A baseball team plays in a stadium that holds 60,000 spectators. With the ticket

label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

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.

R(x) =

 price that maximizes revenue=

 What ticket price is so high=

Oct 22nd, 2017

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

So:

a = 60000 - 3000p

And 

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.

Mar 10th, 2015

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