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Advanced MathQ&A LibraryA 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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