An agency charges $15 per person for a trip for groups of no less than 30 people. But for each person above the 30, the charge (for everyone!) will be reduced by $0.15. Write a revenue function for the agency usin x as number of people above 30. What size group will maximize the total revenue for the agency if the trip is limited to at most 50 people?
I will be happy to critique your thinking.
Since the agency only takes groups of no less than 30 people, the minimum profit for a single trip is $15*30 people.
Let x be the number of people above 30.
The number of people is then 30+x.
The [Total Revenue] is equal to the
[price charged for each person] times the [number of people]
[price charged for each]=$15-.15*x
Notice if x=0, then the price charged is equal to $15 and the number of people in group is 30, which is the minum group size the company will allow for a trip.
Thus our Total Revenue Function is
[Total Revenue] =
[price charged for each person]*[number of people]
=($15-$.15*x)*(30+x)
R(x)= ($15-$.15*x)*(30+x)=450+10.5x-.15*x^2 after foiling everything out.
R'(x)=10.5-2*.15*x is the derivative of the Revenue function and the Revenue will be maximized at a value of x for which R'(x)=0.
Setting R'(x)=0 implies 10.5-2*.15*x=0, which implies that x=35 people. The group size was constrained to be at least 30 people and at most 50 people, and since the number 35 is between 30 and 50, then a group size of 35 people is a valid solution for this problem: according to the revenue function R(x),and the constraints on groups sizes given above above, the company will maximize its profit when it has groups of size 35. Now I must go breath.
To write the revenue function for the agency, we need to consider two scenarios:
1. When the number of people above 30 (x) is 0.
2. When the number of people above 30 (x) is greater than 0.
Let's break it down step-by-step:
1. When x = 0:
- The number of people in the group is 30.
- The charge per person is $15.
- Therefore, the total revenue is 30 * $15 = $450.
2. When x > 0:
- The number of people in the group is 30 + x.
- The charge per person is $15 - $0.15x (since the charge per person reduces by $0.15 for each person above 30).
- Therefore, the total revenue is (30 + x) * ($15 - $0.15x).
Now, to find the size of the group that maximizes the total revenue, we can differentiate the revenue function with respect to x, set it equal to 0, and solve for x.
Revenue function:
R(x) = (30 + x) * (15 - 0.15x)
Differentiating R(x) with respect to x:
R'(x) = 15 - 0.15x + (30 + x) * (-0.15)
Setting R'(x) equal to 0:
15 - 0.15x - 0.15(30 + x) = 0
15 - 0.15x - 4.5 - 0.15x = 0
-0.3x - 4.5 = -15
-0.3x = -10.5
x = -10.5 / -0.3
x ≈ 35
Since x represents the number of people above 30, we can conclude that the group size that maximizes the total revenue for the agency is approximately 30 + 35 = 65 people.
However, since the trip is limited to at most 50 people, we need to consider two cases:
1. When x ≤ 20 (to make sure the total group size, including the initial 30 people, doesn't exceed 50).
2. When x > 20 (to ensure the group size reaches its maximum value).
For the first case, we can substitute x = 0 into the revenue function to find the total revenue:
R(0) = (30 + 0) * (15 - 0.15 * 0)
R(0) = 30 * 15
R(0) = $450
For the second case, we can substitute x = 20 into the revenue function:
R(20) = (30 + 20) * (15 - 0.15 * 20)
R(20) = 50 * 12
R(20) = $600
Therefore, when the trip is limited to at most 50 people, the group size that maximizes the total revenue for the agency is 50 people, and the maximum revenue achievable in this case is $600.