Let's assume the number of parents attending is P and the number of students attending is S.
According to the constraints:
1) P + S <= 210
2) P >= 2S
We want to maximize the amount of money raised, which is given by:
Total revenue = 10P + 5S
Since we want to maximize revenue, let's express P in terms of S from constraint 2:
P >= 2S
P = 2S
Substitute P = 2S into the total revenue equation:
Total revenue = 10(2S) + 5S
Total revenue = 20S + 5S
Total revenue = 25S
Now, substitute the expression for total revenue back into the constraints:
25S <= 210
S <= 8.4
Since we cannot have a fraction of a student, S must be 8.
Therefore, the optimal number of students to attend is 8.
From constraint 2, P = 2S, so P = 16.
Therefore, 16 parents and 8 students should attend to raise the maximum amount of money.
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 $10.00 for parents and $5.00 for students. However, the situation has 2 constraints. The theater can hold no more than 210 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?
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