Mark M. answered 04/14/24
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Let x = number of $0.10 decreases
Average attendance = 27000 + 300x
Ticket price = 10 - 0.10x
Assuming that you want to adjust the ticket price to maximize revenue:
Revenue = R(x) = (27000 + 300x)(10 - 0.10x) = -30x2 + 300x + 270000
The graph of R(x) is a parabola opening downward
Max when R'(x) = -60x + 300 = 0
So, maximum revenue when x = 5
To maximize revenue, decrease the ticket price by 5($0.10) = $0.50
Maximum revenue R(5) = $270,750