1. The problem asks us to find the
maximum amount of adhesive that can be made as
mixture A and
mixture B, separately.
2. To get the maximum amount of a mixture, either resin or hardener must be fully used up (could be both). Then, we have to consider two cases for each mixture.
3. Let's denote the ratio
R :
H as the ratio of
resin to
hardener (both in grams).
4. For mixture A we have
3 : 2.
Case#1: All resin is used up. Then, the ratio is 6 : 4 and 4 \( \leq \) 8. It
works and the total mass is 6 + 4 =
10 grams.
Case#2: All hardener is used up. Then, the ratio is 12 : 8 and 12 \( \geq \) 6. It
doesn't work since we use more resin than we have.
5. For mixture B we have
2 : 4.
Case#1: All resin is used up. Then, the ratio is 6 : 12 and 12 \( \geq \) 8. It
doesn't work since we use more hardener than we have.
Case#2: All hardener is used up. Then, the ratio is 4 : 8 and 4 \( \leq \) 6. It
works and the total mass is 4 + 8 =
12 grams.
6. The answer is
10 grams of mixture A and
12 grams of mixture B.
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There is a faster way by noticing that if the ratio of a mixture is less than 6 : 8, then the hardener will be used all the way without overusing the resin. Similarly, if the ratio of a mixture is more than 6 : 8, then the resin will be used all the way without overusing the hardener. This allows to only have 1 case for each mixture.