SOLUTION: When a school play charges 2$ for admission, an average of 100 people attend. For each 10 cents increase in admission price, the average number decreases by 1. What charge would ma

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Question 1146167: When a school play charges 2$ for admission, an average of 100 people attend. For each 10 cents increase in admission price, the average number decreases by 1. What charge would make the most money?
Found 2 solutions by richwmiller, ikleyn:
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
5.9 61 359.9
61 people at $5.90 yields $359.90
6.0 60 360
$6 with 60 people yields $360
6.1 59 359.90
but $6.10 with 59 people yields $359.90 same as $5.90 for 61 people
Maximum is $6 and 60 people
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
The meaning of this problem is that when the admission price varies as p = 2+0.1x dollars, 

the number of people who attend varies as n(x) = 100 - x.


The revenue function then is

    R(x) = n*p = (100-x)*(2+0.1x)      (1)


It is a quadratic function with negative leading coefficient.


They want you find an admission price which provides the maximum value to the revenue.


For it, you need to find the value of "x" which provides the maximum to the quadratic function (1), and then to calculate the price 2 + 0.1x.


This value of x is exactly half way between the roots of linear binomials 100-x and 2+0.1x.


The roots are  x= 100  and  x= -20;  hence, the value of x which provides the maximum value is   =  = 40.


Then the optimum price is  2 + 40*0.1 = 2 + 4 = 6 dollars for admission.


The attendance then is  p(40) = 100 - 40 = 60  and the revenue is  60*6 = 360 dollars.


Compare it with the revenue of 2*100 = 200 dollars before the optimization.


The plot of the revenue as the function (1) is below



    


    Plot y = (100-x)*(2+0.1x)


----------------

On finding the maximum/minimum of quadratic functions and relevant solved problems, 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
    - A rectangle with a given perimeter which has the maximal area is a square
    - A farmer planning to fence a rectangular garden to enclose the maximal area
    - A farmer planning to fence a rectangular area along the river to enclose the maximal area
    - A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
    - Using quadratic functions to solve problems on maximizing revenue/profit (*)
    - OVERVIEW of lessons on finding the maximum/minimum of a quadratic function

and especially the lesson marked (*) in the list.


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".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


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