price at $10, the average attendance at recent games has been 30,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000.
price that maximizes revenue=
What ticket price is so high=
Revenue = attendance x price
We have, from the description of the problem,
Attendance = c - 3000 x price, where c is a constant. When price = 10, attendance = 30000, so
30000 = c - 3000 x 10
c = 60000
a = 60000 - 3000p
R(p) = a x p
= (60000 - 3000p) x p
= 60000p - 3000p^2
To find the maximum, we need to find a turning point (where dR/dp = 0). Differentiate:
dR/dp = 60000 - 6000p
So p = 10 is a turning point. This is the maximum since the function has a negative p^2 term, so it's an inverted u-shape (with a maximum and no minima).
At p = 10, the revenue is 600000 - 300000 = $300,000.
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