A theater is presenting a program on drinking and driving for students and their parents or other responsible adults. The proceeds

Question

A theater is hosting a program focused on the dangers of drinking and driving, targeting students and their parents or other responsible adults. The proceeds from this event will be donated to a local alcohol information center. Admission prices are set at $12.00 for adults and $6.00 for students. However, there are two constraints to consider: the theater has a maximum capacity of 150 attendees, and for every two adults, there must be at least one student present. What is the optimal number of adults and students that should attend to maximize fundraising efforts?

Anonymous · Accepted Answer

100 adults and 50 students should attendand, The maximum amount raised = $250Explanation: Given:Admission for adults = $2.00Admission for students = $1.00Total persons that can be held in theater = 150For every 2 adults there must be 1 student let the number of adults be 'x' and the number of students be 'y'thus, we can write the above constraints mathematically as:x + y = 150   (1)and,x = 2y  (2)  (for 1 student i.e y = 1, there should be 2 adults i.e x = 2 × 1 = 2)substituting the 'x' from 2 in the equation 1, we get2y + y = 150ory = 50Thus, x = 2 × 50 = 100   (from equation 2)Hence, 100 adults and 50 students should attendand, The maximum amount raised = $2 × 100 + $1 ×50 = $250

Judd Wilkinson · Answer

160 adults and 80 studentsStep-by-step explanation:With the information from the exercise we have the following system of equations:Let x = number of students; y = number of adultsI want to maximize the following:z = 3 * x + 6 * yBut with the following constraintsx + y = 240y / 2

Anonymous · Answer

Maximum people attending programAdult = 140Student = 70Explanation:Provided information,Maximum seating capacity in the theater = 210 peopleFor each pair of adult there must be at-least one student.Thus, maximum revenue can be calculated as follows: Fee for each adult = $10Fee for each student = $5For each combination of two adult and one student revenue = $10 + $10 + $5 = $25Total people in each combination = 3Thus number of combinations possible = Thus, number of adults attending program = 70 2 = 140Number of students = 70 1 = 70Maximum amount = 140 $10 + 70 $5= $1,750Maximum people attending program:Adult = 140Student = 70

Anonymous · Answer

Maximum people attending programAdult = 140Student = 70Explanation:Provided information,Maximum seating capacity in the theater = 210 peopleFor each pair of adult there must be at-least one student.Thus, maximum revenue can be calculated as follows:Fee for each adult = $10Fee for each student = $5For each combination of two adult and one student revenue = $10 + $10 + $5 = $25Total people in each combination = 3Thus number of combinations possible = Thus, number of adults attending program = 70 2 = 140Number of students = 70 1 = 70Maximum amount = 140 $10 + 70 $5= $1,750Maximum people attending program:Adult = 140Student = 70

Allison Lockman

A theater is presenting a program on drinking and driving for students and their parents or other responsible adults. The proceeds

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