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

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 $ 12.00 for adults and $ 6.00 for students.​ However, this situation has two​ constraints: The theater can hold no more than 150 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?

Answers

100 adults and 50 students should attend

and,

The maximum amount raised = $250

Explanation:

Given:

Admission for adults = $2.00

Admission for students = $1.00

Total persons that can be held in theater = 150

For 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 get

2y + y = 150

or

y = 50

Thus,

x = 2 × 50 = 100   (from equation 2)

Hence,

100 adults and 50 students should attend

and,

The maximum amount raised = $2 × 100 + $1 ×50 = $250

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.

Maximum people attending program

Adult = 140

Student = 70

Explanation:

Provided information,

Maximum seating capacity in the theater = 210 people

For each pair of adult there must be at-least one student.

Thus, maximum revenue can be calculated as follows:

Fee for each adult = $10

Fee for each student = $5

For each combination of two adult and one student revenue = $10 + $10 + $5 = $25

Total people in each combination = 3

Thus number of combinations possible = frac{210}{3} = 70

Thus, number of adults attending program = 70 times 2 = 140

Number of students = 70 times 1 = 70

Maximum amount = 140 times $10 + 70 times $5

= $1,750

Maximum people attending program:

Adult = 140

Student = 70

Maximum people attending program

Adult = 140

Student = 70

Explanation:

Provided information,

Maximum seating capacity in the theater = 210 people

For each pair of adult there must be at-least one student.

Thus, maximum revenue can be calculated as follows:

Fee for each adult = $10

Fee for each student = $5

For each combination of two adult and one student revenue = $10 + $10 + $5 = $25

Total people in each combination = 3

Thus number of combinations possible = frac{210}{3} = 70

Thus, number of adults attending program = 70 times 2 = 140

Number of students = 70 times 1 = 70

Maximum amount = 140 times $10 + 70 times $5

= $1,750

Maximum people attending program:

Adult = 140

Student = 70

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