximize the volume. Let x = y > z = L-x, where L is the given sum 100. The volume is V = xyz = x²(L-x) = x²L - x³.

One method to solve this is with elementary calculus such that dV/dx = 2xL - 3x² = 0. The maximum occurs at x = 2L/3, or using L = 100, x = 66.67 (approx). The volume is 148148.148 (approx)

The hard problem doesn't actually say this, but should I assume that the sum of the largest and smallest faces is 100?

[muon2]

Okay, since nobody can answer the second question (or wants to, respectively), here's the answer:

V_max = 2,000/9*√3 ≈ 384.9

Unfortunately I have had to work, so I was going to look at it this weekend. Perhaps I will put up a derivation of the answer if that's of interest.

[muon2]

Here's the hard problem slightly reworded. Consider a rectangular cuboid where the sum of the areas of its smallest and its largest face equal 100. Find the maximum volume of such a cuboid.

Consider the two faces unwrapped into a single rectangle with an area A = 100, and sides L and W. The volume is found by folding the rectangle at a point p along side L so that the volume is a maximum. Since the rectangle consists of the largest and smallest faces the side W must between the other two in length or p <= W <= L-p.

In general the volume after the fold is V = pW(L-p) = WLp - Wp². The volume is maximized when dV/dp = WL - 2Wp = 0, or p = L/2 = L-p, ie folded at the midpoint of the side. However unless W=L/2, W would be either less than both p and L-p or greater than both p and L-p and that is not permitted. The closest one could be is to make the fold so that W equals either p or L-p and the fold would then be made by making a square.

Now return to the problem using sides x,y and z. Two sides must be equal so let V = x²z, with the requirement that x² + xz = A, or z = (A - x²)/x. With this substitution V = x(A - x²). The volume is maximized when dV/dx = A - 3x² = 0, or x = √(A/3). The sides are x = y = √(A/3) and z = 2 √(A/3). The maximized volume is V = √(A/3)[A - A/3] = 2(A/3)√(A/3).

Using A = 100, the answer is (200/3)√(100/3) = (2000/3)√(1/3) = (2000/9)√3 as given above.

[muon2]

Wow! You're really good, muon.
Do you see striking parallels between the answers of question (A) and question (B)?

Thanks. The common issue in both problems is that they lack a traditional constraint on the third side. Instead the constraint is an inequality that forces the third side to be between the other two in length, and that leads to it being the same as one of the other two.

[muon2]

Wow! You're really good, muon.
Do you see striking parallels between the answers of question (A) and question (B)?

Thanks. The common issue in both problems is that they lack a traditional constraint on the third side. Instead the constraint is an inequality that forces the third side to be between the other two in length, and that leads to it being the same as one of the other two.

No, I wasn't getting at that...
In both cases the base of the cuboid is a square, and in both cases one of the wanted sub-polytopes (edge or face, respectively) is twice as big as the other sub-polytopes of the same rank.
Or in other words: In both cases one of the edges is twice as long as the third; in one case the edge of the square base is the longest one, in the other case the third edge is the longest one. Thus, there is a kind of duality inherent in both cuboids, isn't it? ...

I see what you are getting at, but the problem gave no hint that the base of either would be a square. I found that the challenge for both was to prove that the base was a square. Once that was done the problem was straightforward, and the 2 to 1 ratio was actually lost on me until you noted it.