I know that I should use some kind of honeycomb structure but can't work out in which orientation I should arrange it. I've looked at a few websites now and although I have a slightly better idea of the options, I can't work out how to model the problem and therefore calculate how many circles I could cut.
Specifically I would like to know:
~ How could I model this and what mathematics are involved?
~ What is the maximum number of 51 mm-diameter circles I can cut from a 330 mm × 530 mm rectangle?
~ What is the minimum size rectangle from which I could cut 16 circles?
(As you might suspect this is a real-life problem that I must solve, the disks that I will cut are to be used in a physics experiment but the material from which they are made is very expensive. It can however be purchased in any size rectangle, up to 330 mm × 530 mm.)
EDIT
OK so I just discovered this question and the Wikipedia link contained therein. Whilst it is certainly related I am no closer to solving my current queries. (Other than if I were to order a square sheet of the material measuring 204 mm × 204 mm but I'm sure a rectangle would be more efficient.)
For your problem, one choice is to order a rectangle with both dimensions a multiple of 51 mm, use a square pack, and get a density of $\frac{\pi}{4}\approx 0.785$
Another alternative is to use a hexagonal pack. If you have $k$ rows alternating $n$ and $n-1$ you want a sheet $51n \times (1+\frac{k \sqrt{3}}{2})51\ \ $mm, which packs $\lfloor n(k-\frac{1}{2})\rfloor$ circles. For your case, $n=6, k=12$ will fit $66$ circles in $306 \times 537$ mm, with a packing density of about $0.8205$
For $16$ circles, you could also use a $4 \times 4$ hexagonal pack. This would require $230=4.5\cdot 51 \times 184\ \ $mm, giving a density of $0.772$, so you could just buy $204 \times 204\ \ $mm and be better off.