SOLUTION: A new movie theater holds 300 people. They found that when the price was $5 per ticket, the average sales would be 270 people per show. Each time they increase the price by $1, the

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Question 1070691: A new movie theater holds 300 people. They found that when the price was $5 per ticket, the average sales would be 270 people per show. Each time they increase the price by $1, the average attendance drops by 15 people.
A) find an expression, using x, for the number of people that will go to each show
B) find the function R (x) that gives the amount of revenue the theater will make
C) what price should they charge in order to maximize revenue?

Answer by ikleyn(52790)   (Show Source): You can put this solution on YOUR website!
.
A new movie theater holds 300 people. They found that when the price was $5 per ticket, the average sales would be 270 people
per show. Each time they increase the price by $1, the average attendance drops by 15 people.
A) find an expression, using x, for the number of people that will go to each show
B) find the function R (x) that gives the amount of revenue the theater will make
C) what price should they charge in order to maximize revenue?
~~~~~~~~~~~~~~~~~~~~~

1.  Let N(x) be this function: the number of viewers at the ticket price of x dollars.

    Then N(x) = 270 - 15(x-5).    (1)

    It is exact translation of this statement: "Each time they increase the price by $1, the average attendance drops by 15 people."

    So, the question A) is answered, and the answer is: the attendance is 

        N(x) = 270 - 15(x-5) = 345 - 15x.

    where x is the ticket price in dollars.


2.  The revenue R(x) is   R(x) = N(x)*x = x*(345 - 15x) = .     (2)

    See the plot below.   So, the question B) is answered, too.


3.  The maximum of the quadratic function (2) is achieved at x =  =  = 11.5.


     Thus the optimum price is 11.5 dollars per ticket.
     The attendance then is N(11.5) = 270 - 15*(11.5-5) = 172.5 persons. 





        Plot y = 

So,  at given conditions,  the owner of the theater can increase the ticket price to  $11.50.
The attendance will decrease to  172 - 173 viewers,  but the revenue will increase from  300*$5 = $1500  to  172*$11.50 = $1978.

All questions are answered.  The problem is solved.


On finding the maximum/minimum of a quadratic function see the lessons
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
in this site.

For similar solved problems on maximazing revenue/profit see the lesson
    - Using quadratic functions to solve problems on maximizing revenue/profit
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic
"Finding minimum/maximum of quadratic functions".



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