Answer:
The price that maximizes the revenue is $20
Step-by-step explanation:
(a) Express the daily revenue from ticket sales, R as a function of the number of $1.00 price increases,
R(x) = (70000-2500x)(12+x)
(b) What ticket price maximizes the revenue from ticket sales?
The ticket price that maximizes the revenue is associated with the x of the vertex of the parabola.
R(x) = (70000-2500x)(12+x) = 840000 + 70000x - 30000x - 2500x² =
-2500x² + 40000x + 840000
x vertex = -b/2a = -40000/2.(-2500) = 8
12+8 = 20