[bonanova]

The cost of shipping a crate is specified by the sum of its length, width and height.

You'd like to save money by packaging your crate inside a cheaper one.

Prove that this is [or is not] possible.

Assume crates have perpendicular, rectangular, and negligibly thin sides.

[Guest]

The cost of shipping a crate is specified by the sum of its length, width and height.

You'd like to save money by packaging your crate inside a cheaper one.

Prove that this is [or is not] possible.

Assume crates have perpendicular, rectangular, and negligibly thin sides.

Assuming that we are not putting the crate inside another crate, but repackaging its contents, which we estimate to be of infinite granularity :

If the dimensions of the crate are a, b, and c, its cost is a+b+c, and its volume is abc.

We are asked to keep abc fixed and minimize a+b+c. This is achieved when a=b=c, that is when the crate is a perfect cube.

Therefore, if you have a crate of dimensions x, y and z that is not already a cube, create one where all sides are of length cuberoot(xyz).

[Guest]

The cost of shipping a crate is specified by the sum of its length, width and height.

You'd like to save money by packaging your crate inside a cheaper one.

Prove that this is [or is not] possible.

Assume crates have perpendicular, rectangular, and negligibly thin sides.

put your intact crate inside a crate that is smaller in any dimension (LxWxH) than your crate. However, if you disassemble your crate until you have the 6 walls, you can easily get it inside a smaller box.

For example, a 2x2x2 crate would have 6 2x2 panels. As they are negligibly thin, they would fit in a crate that is 2 x sqrt2 x sqrt2.

[bonanova]

Assuming that we are not putting the crate inside another crate, but repackaging its contents, which we estimate to be of infinite granularity :

Don't assume that.

The new containing crate is to have a lower sum of dimensions but contain the first crate thus saving money.

[Guest]

A crate has 3 dimensions but let's work on two dimension: A square has a width of x and height of y. If you tilt this square so that it is in a diamond shape, now this shape has a width of x/sqrt(2) and a height of y/sqrt(2), these are both less than x and y. So if you manage to built a crate that can be stand as a diamond shape, you will pay less.

[bonanova]

You can't put your intact crate inside a crate that is smaller in any dimension (LxWxH) than your crate.

Consider in two dimensions a rectangle can hold a longer rectangle [albeit a narrower one] at an angle.

It's easy to prove the containing rectangle has a greater perimeter in that case.

Can you extend that proof, or come up with another proof, for three dimensions?

[bonanova]

A crate has 3 dimensions but let's work on two dimension: A square has a width of x and height of y. If you tilt this square so that it is in a diamond shape, now this shape has a width of x/sqrt(2) and a height of y/sqrt(2), these are both less than x and y. So if you manage to built a crate that can be stand as a diamond shape, you will pay less.

We assume the crate has rectangular, perpendicular sides.

That is, it's a right rectangular pyramid.

And we want the outer crate to cost less than the inner crate.

[Guest]

Ok, If I understand correctly now, we just want to build a parallelipipedic crate that can hold the 6 sides of another parallelipipedic crate (I'll ingore the pyramid comment as I assume it's a lapsus).

Let's have the sides of the original crate be a, b and c

We can assume in all generality that a <= b <= c.

We thus have 6 panels: 2 of dimensions a * b ; 2 of dimensions a * c ; and 2 of dimensions b * c

If we make a new crate with a bottom side of dimensions b * c, it will obviously be able to hold all 6 panels flat.

But since the panels are of negligible thickness, our new crate can afford to be of "epsilon" (let's just say very small) height.

We thus have a new crate with a shipping cost of b + c.

[Guest]

We assume the crate has rectangular, perpendicular sides.

That is, it's a right rectangular pyramid.

And we want the outer crate to cost less than the inner crate.

Sorry, I made a mistake:

A tilted square has a larger width of sqrt(2) times square's width. not 1/sqrt(2). Thus I would pay more with that diamond shaped crate, so thanks!

And I'm sure that the answer is

"no, impossible", but proof is hard for me.

[Guest]

And I'm sure that the answer is "no, impossible", but proof is hard for me.

then I'll have to agree with you. Any rectangle A inscribed in a rectangle B will have perimeter that is smaller than B's.

[Guest]

The perimeter of the enclosed crate is maximized as the lengths of two of the sides approach zero while the third side increases in length.

in my mind it suffices to say that the length of the diagonal of a cube with sides of length s (sqrt(3)*s) is less than 3*s. A cheaper external crate is impossible.

[bonanova]

You all have it.

It is impossible,

The proof in two dimensions is straightforward - and rectangle inscribed inside another rectangle has a smaller perimeter.

But it's not so easy to carry out that proof in three dimensions.

If anyone is interested I'll post a proof that I read.

It requires a figure; it starts with identical nested crates of sides a, b, c, and increasing the containing crate's dimensions by epsilon.

You can then calculate the wiggle room that gives to the inside crate's dimensions.