Fundraiser Ticket Pricing and Revenue
A school theater group sells tickets to a show. At $8 per ticket they sold 300 tickets. A survey suggests that for every $1 increase in price, attendance drops by 20 people. The group has a fixed cost of $1,500 for the production and a variable cost of $2 per attendee for programs and refreshments. They want to choose a price that maximizes profit rather than just revenue, and they are considering prices of $8, $10, and $12.
1. Write attendance and profit as functions of the price increase x (in dollars above $8). Then evaluate profit at x = 0, 2, and 4.
Read the worked answer
Attendance: A(x) = 300 − 20x. Price: P(x) = 8 + x.
Revenue = (8 + x)(300 − 20x). Variable cost = 2 × (300 − 20x).
Profit = (8 + x)(300 − 20x) − 2(300 − 20x) − 1500 = (6 + x)(300 − 20x) − 1500.
- x = 0: (6)(300) − 1500 = 1800 − 1500 = $300
- x = 2: (8)(260) − 1500 = 2080 − 1500 = $580
- x = 4: (10)(220) − 1500 = 2200 − 1500 = $700
2. Find the price increase x that maximizes profit, and identify the corresponding ticket price. Show that this is a maximum, not a minimum.
Read the worked answer
Profit(x) = (6 + x)(300 − 20x) − 1500 = 1800 + 300x − 120x − 20x² − 1500 = 300 + 180x − 20x².
This is a downward-opening parabola (coefficient −20 < 0), so the vertex is a maximum. Vertex at x = −b/(2a) = −180/(2 × −20) = 180/40 = 4.5.
Optimal price = 8 + 4.5 = $12.50. Profit at x = 4.5: 300 + 180(4.5) − 20(4.5²) = 300 + 810 − 405 = $705.
Since a < 0, the parabola opens downward, confirming a maximum. The $12 price (x = 4) gives $700, close but not optimal.