What ticket price would maximize revenue? $

Question

A baseball team plays in a stadium that holds 66000 spectators. With the ticket price at $8 the average attendance has been 28000. When the price dropped to $7, the average attendance rose to 33000. Assume that attendance is linearly related to ticket price.

Answer 1

If we assume a linearly related ticket price, we can set up answers for both price and quantity:

Price = 8 - n (Our starting point is $8 and it drops $1)

Quantity = 28000 + 5000n (Our starting point is 28000 people and it increases 5000)

Revenue = Quantity * price = (28000 - 5000n) * (8 - n) = 224000 - 28000n + 40000n - 5000n² = 224000 + 12000n - 5000n²

To find max revenue, we must take the derivative and set equal to 0:

R' = 12000 - 10000n = 0 or n = 1.2

Therefore when the Price = 8 - 1.2 = $6.80, the quantity = 28000 + 5000*1.2 = 34000 and the revenue is maximized,

Answer 2

The price of the ticket y varies indirectly with attendance x

.

y=k/x

8=k/28000

k=204000

y=204000/66000

y=$3.00 approx.