Question

A baseball team plays in a stadium that holds 54000 spectators. With the ticket price at $8 the average attendance per game has been 20000. When the price dropped to $6, the average attendance rose to 27000. Assume that attendance is linearly related to ticket price.
Find a function that models the per game revenue in terms of the price
p


R
(
p
)
=



What ticket price would maximize revenue? (round to the nearest cent)

Answer 1

9514 1404 393

Answer:

  R(p) = -3500p^2 +48000p . . . revenue function

  $6.86 . . . price for maximum revenue

Step-by-step explanation:

The 2-point form of the equation for a line can be used to find the attendance function.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  y = (27000 -20000)/(6 -8)(x -8) +20000

  y = -3500(x -8) +20000

  y = 48000 -3500x . . . . y seats sold at price x

The per-game revenue is the product of price and quantity sold. In functional form, this is ...

  R(p) = p(48000-3500p)

  R(p) = -3500p^2 +48000p . . . per game revenue

__

Revenue is maximized when its derivative is zero.

  R'(p) = -7000p +48000

  p = 48/7 ≈ 6.86

A ticket price of $6.86 would maximize revenue.