A university is trying to determine what price to charge for tickets to football games. At a price of $18 per ticket, attendance averages 40,000 people per game. Every decrease of $3 adds 10,000 people to the average number. Every person at the game spends an average of $4.50 on concessions. What price per ticket should be charged in order to maximize revenue? How many people will attend at that price?
Accepted Solution
A:
They should charge $12.75 per ticket. If they do, 57,500 people will attend.
We start the equation with (18-3x). We don't know how many times we're subtracting $3 from the price. This is multiplied by (40,000+10,000x). We will add 10,000 people the same number of times that we subtract $3 from the price. 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 $4.50, the average amount each person spends on concessions. This gives us: [tex](18-3x)(40,000+10,000x)+(40,000+10,000x)(4.50)=0[/tex] We set it equal to 0 because when we are solving a quadratic equation, the solutions are where it intersects the x-axis, where y=0.
We multiply our terms on both pieces next: [tex]18*40,000+18*10,000x+(-3x)*40,000+(-3x)*10,000x
\\+40,000*4.50+10,000x*4.50
\\
\\=720,000+180,000x+(-120,000x)+(-30,000x^2)
\\+180,000+45,000x
\\
\\=-30,000x^2+105,000x+900,000[/tex]
When we find the maximum of a quadratic, we must find the axis of symmetry. That is done using x=-b/2a: