# 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 \$14.00 for adults and \$7.00 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?

Question
Asked Jun 11, 2019
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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 \$14.00 for adults and \$7.00 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?

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Step 1

Let x= number of adults

Let y= number of parents

To maximize the revenue Z=14x+7y,

Subject to a contraints

x+y<=240,

x/2<=y,

restrictions are x<=0, y<=0.

Step 2

Graph these 4 inequalities with x on the x-axis and y on th y-axis.

In the graph the common solution of the 4 inequalities form a triangle in the first quadrant with one vertex at the origin (0,0), second vertex at(0,240), and third vertex  y=240-x, and  y=x/2

(3/2)*x=240,

x=160, and y=80.

So, the third vertex is (160,80).

Step 3

The extrema (maximum and minimum ) of the revenue function will occur at the vetrices of the triangle,

So, plug those values into the r...

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