SOLUTION: The average nightly attendance at a certain movie theater is 100 when the ticket price is $4.00. The manager estimates that for each 10 cents reduction in price, the average attend
Algebra ->
Customizable Word Problem Solvers
-> Evaluation
-> SOLUTION: The average nightly attendance at a certain movie theater is 100 when the ticket price is $4.00. The manager estimates that for each 10 cents reduction in price, the average attend
Log On
Question 406843: The average nightly attendance at a certain movie theater is 100 when the ticket price is $4.00. The manager estimates that for each 10 cents reduction in price, the average attendance will increase by 10. Find the revenue function R(x), where the number of 10 cents reductions in price. What ticket price yields the maximum revenue, and what is the maximum revenue?
My teacher said it's $2.50 and $625 in revenue, but I have no idea how he got that.
Thanks Answer by ewatrrr(24785) (Show Source):
Hi
The average nightly attendance at a certain movie theater is
100 when the ticket price is $4.00.
The manager estimates that for each 10 cents reduction in price,
the average attendance will increase by 10
R(x) = 400 + (4-.10x)(100+10)
R(x) = -x^2 + 30x + 800 |Completing the square to put into the vertex form
the vertex form of a parabola, where(h,k) is the vertex
R(x) = -[(x-15)^2-625] + 800
R(x) = -(x-15)^2 + 1425 Vertex is (15,1425) it is a maximum point (a<0)
15*.10 = 1.50 reduction in ticket price.
Ticket = $4.00 - $1.50 = $2.50 and Revenue at that price is $1425
