The marching band is holding a fundraiser. The band is selling t-shirts for $20 and yearbooks for $24. The goal is to sell at least $2,400 in merchandise. Which of the following is a solution to this scenario

Answer 1

There are multiple solutions to this scenario, but one possible solution is:

- Let x be the number of t-shirts sold.
- Let y be the number of yearbooks sold.
- The total revenue generated can be expressed as 20x + 24y.
- The goal is to sell at least $2,400, so we have the inequality 20x + 24y ≥ 2400.
- We want to maximize the revenue, so we want to find the values of x and y that satisfy the inequality and make 20x + 24y as large as possible.
- One way to do this is to use linear programming or graphing techniques, but for simplicity we can try a few values for x and y and see which ones work. Starting with x = 0 and y = 100 (selling 100 yearbooks only), we get 24y = 2400, which satisfies the inequality. However, the revenue is only 24(100) = $2,400, which is the minimum target. We want to do better.
- Let's try increasing the number of t-shirts sold while keeping the number of yearbooks sold the same. If we sell x = 50 t-shirts and y = 100 yearbooks, we get 20x + 24y = 20(50) + 24(100) = $3,000, which is higher than the target. This is a viable solution.
- We could also try selling more t-shirts and fewer yearbooks. For example, if we sell x = 120 t-shirts and y = 50 yearbooks, we get 20x + 24y = 20(120) + 24(50) = $3,240, which is higher than the target. This is another viable solution.
- In fact, there are many other combinations of x and y that would work, such as x = 60 and y = 90, x = 80 and y = 75, x = 100 and y = 62, etc. The key is to find values of x and y that satisfy the inequality and maximize the revenue.
- Therefore, one possible solution to the scenario is to sell 50 t-shirts and 100 yearbooks, or to sell 120 t-shirts and 50 yearbooks (or any other combination that meets the target and maximizes the revenue).