calculus test prep

Question

A university is trying to determine what price to charge for tickets to football games. At a price of $2525 per ticket, attendance averages 40 comma 00040,000 people per game. Every decrease of $55 adds 10 comma 00010,000 people to the average number. Every person at the game spends an average of $5.005.00 on concessions. What price per ticket should be charged in order to maximize revenue? How many people will attend at that price?

Answer 1

 

At $25 , the demand is 40,000  so ($25, 40,000) is a point on the demand line

 

Every decrease of $55 adds 10,000 to the demand,

so the slope is -10,000/55 = -181.81

 

y = M x + b

 

40000 = (-181.81)*25 + b

 

40000 + 181.81 * 25 = b

 

b = 44545.45

 

The equation of the demand line is y = (181.81)X + 44545.45

 

The revenue function is R(x) = x [ (-181.81)X + 44545.45 ]

 

The derivative is R'(x) = -181.81X +  (-181.81)X + 44545.45 

 

                                  = -363.63X + 44545.45

 

The max revenue occurs when R'(x)=0, so

 

       Xmax = -44545.45 / -363.63 = 122.50