Any Brainiacs Out There? Math Question.

[flobadob]

If I have a large box, and I am going to fill it with smaller boxes, is there any way to work out the best way to pack the smaller boxes so as to maximise the space in the larger box and find out how to fit the most amount of small boxes possible.

For reference the large box has internal dimensions of 40cm Height, 38cm Width and 58cm Length. The smaller boxes have external dimensions of 12cm Height, 12cm width and 16cm Length.

At the moment the best I can do is fit 39 of the smaller boxes into the larger box and still be able to close the lid.

But is there a formula out there that will allow me to safely squeeze in a couple more??

[THECORBYLOON]

The volume of the large box is 40cm × 38cm × 58cm which is 88,160 cubic centimetres.

The external volume of the smaller boxes is 12cm × 12cm × 16cm which is 2,304 cubic centimetres.

Dividing the smaller capacity into the larger capacity shows 38.26 boxes would fit.

Allowing for errors in measurement, it appears your figure of thirty-nine boxes is the most you can fit.

[gingejbee]

If there was room to fit in a couple more, wouldn't it be obvious from the amount of room you have left?...given that you are (mathematically) at maximum capacity already?

[Zacs-Master]

There's no complex maths needed other than to take the volume of the large box and divide it by the volume of the small boxes.

each box is 2304 centimeters3

the large box is 88160 centimeters3

you've done well to fit 39 in as the answer is 38.26.

[Zacs-Master]

As corby said (LOL)

[bobbinwales]

Defo can't be 39. 38 is largest theretical number if boxes can be split... but with full boxes thered be gaps you can't fill without breaking some small box's up??

[Barsel]

If these boxes are empty, just put one inside the other like Russian dolls and you'll get loads in!

[gingejbee]

bobb...as TCL said..."allowing for errors in measurement"...that's why 39 fitted in presumably.

[ToraToraTora]

something is wrong because dimensions do no allow the whole space to be taken up by the smaller boxes. eg the long side of 58 will fit say 3 of the other long side(48) leaving 10cm that is unusable. Now there will no doubt be an optimum configuration but I can't see how you can get 39 in.

[bobbinwales]

//bobb...as TCL said..."allowing for errors in measurement"...that's why 39 fitted in presumably.//
I wernt thinking of tiny gaps...I were thinking of big gaps, like Toratoratora as since mentioned.
As a example if the length is 58cm the nearest you can get with these is 56cm (2@16cm + 2@24) so theres a unfilled gap of 2cm (x 38 wide).
And widthways the closest you can get then to 38cm is 3x12 =36 cm so thats another big gap.

[bobbinwales]

^2@24 above should say 2@12 (=24).
Anyway the only way to use up all the height of 40cm is to have some some a different way up to have 2x12 + 1x16.
But your still left with problem that other rows cant be filled fully

[bobbinwales]

This link just shows how difficult these can be
https://en.wikipedia.org/wiki/Bin_packing_problem

[Zacs-Master]

That's about items of different sizes, bob.

[bobbinwales]

But the point is they don't fit exactly...there'll be fair size gaps so answer won't be 38 and there's no simple formula.

[bobbinwales]

Wonder why filters filter'd out ' 2 @ 12'

[bobbinwales]

I think there's only me and toratoratora seeing this isn't simply a 'case' of dividing big volume by little volume.
Hopefully claretgold will see this or zebu or someone else who done more than gcse maths

[JJ109]

Again, I don't have enough knowledge on this subject. I can use logic and get 32 boxes..... but I don't know if this is the optimum.

[douglas9401]

Shouldn't a boxing question be in 'Sport'?

[ClareTG0ld]

I don't think I can add anything to, say, bobbinwales's link, or the observation that 39 sounds likely to be the upper limit if the dimensions are slightly rounded.

Maybe, at a pinch, you could go up to 40, but I doubt it.

[JJ109]

I'll explain my 32 anyway
The height of the large box is 40. Therefore, no matter which way you lay the boxes on top of each other, you are only going to get three layers. So best to use 16 + 12 + 12 = 40
For the layer of height 16, you can get 3 boxes (12 cms) wide and 4 boxes (12 cms)lengthwise. Therefore on this layer there are 4 * 3 = 12 boxes

In the other two layers where the height is 12 cms. You can
a) lay 3 * 16 length ways (48cm) by 3 boxes wide (36 cm) = 9 boxes
b) lay 2 boxes * 16 width ways and 4 lengthways = 8 boxes
c) Lay 2 boxes length ways (32 cms) by 3 boxes width ways (36 cm) = 6 boxes pLUS....
2 boxes 12 cm lengthways by 2 boxes 16 cm widthways = 4 boxes
For answer c add 6 and 4 =10 boxes on this layer, which is the best

Therefore answer = 12 layer 1, 10 layer 2, 10 layer 3 = 32

[Peter Pedant]

thanks Bo Bo
I thought someone must have written an algorithm for this
( list of actions, poss repeated, which allows you to find the max capacity) and you have shown where

The difficulty dear reader is the same as the shortest distance issue ( finding the shortest dist in a network between A and B)

you see if you have three boxes A B and C - start with C then do B....but that gives no info on whether A is better to start with. So you have to go thro all combinations...

Big networks, big lulu to be had if you can write an efficient (P) algorithm.

BUT - - for the shortest distance algo - if you do a network in string ( no one does) and then pick up the two points. The shortest distance falls out in space. Cute huh?