Carcass wrote:
OE
We are trying to arrive at 140 mg of iron and 160 mg of calcium efficiently, so we might as well not consume any of Tablet Y, which is highest in zinc, the one element we have no interest in. Notice that 3 of Tablet X and 2 of Tablet Y would contain 3 × 40 + 2 × 20 = 120 + 40 = 160 mg of calcium exactly. This same combination contains 3 × 17 + 2 × 45 = 51 + 90 = 141 mg of iron, which is greater than 140, as required. This combination of 3 + 2 = 5 tablets achieves the desired quantities with very little wastage. No combination of four tablets will have sufficient amounts of both iron and calcium, so 5 is the minimum