8. A baseball team charges $\$ 30$ per ticket and averages 20,000 people in attendance per game. Each person spends an average of $\$ 8$ on consessions. For every drop of $\$ 1$ in the ticket price, the attendance rises by 800 people. What ticket price should the team charge to maximize total revenue?
$\$ 24.50$
$\$ 27.50$
$\$ 22.50$
$\$ 25.50$

Step 1 ：Let's denote the drop in the ticket price from $30 as x. So, the ticket price is \(30 - x\) and the number of people in attendance is \(20000 + 800x\).

Step 2 ：The total revenue R of the team is the sum of the revenue from ticket sales and the revenue from concessions. The revenue from ticket sales is the product of the ticket price and the number of people in attendance. The revenue from concessions is the product of the average amount spent on concessions and the number of people in attendance. So, we can express the total revenue R as a function of x: \(R(x) = (30 - x)(20000 + 800x) + 8(20000 + 800x)\).

Step 3 ：To find the ticket price that maximizes total revenue, we need to find the maximum of this function. We can do this by taking the derivative of the function, setting it equal to zero, and solving for x. The derivative of the function is \(R'(x) = 10400 - 1600x\).

Step 4 ：Setting the derivative equal to zero gives us the critical points: \(x = 13/2\).

Step 5 ：Substituting x back into the equation for the ticket price gives us the optimal price: \(30 - x = 47/2\).

Step 6 ：Final Answer: The team should charge \(\boxed{\$ 23.50}\) per ticket to maximize total revenue.