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?
~~~~~~~~~~~~~~~~~~~~~


<pre>
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) = {{{-15x^2 + 345x}}}.     (2)

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


3.  The maximum of the quadratic function (2) is achieved at x = {{{-b/(2a)}}} = {{{-345/(2*(-15))}}} = 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. 


{{{graph( 330, 330, -2.5, 26.5, -200.5, 2400.5,
          -15x^2 + 345x
)}}}


        Plot y = {{{-15x^2 + 345}}}
</pre>

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


All questions are answered.  &nbsp;The problem is solved.



On finding the maximum/minimum of a quadratic function see the lessons  

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-complete-the-square-of-a-quadratic-function-to-find-its-minimum-maximum.lesson>HOW TO complete the square to find the minimum/maximum of a quadratic function</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Briefly-on-How-to-complete-the-square-of-a-quadratic-function-to-find-its-minimum-maximum.lesson>Briefly on finding the minimum/maximum of a quadratic function</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/HOW-TO-complete-the-square-to-find-the-vertex-of-a-quadratic-function.lesson>HOW TO complete the square to find the vertex of a parabola</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/Briefly-on-finding-the-vertex-of-a-parabola.lesson>Briefly on finding the vertex of a parabola</A>

in this site.


For similar solved problems on maximazing revenue/profit see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/Using-quadratic-functions-to-solve-problems-on-maximizing-profit.lesson>Using quadratic functions to solve problems on maximizing revenue/profit</A>

in this site.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


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