Answer to Question #174434 in Calculus for Jc young

Question

The number of people attending a music festival is represented by 12000(330-x) where X is the cost of tickets in dollars what ticket price will maximize the total revenue of the concert?

a) what is the optimal ticket price?

b) what will be the total revenue of the festival at that ticket price

Answer 1

a)

The total revenue is

$f(x)=12000(330-x)x=12000*330x-12000x^2$

$f'(x)=12000*330-24000x$

Maximum of revenue is when $f'(x)=0$

$12000*330-24000x=0$

$x=\frac{12000*330}{24000}=165$ is the optimal ticket price.

b)

the total revenue of the festival at that ticket price:

$f(165)=12000*(330-165)*165=12000*165^2=$

$=326700000$