Question

A baseball team plays in a stadium that holds 58000 spectators. With the ticket price at $12 the average attendance has been 25000. When the price dropped to $9, the average attendance rose to 29000. Assume that attendance is linearly related to ticket price.
Required:
a. Find the demand function p(x), where x is the number of the spectators.
b. How should ticket prices be set to maximize revenue?

Answer 1

Answer:

We need to assume that the relationship is linear.

a) Remember that a linear relation is written as:

y = a*x + b

then we will have:

p(x) = a*x + b

where a is the slope and b is the y-intercept.

If we know that the line passes through the points (a, b) and (c, d), then the slope can be written as:

y = (d - b)/(c - a)

In this case, we know that:

if the ticket has a price of $12, the average attendance is 25,000

Then we can define this with the point:

(25,000 , $12)

We also know that when the price is $9, the attendance is 29,000

This can be represented with the point:

(29,000, $9)

Then we can find the slope as:

a = ($9 - $12)/(29,000 - 25,000) = -$3/4,000 = -$0.00075

Then the equation is something like:

y = (-$0.00075)*x + b

to find the value of b we can use one of the known points.

For example, the point (25,000 , $12) means that when x = 25,000, the price is $12

then:

$12 = (-$0.00075)*25,000 + b

$12 = -$18.75 + b

$12 + $18.75 = b

$30.75 = b

Then the equation is:

p(x) = (-$0.00075)*x + $30.75

b) We want to find the ticket price such that it maximizes the revenue.

The revenue will be equal to the price per ticket, p(x) times the total attendance, x.

Then the revenue can be written as:

r(x) = x*p(x) = x*( (-$0.00075)*x + $30.75 )

r(x) =  (-$0.00075)*x^2 + $30.75*x

So we want to find the maximum revenue.

Notice that this is a quadratic equation with a negative leading coefficient, thus the maximum will be at the vertex.

Remember that for an equation like:

y = a*x^2 + bx + c

the x-value of the vertex is:

x = -b/2a

Then in our case, the x-value will be:

x = -$30.75/(2*(-$0.00075)) = 20,500

Then the revenue is maximized for x = 20,500

And the price for this x-vale is given by:

p( 20,500) =  (-$0.00075)*20,500 + $30.75 = $15.375

which should be rounded to $15.38