Consider an infinite chessboard. Each square contains either a 1 or an X in some pattern. (X can be any real number but for a given board, all the X's are the same.)

Each square with an X on it has weight equal to zero.

Each square with a 1 on it has a weight of 1 + N*X where N is the total number of X's on the 8 surrounding squares.

For a given value of X, find a way of tiling the board with the highest average weight per square.

*Inspired by various Tower Defense games.*

Well, thanks for the push, Jer.

I don't know if this is what you had in mind, but

..1 1 1 1..

..1 X 1 X..

..1 1 1 1..

..1 X 1 X..

Has an average weight of (8x+3)/4

This is better than 1 if X > 1/8.

This is better than (6x + 1)/2 if X < 1/2

So, the revised Final Solution (??), stated without confidence that there are no further refinements available, is

X <= 1/8 -- Solid 1's = average weight of 1

X between 1/8 and 1/2 -- Plaid 1's (see above) = average weight of 3/4 + 2X

X >= 1/2 Solid alternating rows of 1's and X's = avg weight 1/2 + 3X