First show that as much as possible of the highest value/pound item must be
included in the optimal solution.
Let wh, vh be the weight available and value of the item with the highest
value/pound ratio (item h).
Let L(i) be the weight of item i contained in the thief's
If some of item h is left, and L(j) 0 for some j h, then
replacing j with h will yield a higher value.