Question

A baseball team plays in a stadium that holds 50,000 spectators. With the ticket price at $10, the average attendance at recent games has been 21,000.
A baseball team plays in a stadium that holds 50,000 spectators. With the ticket price at $10, the average attendance at recent games has been 21,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000.
a. Find a function that models the revenue in terms of ticket price (let x represent the price of a ticket and R represent the revenue.)
b. Find the price that maximizes revenue from ticket sales.
c. What ticket price is so high that no revenue is generated?

Answer 1

Answer:

A:

Let x represent the price of a ticket and R represent the revenue.

Let r be the number of 1 $ reduction.

So,  $x=10-r$

=> $r= 10-x$     .......(1)

As given that for every dollar the ticket price is lowered, attendance increases by 3000.

So, if we reduce the ticket price by r dollars, the attendance will increase by 3000r.

Revenue = price of ticket X number of ticket sold

R = $(10-r)\times(21000+3000r)$    .....(2)

Substituting the value of r from (1) in (2) we get;

$x\times(21000+3000(10-x))$

Solving this we get;

$R= 51000x- 3000x^{2}$

B:

To maximize revenue, we have to find the derivative for $R= 51000x- 3000x^{2}$

$dR/dp=51000-6000x$

Setting this to zero:

$51000-6000x=0$

=> $6000x=51000$

=> $x = 51000/6000$

x = 8.5

To maximize the revenue, the price should be $8.50.

C:

When R=0

$0=51000x-3000x^{2}$

=>$0=3000x(17-x)$

=> Either x=0 or x=17

x = $0 means there is no price for ticket, that is not possible, so we will neglect it.

And x = $17 means that the ticket price is so high, and no revenue will be generated.