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 $6 per parent, and $3 per student. However, this situation has two constraints: The theater can hold no more than 210 people and every two parents must bring one student. How many parents and students should attend to maximize profits?

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18 December, 00:12

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 $6 per parent, and $3 per student. However, this situation has two constraints: The theater can hold no more than 210 people and every two parents must bring one student. How many parents and students should attend to maximize profits?

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Elias Randolph · Answer 1

The answer is 140 parents and 70 students

Explanation:

To start let's analyze the second constraint. Every two parents must bring one student. It means that the number of parents is always twice the number of students.

#parents = 2*#students.

If we want to maximize profits the best would be to fill the auditorium to its maximum capacity, so we have to use the first constrain, it means 210 people between students and parents.

210 = #students + #parents.

replacing the first equation into the second,

210 = #students + 2*#students = 3*#students,

we have that the number of students is

#students = / frac{210}{3} = 70,

and the number of parents is twice the number of students,

#parents = 2*70 = 140.