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how many product can fit inside one box? and how to find box cubic feet

Product size $= 6 \times 5$ Inches

Box size $= 10 \times 10$ inches

here's what I did so far:

$6 \times 5 = 30$ (product volume)

$10 \times 10 = 100$ (box volume)

$100 / 30 = 3$

so $3$ products can fit in this box (assume product is liquid) but if you take a look below, product is solid so only 2 can fit. i can see this in small numbers by drawing it out but how to put this in calculation, that shape is solid?

enter image description here

Star Gates
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  • Your computation is not useless: it gives a "great maximum number" of objects one can place... But I think you should ask your instructor to ask you first the same problem in 2D by suppressing the third dimension with a $10 cm \times 10 cm$ and items $6cm \times 3cm$ (sorry for the centimeters, but I am not used to feet and inches) to see really where the problem is... – Jean Marie May 27 '21 at 21:10
  • Can I ask you your level of study ? – Jean Marie May 27 '21 at 21:15
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    o this isnt for school or anything. i have a item and a box so i was just wondering. I think i did it correct but I wasn't how to get the cubic feet. i think i have to divide the volume by some number – Star Gates May 27 '21 at 21:26
  • With my example of a $10 \times 10$ box with, say objects of size $6 \times 5$, you can theoretically place 100/30=3,33 such objects that we will round to 3 objects, but in fact you can place only 2 objects ! Try it... – Jean Marie May 27 '21 at 21:32
  • i am following you until 3 object able to fit in 10x10 box. that make sense but how in fact 2 object can fit? – Star Gates May 27 '21 at 21:40
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    If, for example you place the two objects along their longest size $6$, they fit, it is like having a rectangle $10 \times 6$. What is the remaining place ? A rectangle $4 \times 10$. You cannot enter a third object in this place (although, $4 \times 10$ is greater than $6 \times 5$). And you can try in many other ways, it will never work... – Jean Marie May 27 '21 at 21:45
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    aw i see it after drawing it out – Star Gates May 27 '21 at 21:52
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    You haven't made a mistake, it is what I have said you: you have like this a maximum number, but you are reasoning on quantities that could be thought as liquids. You do not take into account the shape of the boxes. If for example for the same surface, you have had objects $3 \times 10$, you could accommodate 3 of them in your $10 \times 10$ box ! – Jean Marie May 27 '21 at 21:57
  • i got it now. last question, do you if there is any way to put shape is sold and not liquid in calculation? because for this example i can draw it out and see it but for large numbers and 3d shapes, i would never had guessed it – Star Gates May 27 '21 at 22:03
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    In general, no systematic way for humans... just trial and error. It is the domain of computer programs optimizing occupancy. – Jean Marie May 27 '21 at 22:06
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    @Jeans got it. thanks a lot for helping me out with this. btw some one need to come up a some formula for this – Star Gates May 27 '21 at 22:11
  • https://math.stackexchange.com/questions/1314167/how-many-rectangles-or-squares-of-fixed-and-equal-sizes-can-fit-inside-a-squar might be of interest. – Barry Cipra May 27 '21 at 23:36
  • See also https://math.stackexchange.com/questions/99796/given-the-dimensions-of-two-rectangles-find-how-many-smaller-rectangles-fit-the (especially a link in the answer there). – Barry Cipra May 27 '21 at 23:42

1 Answers1

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A little late probably but one can easily solve that by dividing all the dimensions of the bigger box(W, H) seperately through the corresponding dimension of the smaller box(w, h)after that round it down and then multiply the results.

(note: you eventually have to sort the dimensions)

In this case: W = 10, H = 10, w = 6, h = 5

fitW = floor(W / w) = 1

fitH = floor(H / h) = 2

result = fitW * fitH = 2 * 1 = 2

this works analogue for 3 dimensional boxes

Dabso
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  • This assumes the smaller boxes are all oriented the same way. You can for example fit four 2x1 boxes inside a 3x3 box, not just three. It gives the right answer for the case in the OP though, because reorienting a small box does not help there. – Jaap Scherphuis Oct 07 '22 at 09:58