can anyone please help me set this problem up, I can do the math but I have tried for hours...
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?
current rate = $300
number of members = 50
let the number of additonal member be n
so actual number of members = n+50
cost per member = 300 - 2n
Revenue = number of members x membership fee
= (n+50)(300 - 2n)
= 300n - 2n^2 + 15000- 100n
= -2n^2 + 200n + 15000
= -2(n^2 - 100n + 2500 - 2500 ) + 15000
= -2( (n-50)^2 - 2500) + 15000
= -2(n-50)^2 + 5000 + 15000
= -2(n-50)^2 + 20000
This is a parabola opening downward with a vertex of (50,20000)
So the maximum of Revenue is $20,000 when n = 50
So 50 addition to the original 50 should join.
The maximum revenue is obtained when 100 members are in.
check:
at 100 members cost is 300 - 2(50) = 200
revenue is 100(200) = 20000
at 99 members cost per member is 300-2(49) = 202
revenue is 99(202) = 19998 which is less than 20000
at 101 members cost per member is 300-2(51) = 198
revenue is 101(198) = 19998 , which is again less than 20000
My answer is correct.
To set up the problem, let's define the variables:
x = Number of memberships
R = Revenue generated
Given information:
- Membership cost = $300
- At least 50 people must join
- Fare reduction = $2 per additional member over 50
- Maximum number of passengers allowed = 125
To determine the revenue generated, we need to consider two scenarios:
1. When the number of memberships is less than or equal to 50
In this case, the revenue is calculated by multiplying the number of memberships by the membership cost:
R = 300x
2. When the number of memberships is greater than 50
In this case, the revenue is calculated by subtracting the fare reduction from the membership cost for each additional member over 50, and then multiplying it by the total number of additional members:
R = (300 - 2(x - 50)) * 50
To find the maximum revenue, we need to consider the limit of passengers. The total number of passengers is the sum of the first 50 members and the additional members allowed (min of x - 50 or 75). Therefore, the total number of passengers is given by:
Total number of passengers = 50 + min(x - 50, 75)
Since the maximum number of passengers allowed is 125, we have the following constraint:
Total number of passengers ≤ 125
Now we can set up the problem using the above equations and the constraint, and solve for x to find the optimal number of memberships.
To solve this problem, we need to find the number of memberships that will maximize the revenue for Air Fare Club. To do this, we'll need to set up an equation and find the maximum point.
Let's start by breaking down the information given:
- Membership fee: $300.00
- Minimum number of people required to join: 50
- Fare reduction per member over 50: $2
- Maximum number of passengers allowed: 125
Now, let's set up the equation to represent the revenue based on the number of memberships. We can assume that the number of memberships is represented by 'x'.
Revenue = (300 - 2(x - 50)) * min(x, 125)
The first part of the equation represents the fare after reductions, which decreases by $2 for each additional member over 50. The second part of the equation represents the number of passengers, which is capped at 125. By multiplying these two parts, we get the total revenue generated.
To find the number of memberships that will maximize the revenue, we need to find the value of 'x' that gives us the highest revenue. We can do this by finding the maximum point of the revenue function.
One way to find the maximum point is to take the first derivative of the revenue function and set it equal to zero. Then, solving for 'x' will give us the number of memberships that maximize the revenue.
However, since you mentioned you can do the math, I'll leave the differentiation and finding the maximum to you. Simply differentiate the revenue function with respect to 'x', set it equal to zero, and solve for 'x'. The value of 'x' obtained will be the number of memberships that maximize the revenue.