A university is trying to determine what price to charge for tickets to football games. At a price of $25 per ticket. attendance averages 40,000 people per game. Ever),I decrease of $5 adds 10,000 people to the average number. Every person at the game spends an average of $5.00 on concessions. What price per ticket should be charged in order to maximize revenue? How many people will attend at that price? II" What is the price per ticket? SD . Ice or $25 per ticket. 1 VIEW an example I A" pans Showmg X the average number Every no he charged ill order to A university is trying to determine what price to charge for tickets to football games At a price of $27 per ticket. attendance averages 40.000 people per game. Every decrease at $3 adds 10.000 people to the average number. Every person at the game spends an average of $3.00 on concessions. What price per ticket shoud be charged in order to maximize revenue? How many Wattle will attend at that price? (3) Make a list of the variables and constants in the problem Let I represent the decrease in ticket price from $27. If the ticket price increases. then x is negative. Write the attendance. price per ticket. and revenue in terms of this variable. Start with the average attendance Notice that the ratio of the change in attendance to the change in price is a constant. which means the slope of the equation must be a constant. Therefore. the attendance must have the formy = mx+b The attendance increases by 10.000 tor a decrease at $3 in the ticket price Also. when the ticket price is 327, then the attendance is 40.000 Use this information to write a linear equation for the attendance in terms of it. 10.000 3 y=40.000+ I Write the price per ticket. p. in terms of x as well. p = 2'! - x The revenue has two parts. the price per ticket multiplied by the number of people and the average concession spending multiplied by the number of people. 1 0.000 Use y=40.000+ 3 it and p = 27 - xto write the revenue function. R(x). R(It) = (MPH 001313\") 10.000 10.000 "('0 = 40,000+ 3 it (ZTvxh 40.000+ 3 I {300) round 2 \"( = ' 3 1 i60.000x+1.200.000 From the problem statement. the revenue is the quantity to be maximized. 10000 2 _ _ _ 3 X 1 60.000x + 1.200.000. and Is already written In terms of a single variable. To maximize the function. rst nd the derivative R'(x). The value to be maxtmtzed is R(x) : - 0 3 x + 60,000 R'rx) : Since who is dened for all x. the only critical values occur where R'bt) = 0. Solve this equation. 20.000 3 0 it + 60.000 x = 9.00 There is only one critical value. Use the second derivative. R"(x). to determine if it is a maximum or minimum. , 20,000 MK) : 7 3 x+60.000 \"\"00 : 7 20:00 Sincethe second derivative is negative at it = 9.00mi: cntioal point is a maximum. Find the new ticket price using the equation found previously. p = 271 p =1300 Find the attendance using the corresponding equation. 10.000 3 K 3' 00 31 = 70.000 Therefore. the revenue is maximized when the price per ticket Is $18.00. At that price. the average attendance will be 70,000 people. :his View an example Textbook ( (than all 3