Calculus/Calculus, maximum revenue
a theatre determines that if the admission price is $10, it averages 100 people in attendance . but for every increase of $2, it loses 5 customers from the average number.every customer spends an average of $2 on concessions. what admission price should the theater charge in order to maximize the revenue?
We may express the admission price as 10 + 2x , where x is the number of times the price is raised by $2.
The number of customers will then be 100 - 5x .
Revenue is then R(x) = (100 - 5x)(10 + 2x) + (100 -5x)(2) , where (100 -5x)(2) is added because each customer spends $2 on concessions.
Simplifying this expression for revenue gives
R(x) = (100 - 5x)(12 + 2x) and
R(x) = (10)(20 - x)(6 + x)
You can use calculus to find the value for x that gives the maximum , or just recognize that the graph of this function is a parabola that opens down and has its highest point when x is the average of the roots .
0 = (10)(20 - x)(6 + x) gives x = 20 and x = -6 for the roots.
Thus the revenue is a maximum when x = (20 - 6 ) / 2 = 14/2 = 7
The admission price will be 10 + (2) (7) = 10 + 14 = 24
The maximum revenue occurs when the admission price is $24