Maximize revenue

Question

A baseball team plays in a stadium that holds 68000 spectators. With the ticket price at $9 the average attendence has been 28000. When the price dropped to $8, the average attendence rose to 34000. Assume that attendence is linearly related to ticket price.


What ticket price would maximize revenue?

Answer 1

Each time the price is decreased by $1, attendance increases by 6000. 

 

Let x = number of $1 decreases

 

Let R(x) = revenue 

 

             = (price per ticket)(number of tickets sold in thousands)

 

             = (9 - x)(28 + 6x)

 

             = -6x² + 26x + 252

 

Maximum revenue when R'(x) = 0

 

-12x + 26 = 0

 

             x = 2.166666 ≈ $2.17

 

To maximize revenue, the price per ticket should be reduced by $2.17 from the original $9 price.

 

So, $6.83 is the optimal price.