The Answer is $\boxed{49}$ But How can I prove this? :/ $A={2,4,7,8,11,14,16}$ $B={3,5,9,13,15,17,19}$

We have 8 prime numbers and to get the maximum possible value of this multiplication we have to have equal or close to equal number of integers in both $A$ and $B$. Writing $2$ and $3$ in different pairs gives us $4$ powers of $2$ in $A$ and $2$ power of $3$ in $B$. As we have a smaller number of integers in $B$ we have to have $5$ in $B$ to get $3 \cdot 5 = 15$ in it. Then having $7$ in $A$ would give us a greater value of our multiplication. Cause we will have $7 \cdot 2 = 14$ in $A$. ( we cannot have it in $B$ as it wouldn't give us any multiple of $7$). In this case we have $6$ elements in $A$ and $4$ in $B$ and $4$ prime numbers left. As we said before we have to have equal number of integers in both $A$ and $B$. So we give $1$ prime to $A$ and other $3$ to $B$. And we get $7 \cdot 7 = \boxed{49}$

Am I missing something or shouldn't it be $5 \cdot 13 = 65$? Just use: A = 7, 11, 13, 17, 19 and B = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20.