Revenue Optimization and Quadratic Functions

The video tutorial explains how to determine the ticket price that maximizes revenue for a stadium with a seating capacity of 1000. Initially, with a ticket price of $30, the attendance is 400 people. For each $1 decrease in price, attendance increases by 50. The tutorial guides viewers through setting up a revenue equation as a function of price decreases, solving it using the vertex formula to find the optimal ticket price. The solution reveals that reducing the ticket price to $19 maximizes revenue. The video concludes with an invitation to explore similar quadratic problems.

What is the initial ticket price mentioned in the problem?

• $20

• $25

• $30

• $35

How many people attend when the ticket price is $30?

• 400

• 600

• 300

• 500

What happens to attendance for each $1 decrease in ticket price?

• Increases by 60

• Increases by 30

• Increases by 40

• Increases by 50

What is the seating capacity of the stadium?

• 800

• 1100

• 900

• 1000

What is the formula for revenue in this context?

• Number of tickets sold times the price per ticket

• Number of tickets sold plus the price per ticket

• Number of tickets sold minus the price per ticket

• Number of tickets sold divided by the price per ticket

What does the variable X represent in the revenue equation?

• Number of $5 price decreases

• Number of $2 price decreases

• Number of $1 price decreases

• Number of $10 price decreases

What is the initial revenue when the ticket price is $30?

• $14,000

• $10,000

• $16,000

• $12,000

What is the effect of decreasing the ticket price by $2?

• Attendance increases by 200

• Attendance increases by 150

• Attendance increases by 100

• Attendance increases by 50

What is the expanded form of the revenue equation?

• R = 12,000 + 1,100x + 50x^2

• R = 12,000 - 1,100x + 50x^2

• R = 12,000 + 1,100x - 50x^2

• R = 30x + 400

What is the coefficient of x in the expanded revenue equation?

• 1200

• 1100

• 1000

• 1300