A theater is presenting a program for students and their parents on drinking and driving. The proceeds will be donated to a local alcohol center: Admission is $14.00 for parents and $7.00 for students. However, the situation has 2 constraints. The theater can hold no more than 180 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?

Answer 1

Let's use algebra to solve this problem:

Let x be the number of parents attending and y be the number of students attending.

From the given information, we have the following constraints:
1. x + y ≤ 180 (constraint on theater capacity)
2. x/2 ≤ y (every two parents must bring at least one student)

We want to maximize the amount of money raised, which is given by the total revenue R:
R = 14x + 7y

We can rewrite the second constraint as x - 2y ≤ 0.

Now, we can set up the linear programming problem:

Maximize R = 14x + 7y
Subject to:
x + y ≤ 180
x - 2y ≤ 0
x, y ≥ 0

Let's graph the feasible region and find the vertices to determine the maximum revenue:

The vertices of the feasible region are:
(0, 0), (0, 90), (180, 0), (120, 60)

Calculate the revenue at each vertex:
(0, 0): R = 14(0) + 7(0) = $0
(0, 90): R = 14(0) + 7(90) = $630
(180, 0): R = 14(180) + 7(0) = $2520
(120, 60): R = 14(120) + 7(60) = $2040

The maximum revenue of $2520 will be raised when 180 parents and 0 students attend the program.