What is the maximum volume of a set $M$ contained in $[0,6]^2$ such that $M\cap(M+(1,2))=\emptyset$ where $M+(1,2)$ stands for $\{x+(1,2) | x\in M\}$ ?
The question was given in a basic measure theory course so the answer should rely on related techniques. I would appreciate any suggestions.
Thanks!
We may think of the square $[0,6]^2$ as being a collection of disjoint blocks of closed sets $[a,b]\times [c,d]$ and choose which of them will lie in $M$. Other partitions will at best lead to the same maximum volume, and at worst, lead us to a smaller volume. (This can be seen from the definition of Lebesgue measure: the infemum of the sum of the measures of closed sets that cover $M$.)
By the nature of the problem, the partition with the most convenient blocks have blocks of size 1 by 2. We then see easily that a best choice for $M$ is $M = \big([0,6]\times[0,2]\big)\cup\big([0,1]\times[0,6]\big)\cup\big([2,6]\times[4,6]\big)$, which has measure $24$.