How to place the biggest number of cans in this box

Question

A rectangular box has dimensions of 108 cm x 144 cm for its bottom. I want to place the biggest number of cans of 12 cm of diameter in it. How can I place this biggest number?

It couldn't be as simple as 9 boxes x 12 boxes, isn't it?

Answer 1

You are talking about packing circles in 2D. The most efficient method is to use a hexagonal lattice instead of a cubic lattice. Try packing your cans so there centres form equilateral triangles.

Answer 2

There would be the "obvious" rectangular lay out. 9x12 = 108 cans. the other way to go would be to pack them "hexagonally", that is, fill in one row along the edge of the box and put the next row in in the divot between the cans. You have more wasted space between by the walls, but you should be able to fit an additional row of cans. 9+8+9+8+9+8+9+8+9+8+9+8+9 = 111 cans.

Asked for more input and realized I have overlooked possibly going 12+11+12+11+12+11+12+11+12+11 = 115

When you box the cans hexagonally, what is the height of the second row of cans? R (1+sin 60) and the $n^{th}$ row? R(1+(n-1) sin 60)

If (1+9 sin 60) < 10 then you can fit in one more row going the long way down the box.

If (1+12 sin 60) < 12 you can fit one more row going the short way across the box.

And if (1+13 sin 60) < 12 you could fit two more rows going the short way across the box.