Two Cubes Calendar

It's done with two blocks with the numbers $(0,1,2,3,4,5)$ and $(0,1,2,6,7,8)$ , where $6$ is used as a $9$ . The total is $39$ , not $48$ .

Otherwise, how do you display $(06, 07, 08, 09)$ ?

The total number of edges is $12$ so we need every numeral from $0$ to $9$ at least once. Moreover, it should be possible to make numbers $11$ and $22$ which requires at least one additional $1$ and at least one additional $2$ . However, we will not be able to make all of the dates from $01$ to $09$ because the another cube has only $1$ edges. So, we need to use $6$ as $9$ and have $0$ on the both cubes. Therefore, the necessary numbers are $(0,1,2,?,?,?)$ and $(0,1,2,?,?,?)$ where $?$ is a non-repeated element of the set $(3,4,5,6,7,8)$ . Therefore the minimal sum is

$0+1+2+0+1+2+3+4+5+6+7+8=39$

With addition of $9$ it is the answer $48$

The highest possible sum is bigger by $3$ when you consider to count $6$ as $9$