Calculus question please help

Question

A baseball team plays in a stadium that holds 54000 spectators. With the ticket price at $12 the average attendance has been 23000. When the price dropped to $11, the average attendance rose to 27000.a) Find the demand function , where  is the number of the spectators. (Assume that  is linear.)How should ticket prices be set to maximize revenue?

Answer 1

The revenue maximizing price is $8.88

You have 2 points on a line, (12, 23) and (11,27) where (x,y) has x in dollars for price and y is in thousands of spectators

Two points determine a line. Its slope is "rise over run," the difference in y coordinates divided by the difference in x coordinates: (27-23)/(11-12) = 4/-1 = -4

Use either point such as (12,23) with slope = m= -4

(y-y₁) = m(x-x₁) is the point slope form of the linear equation.

(12,23) = (x₁,y₁)

(y-23)/(x-12) = -4

y-23 = -4(x-12)

y-23 = -4x+48

y= -4x +48+23

y= 71-4x

attendance in thousands = 71 -4(Price of tickets in dollars)

check the answer

for $11, attendance = 71-4(11) = 71 - 44= 27 thousand

for $12 attendance = 71- 4(12) = 71 -48 = 23 thousand

for $8.88 attendance = 71-4(8.88) = 71- 35.52 = 35 thousand & 4 hundred & eighty

In economics demand curve is usually represented with P and Q, replacing x and y

Q=71-4P

Revenue is quantity times price = R = QP = (71-4P)P = 71P - 4P^2

take the first derivative & set equal to 0

R' = 71 -8P = 0

8P= 71

P = 71/8 = $8.88 is the revenue maximizing price, which also leaves the stadium less than full with less than 54,000 tickets sold

maximum R = 71(71/8) - 4(71/8)^2

= 630,130 - 315,062.50 = $315,062.50

(Revenue maximum is not the same as profit maximizing,

as it fails to consider costs)