Question

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 comma 000 people per game. Every decrease of ​$3 adds 10 comma 000 people to the average number. Every person at the game spends an average of ​$6.00 on concessions. What price per ticket should be charged in order to maximize​ revenue? How many people will attend at that​ price?

Answer 1

ticket price is 27 -3(3)= $18 , and Number of people = 40,000+10,000(3)=70,000 .

Explanation:

We start the equation with (27-3x). We don't know how many times we're subtracting $3 from the price. This is multiplied by (40,000+10,000x).

⇒
`(27-3x)(40,000+10,000x)`

This represents the amount of money for tickets multiplied by the amount of people. We still need to add the amount of concessions. Since (40,000+10,000x) represents the number of people, we multiply that by $6.00.This gives us:

⇒
`(27-3x)(40,000+10,000x) + (40,000+10,000x)6`

⇒
`1,08,000+2,70,000x-1,20,000x-30,000x^2 +240,000 + 15,000 x`

⇒
`1,08,000+1,65,00x-30,000x^2 +240,000`

Differentiating this w.r.t x :

⇒
`1,65,00=2x(30,000)`

⇒
`1,65,00=x(60,000)`

⇒
`x=2.75`,

⇒
`x=3`, { Rounding off }

So, ticket price is 27 -3(3)= $18 , and Number of people = 40,000+10,000(3)=70,000 .