Fill The Bags

Let the two bags are denoted by set $A$ and $B$ , where the sum of its member is at 10 the maximum.

Since there are two 6kg items, to maximize it, we have to first pick them into different bags, which means $A = \{6\}$ and $B = \{6\}$ . This will leave the remaining bag to contains space for 4kg each, precluding the possibility to select a 5kg items. Since the remaining items is at 1kg, 2kg and 4kg, we can split them into to bags as $A = \{6, 2, 1\}$ and $B = \{6, 4\}$ (the bag may be swapped). As they are both less than 10kg for each bag, this produces the maximum weight of 19kg for both bags combined.

It is noted that this solution omits the possibility of selecting 6kg and 5kg into different bags (or $A = \{6\}$ and $B = \{5\}$ ). If we follow this approach, another 6kg items cannot be fitted into any bag, and although we can put a 1kg, 2kg and 4kg into the bag (with various combination, such as $A = \{6, 1, 2\}$ and $B = \{5, 4\}$ or $A = \{6, 4\}$ and $B = \{5, 1, 2\}$ ), its sum will be 18kg combined, less than the solution above.