A theater is presenting a program on drinking and driving for students and their parents or other responsible adults. The proceeds will be donated to a local alcohol information center. Admission is $6.00 for adults and $3.00 for students. However, this situation has two constraints: The theater can hold no more than 240 people and for every two adults, there must be at least one student. How many adults and students should attend to raise the maximum amount of money?

160 adults and 80 students

Step-by-step explanation:

With the information from the exercise we have the following system of equations:

Let x = number of students; y = number of adults

I want to maximize the following:

z = 3 * x + 6 * y

But with the following constraints

x + y = 240

y / 2 <= x

As the value is higher for adults, it is best to sell as much as possible for adults.

So let's solve the system of equations like this:

y / 2 + y = 240

3/2 * y = 240

y = 240 * 2/3

y = 160

Which means that the maximum profit is obtained when there are 160 adults and 80 students, so it is true that added to 240 and or every two adults, there must be at least one student.

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about the problem:

srry i'm not done ! i'll