maximizing number of 4s times number of 7s in decimal representation $F_4(X)$ be the number of digits 4 in the decimal representation of $X$, and $F_7(X)$ be the number of digits 7 in the decimal representation of $X$.  We have to find  largest product $F_4(X)\cdot F_7(X)$, where $L \leq X \leq R$. 
$$\max\{F_4(X)\cdot F_7(X) : L ≤ X ≤ R\}$$
can a general solution be acheived?
eg: 
$L=47$ AND $R=74$
$$\max\{F_4(X)\cdot F_7(X)\}=1$$

For any range with a fixed number of unrestricted digits like $0 \le X \le 99999999$, the problem is trivial: since we can choose any 8 digits, it is easy to maximize $F_4(X) \cdot F_7(X)$ by choosing four 4's and four 7's.
Even if some initial digits are unchangeable, as in $4440000 \le X \le 4449999$, the problem is very simple.  Here, the first 3 digits are fixed, but the last 4 digits may be chosen at will, and it is easy to see that we should additionally choose one/zero 4's and three/four 7's.
Finally, note that any range $L \le X \le R$ can be decomposed into $O(\log R)$ separate intervals of the type in the previous paragraph.  For instance, if $47 \le X \le 74$, then $X$ fits one of the following templates: 47, 48, 49, 5x, 6x, 70, 71, 72, 73, 74.  Since we can maximize over each one, we can maximize over their union.
Of course some optimizations are possible, but this is already an linear-time algorithm in the number of digits of $L$ and $R$.  One could even use this technique to compute things like $\sum\limits_{L \le X \le R} F_4(X) \cdot F_7(X)$.
