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April 27th, 2016, 01:57 AM  #1 
Newbie Joined: Apr 2016 From: Ireland Posts: 1 Thanks: 0  Dimensions of small boxes in the large box
Hi, Dimensions of large box: 38x27x37 Quantity of small boxes in large box: 100 Packed tight, no free space. How to calculate the size of a small box? Thanks 
April 27th, 2016, 03:51 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,878 Thanks: 1502 
One possible set of dimensions for a single small box ... 38/5 x 27/5 x 37/4 , arranged in a 5 x 5 x 4 fashion ... there are more possibilities. 
April 27th, 2016, 03:53 AM  #3 
Senior Member Joined: Apr 2014 From: UK Posts: 903 Thanks: 331 
Ignoring the thickness of the box walls, the volume of the large box is 38x27x37 'units', which is 37962 'cubic units'. Each of the smaller boxes is therefore 379.62 'cubic units'. We can't say what the dimensions of the small boxes are without more information. 
April 28th, 2016, 02:45 AM  #4 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Although it is true that there are probably multiple answers, there are two extra constraints... a "tightlypacked" box requires that the cubical boxes be placed next to each other so that they tesselate. This, together with the constraint of "no free space", means that the each dimension of the small box must be a factor of the corresponding dimension of the large box. This can constrain the number of possible dimensions of the small box considerably. So... let L, W and H be the three dimensions of the large box and l, w and h be the dimensions of the small box. We have $\displaystyle L = 38 = al$ $\displaystyle W = 27 = bw$ $\displaystyle H = 37 = ch$ where a, b and c are integers. Then $\displaystyle LWH = 100 lwh = abc lwh$ so $\displaystyle abc=100$ So... we pick any integers a, b and c whose product gives 100 and then solve each equation above to give each individual dimension. We can perform prime factor decomposition on 100 to give $\displaystyle 100 = 2 \times 2 \times 5 \times 5$ We only have four numbers here and we need to populate three integers, so one number (a, b or c) must be the product of 2 individual prime factors, giving three possible combinations: $\displaystyle 2 \times 2$, $\displaystyle 2 \times 5$ or $\displaystyle 5 \times 5$ (or, 4, 10 and 25) So the possible choices of a, b and c are: $\displaystyle a = 2, b = 2, c = 25$ $\displaystyle a = 2, b = 5, c = 10$ $\displaystyle a = 5, b = 5, c = 4$ not including the combinations where we just shuffle the numbers around between a, b and c. If we allow that, we get $\displaystyle a = 2, b = 2, c = 25$ $\displaystyle a = 2, b = 25, c = 2$ $\displaystyle a = 25, b = 2, c = 2$ $\displaystyle a = 2, b = 10, c = 5$ $\displaystyle a = 2, b = 5, c = 10$ $\displaystyle a = 5, b = 2, c = 10$ $\displaystyle a = 5, b = 10, c = 2$ $\displaystyle a = 10, b = 2, c = 5$ $\displaystyle a = 10, b = 5, c = 2$ $\displaystyle a = 4, b = 5, c = 5$ $\displaystyle a = 5, b = 4, c = 5$ $\displaystyle a = 5, b = 5, c = 4$ So, the possible dimensions of the small box are $\displaystyle l = 19, w = 13.5, h=1.48$ $\displaystyle l = 19, w = 1.08, h=18.5$ $\displaystyle l = 0.76, w = 13.5, h=18.5$ $\displaystyle l = 19, w = 2.7, h=7.4$ $\displaystyle l = 19, w = 5.4, h=3.7$ $\displaystyle l = 7.6, w = 13.5, h=3.7$ $\displaystyle l = 7.6, w = 2.7, h=18.5$ $\displaystyle l = 3.8, w = 13.5, h=7.4$ $\displaystyle l = 3.8, w = 5.4, h=18.5$ $\displaystyle l = 9.5, w = 5.4, h=7.4$ $\displaystyle l = 7.6, w = 6.75, h=7.4$ $\displaystyle l = 7.6, w = 5.4, h=9.25$ 12 possible combinations... not that many actually! Just so you know for the future... it can be worth doing the above exercise each time you have this sort of question because if the prime factor decomposition of the number of boxes gave you three integers (rather than four in the above example), that would mean much fewer combinations (maybe even 1 possible combination if you're lucky to get a prime cubed as the number of small boxes in the large box) Last edited by Benit13; April 28th, 2016 at 02:52 AM. 
April 28th, 2016, 03:06 AM  #5 
Senior Member Joined: Apr 2014 From: UK Posts: 903 Thanks: 331 
Lets say I have a box which is 8x8x8 and it contains 8 boxes (an exact cube amount), how many combinations are there? There is more than 1 solution. Here's 3 solutions: 4x4x4, 8x4x2, 8x8x1. You combinations don't seem to account for solutions where some of the boxes are orientated differently to fill the space. 
April 28th, 2016, 03:19 AM  #6 
Senior Member Joined: Apr 2014 From: UK Posts: 903 Thanks: 331 
There are more than 12 solutions even without changing orientation, 38x27x0.37 springs to mind.


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