A theater is presenting a program on drinking and driving for students and their parents. The proceeds will be donated to a local alcohol information center. Admission is $18 for parents and $9 for students. However, this situation has two constraints: The theater can hold no more than 120 people and every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?
number of adults --- x
number of students ---y
condition#1: x+y ≤ 120
condition #2 : 2y ≥ x ----> y ≥ (1/2)x
Revenue equation:
R = 18x + 9y
I can "drag" this line out as far as the intersection of
y = (1/2)x and x+y = 120
then (1/2)x = 120-x
x = 240 - 2x
3x=240
x=80
then y = 120-80 = 40
so 80 parents and 40 students should attend for a max revenue of 18(80) + 9(40) or $1800
To find the optimal number of parents and students that should attend to raise the maximum amount of money, we can approach this problem step by step.
Let's assume:
Number of parents attending: P
Number of students attending: S
According to the given constraints, we can determine the following relationships:
1. Number of People attending: P + S ≤ 120
2. Every two parents must bring at least one student:
P ≥ 2S
3. Admission Costs:
Total Revenue = (Admission Price for Parents * Number of Parents) + (Admission Price for Students * Number of Students)
= (18 * P) + (9 * S)
To maximize the total revenue, we need to maximize the objective function: Total Revenue.
Let's solve the problem step by step using the given information:
Step 1: From constraint 2, we have P ≥ 2S. Since we need to find the maximum number of parents and students, let's consider P = 2S.
Step 2: Substitute P = 2S in constraint 1: 3S ≤ 120 (since P + S = 3S)
=> S ≤ 40
Step 3: As we want to maximize the revenue, let's determine the total revenue function:
Total Revenue = (18 * P) + (9 * S)
Substitute P = 2S in the total revenue function:
Total Revenue = (18 * 2S) + (9 * S)
= 36S + 9S
= 45S
Step 4: As per constraint 2, P ≥ 2S. Substituting P = 2S, we have:
P ≥ 2S
Substitute P = 2S in constraint 1: 3S ≤ 120 (since P + S = 3S)
So, S ≤ 40
Step 5: To find the maximum value for S that satisfies the constraints, substitute S = 40 in the total revenue function:
Total Revenue = 45S
= 45 * 40
= 1800
Step 6: Using the value of S = 40, substitute it in constraint 2: P ≥ 2S
P ≥ 2 * 40
P ≥ 80
So, to maximize the total revenue and satisfy the given constraints, the number of parents (P) should be at least 80, and the number of students (S) should be 40.
Therefore, the optimal number of parents and students to attend is:
Number of parents (P) = 80
Number of students (S) = 40
To determine the number of parents and students needed to raise the maximum amount of money, we can approach the problem using algebra and logical reasoning.
Let's assume that the number of parents is "P", and the number of students is "S". We need to find the values for P and S that satisfy the given constraints and maximize the amount of money raised.
Constraint 1: The theater can hold no more than 120 people.
This means the total number of attendees (parents and students combined) cannot exceed 120.
Mathematically, this can be represented as: P + S ≤ 120.
Constraint 2: Every two parents must bring at least one student.
This implies that the number of parents must be at least half the number of students, in order to meet this requirement.
Mathematically, this can be presented as: P ≥ S/2.
Now, we need to maximize the amount of money raised. Since the admission fee for parents is $18 and for students is $9, the total amount of money raised can be calculated as: Total Money = 18P + 9S.
To solve this problem and find the values for P and S, we can use a technique called "Linear Programming." We can graphically represent the solution space and find the optimal solution by identifying the corner points of the feasible region.
However, let's simplify this problem further without using graphical representation.
The second constraint tells us that every two parents must bring at least one student. It means that for each pair of parents (2 parents), there must be at least one student. Therefore, the number of students should be at least half the number of parents.
To maximize the amount of money raised, it is ideal to have the highest number of parents and students while still satisfying the constraints.
Based on this, we can start by assuming the maximum number of parents and calculate the corresponding number of students.
Let's assume P = 120 (maximum number of parents).
Using the second constraint, S = P/2 = 120/2 = 60 (number of students).
Now, let's calculate the total amount of money raised:
Total Money = 18P + 9S
= 18(120) + 9(60)
= 2160 + 540
= 2700.
Therefore, to raise the maximum amount of money, the theater should have 120 parents and 60 students in attendance. This would result in a total revenue of $2700.