Ticket Pricing and Revenue
A school drama club sells tickets to a 400-seat auditorium. Last year tickets cost $12 and the club sold 280. A survey suggests that for every $1 increase in price, attendance falls by 20, and for every $1 decrease, attendance rises by 20, up to the 400-seat capacity. Costs are $1,500 fixed plus $3 per ticket sold. The club also offers a $2 student discount; last year 40% of attendees used it. The club wants revenue net of costs to reach at least $2,600 and asks whether a price change can achieve this without exceeding capacity.
1. Model net income as a function of the price change x (in dollars) and find the price that maximizes net income. State the maximum and whether it meets the $2,600 goal.
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Attendance. A(x) = 280 − 20x, valid while 0 ≤ A ≤ 400, i.e. −6 ≤ x ≤ 14.
Average ticket price. The $2 discount applies to 40% of attendees, so the average price is (P − 0.4 × 2) = P − 0.8, where P = 12 + x. Average = 11.2 + x.
Net income. N(x) = (280 − 20x)(11.2 + x) − 1,500 − 3(280 − 20x). = (280 − 20x)(8.2 + x) − 1,500. Expand: 2,296 + 280x − 164x − 20x² − 1,500 = −20x² + 116x + 796.
Vertex. x = −b/(2a) = −116 ÷ (2 × −20) = 2.9. N(2.9) = −20(8.41) + 116(2.9) + 796 = −168.2 + 336.4 + 796 = $964.20.
Judgment. The maximum net income is about $964, well below $2,600. No price change within the valid range meets the goal; the club must cut costs or raise the base price.
2. The club considers dropping the student discount entirely to simplify sales. At the optimal price from the previous task, does removing the discount help or hurt net income? Quantify the difference.
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Without discount. Average price = 12 + x. N(x) = (280 − 20x)(12 + x) − 1,500 − 3(280 − 20x) = (280 − 20x)(9 + x) − 1,500. Expand: 2,520 + 280x − 180x − 20x² − 1,500 = −20x² + 100x + 1,020. Vertex: x = −100 ÷ −40 = 2.5. N(2.5) = −20(6.25) + 100(2.5) + 1,020 = −125 + 250 + 1,020 = $1,145.
Comparison. $1,145 − $964.20 ≈ $180.80 better without the discount. The discount lowers the effective price by $0.80 per attendee on 40% of sales, costing about $0.32 per attendee, which outweighs the small attendance benefit. Removing it helps, though the $2,600 goal is still unmet.
3. Find the price range (in dollars, to the nearest cent) for which net income is at least $2,600, or show that none exists. Use the no-discount model.
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Set the inequality. −20x² + 100x + 1,020 ≥ 2,600 → −20x² + 100x − 1,580 ≥ 0 → divide by −20 (flip sign): x² − 5x + 79 ≤ 0.
Discriminant. b² − 4ac = 25 − 316 = −291 < 0.
Interpretation. The quadratic x² − 5x + 79 is always positive (opens upward with no real roots), so the inequality has no solution. The maximum net income is $1,145, far below $2,600.
Conclusion. No ticket price achieves $2,600 net income under these assumptions. The fixed cost of $1,500 and per-ticket cost of $3 make the goal infeasible at this demand level; the club would need roughly 2.3 times the maximum possible net income, so it must change costs or demand, not price.