Partition a square with convex polygons. What is the maximal number of edges? Given a square that is partitioned into convex polygons such that $n$ regions are created. What is the maxmimal number of edges in such a partition?
Example. 
The two squares below both have $n=4$ regions, but the left one has 8 edges and the other one has 12 edges. So in this case 12 edges is maximal.

I do have have an observation: every face is bounded by at least 3 edges. However, how do I move on from there?
 A: If there are $n$ convex regions inside the square, there are $n+1$ regions total counting the outside region (which is not convex). 
Except for the four corners of the square, each vertex must have degree at least $3$: a vertex of degree $2$ subdivides an edge between two regions, and if they're both convex, then the boundary between then can only be a straight line (so the vertex wouldn't exist).
So if there are $v$ vertices, then the degree sum is at least $3v - 4$, so there are at least $\frac32v - 2$ edges.
By Euler's formula, $v - e + f = 2$, so $v = e-f+2 = e-(n+1)+2 = e-n+1$. Substituting this into $e \ge \frac32v - 2$ yields
$$
    e \ge \frac32(e-n+1)-2 \implies e \le 3n+1.
$$
So there can be at most $3n+1$ edges.
One way to achieve this is to put down $n-1$ vertical edges, separating the square into $n$ thin strips. Then the top and bottom side of the square are made up of $n$ edges each, and the left and right sides are also each one edge, giving us $(n-1)+2n+2 = 3n+1$ edges.
