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?

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:

[tex]x=\frac{-105,000}{2*(-30,000)}=\frac{-105,000}{-60,000} \\ \\=1.75[/tex]

This means the new price will be 18-3*1.75=12.75.

The new attendance will be 40,000+10,000*1.75=57,500.