The biggest little polygon is the n-gon with maximum diameter 1 that has the greatest volume.

I've often wondered the same problem about polyhedra, and posted a biggest little polyhedron question on StackExchange, where it got no answers.

Mathematica 10 has Volume[] and ConvexHullMesh[]. Calculating the volume of the convex hull of 20 random points is as simple as

Volume[ConvexHullMesh[RandomReal[{-1, 1}, {20, 3}]]]

so I decided to solve the problem myself. First, I used some code like the below to find the solution randomly, with a bit of anealling.

numpts = 5; maxvol = .05; count = 0; div = 1; pts = RandomReal[{-1, 1}, {numpts, 3}]; Monitor[Do[ kk = RandomReal[{-1, 1}, {numpts, 3}]/div + pts; maxlen = Max[EuclideanDistance[#[[1]], #[[2]]] & /@ Subsets[kk, {2}]]; vol = Volume[ConvexHullMesh[kk/maxlen]]; If[vol > maxvol, count = 0; maxvol = vol; pts = kk/maxlen, count = count + 1]; If[count > 1000, count = 0; div = 1.3 div; Print[{maxvol, pts}]], {gg, 1, 100000}], {count, maxvol, pts}]

The best solution for 4 points is the regular tetrahedron with edge lengths 1. The best solutions I could find for 5 through 12 points were considerably stranger. Red lines indicate a distance of 1. Volumes appear underneath.

Most of these solutions could doubtlessly be improved or made more exact. If anyone can provide any improvements, please post them.