Problem 21 - College Algebra

Question

A theater is presenting a program for students and their parents on drinking and driving. The proceeds will be donated to a local alcohol information center. Admission is \(\$ 2.00\) for parents and \(\$ 1.00\) for students. However, the situation has two constraints: The theater can hold no more than 150 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?

Step-by-Step Solution

Answer

To maximize the proceeds, 100 parents and 50 students should attend, raising a total of $250.

Step 1: Define the Variables

Let's define P as the number of parents and S as the number of students.

Step 2: Set up the Objective Function

The objective is to maximize the total revenue (R). The revenue collected from the parents would be 2*P dollars and from the students it would be S dollars. So, the objective function becomes \(R = 2P + S \).

Step 3: Set up the Constraints

The constraints given are: First, the theater can hold a maximum 150 people, i.e., \( P + S <= 150 \). Second, for every two parents, there should be at least one student. Hence, the number of students should be at least half of the number of parents, i.e., \( S >= 0.5P \). Let's convert it to \( S - 0.5P >= 0 \) which is more suitable for our solution process.

Step 4: Apply Linear Programming Techniques

By applying linear programming methods to the objective function and constraints, we find two points of interest where constraints intersect: (0,150) and (100,50). These represent the possible maximums for the objective function.

Step 5: Substitution and Comparison for Maximization

Substituting these points into the objective function: For (0,150), the revenue would be \( 2*0 + 150 = 150) dollars. For (100,50), the revenue would be \( 2*100 + 50 = 250 \) dollars. By comparison, we see that (100,50) gives us the maximum revenue.