To maximize the weight of a flour sack, minimize sugar weight. Let S be sugar weight, F be flour weight. We have two inequalities: 3S + 4F ≤ 50 and 2F ≤ 13 + 3S. Solving, we find the maximum F is 11.25 pounds.
We're given two inequalities that constrain the weights of sugar (S) and flour (F):
Three sacks of sugar and four sacks of flour weigh no more than 50 pounds: 3S + 4F ≤ 50.
Two sacks of flour weigh no more than 13 pounds more than three sacks of sugar: 2F ≤ 13 + 3S.
To maximize the weight of a flour sack (F), we want to minimize the weight of a sugar sack (S) while still satisfying the inequalities.
Here's how we can solve for the maximum F:
- Isolate F: 2F ≤ 13 + 3S --> F ≤ 6.5 + 1.5S.
- Replace F with the maximum allowed value: 3S + 4(6.5 + 1.5S) ≤ 50.
- Combine terms and solve for S: 3S + 26 + 6S ≤ 50 --> 9S ≤ 24 --> S ≤ 2.67 (rounded to two decimal places).
- Using the minimum S value: F ≤ 6.5 + 1.5S ≤ 6.5 + 1.5 * 2.67 ≈ 11.25 pounds.
Therefore, the largest possible weight of a sack of flour is 11.25 pounds.