air fare club offers membership at $300.00, at least 50 people must join.For every member over 50 their fare will be reduced by $2 for every member. Due to space limits only 125 passengers will be allowed. how many memberships will maximize the revenue?

done

http://www.jiskha.com/display.cgi?id=1350928794

revenue = #members * rate

y = x(300-2(x-50)) = 2x(200-x)

max value occurs midway between the roots (0,200) at x=100

Thank you, spent way too much time on this one problem, you all are great :)

To find the number of memberships that will maximize revenue, we need to consider two factors: the revenue generated by membership fees and the revenue generated by fare reductions.

Let's break down the problem step by step:

1. Calculate the revenue generated by membership fees:
The membership fee is $300 per person, and at least 50 people must join. Therefore, the minimum revenue from membership fees is 50 * $300 = $15,000.

2. Determine the number of passengers eligible for fare reduction:
Since the maximum capacity is 125 passengers, the number of passengers eligible for fare reduction will be 125 - 50 = 75 passengers.

3. Calculate the fare reduction for each member over 50:
The fare reduction is $2 for every member over 50. Therefore, for each member over 50, the fare reduction applied to the membership fee would be $2 * (number of members over 50).

4. Calculate the total revenue generated by fare reductions:
Since there are 75 passengers eligible for fare reduction, the maximum fare reduction that can be applied is $2 * 75 = $150.

5. Determine the number of memberships that will maximize revenue:
To maximize revenue, we need to find the number of memberships that balances the revenue from membership fees and the revenue from fare reductions. In this case, the revenue from membership fees is fixed at $15,000, and the revenue from fare reductions has a maximum potential of $150. Therefore, the number of memberships that will maximize revenue is 50 (the minimum required memberships) + 75 (the maximum number of passengers eligible for fare reduction) = 125 memberships.

Thus, the maximum revenue will be achieved by having 125 memberships.