The problem is to pack $n$ non-overlapping disks (not necessarily of the same size) of greatest total area into a unit square. The case $n=1$ is obvious: just place a disk of radius $\frac12$ centrally in the square, to occupy an area $\frac14\pi\approx78.54\%$ of the square. For $n=2$, we can do no better than follow up the previous case and add a disk (of radius $\frac32-\surd2$) in a corner to touch the existing disk and two sides of the square, thus altogether covering an area $(\frac92-3\surd2)\pi\approx80.85\%$ of the square. Similarly, for $n=3,4,5$, we can do no better than place further disks of that size in the remaining corners of the square.
For $n=6,...,13$, it seems optimal to build on the case $n=5$ and place further disks in the eight regions bounded by a side of the square, the disk of radius $\frac12$, and a disk of radius $\frac32-\surd2$. Clearly, we can keep going like this, packing the next disk into whatever remaining space will accommodate the largest disk—the greedy method. But the question arises:
Is there an $n$ for which the packing of maximal total area is not the greedy packing described above?
