Ticket pricing for a 240-seat cinema: revenue versus profit
A 240-seat independent cinema in Sheffield charges £9.00 per ticket and currently sells about 1,080 tickets a week. The owner's customer survey suggests that for every £0.50 rise in price, weekly attendance falls by 40, and that this linear relationship is reasonable up to a price of £15.00. Weekly fixed costs are £6,800, and the booking platform charges the cinema £1.20 for every ticket sold. The owner writes the price as £(9 + 0.50x) and attendance as 1080 − 40x, where x is the number of 50-pence increases. An adviser proposes a different model in which attendance falls by 3% for each 50-pence increase rather than by a constant 40 tickets. Before reprinting the price board the owner wants three numbers: the revenue-maximising price, the profit-maximising price, and a comparison of the two attendance models at a £12.00 ticket - where the two disagree most.
1. Show that weekly revenue is R(x) = 9720 + 180x − 20x² and find the price that maximises revenue. Include what happens if prices must move in 50-pence steps.
Read the worked answer
n = 1080 − 40x and p = 9 + 0.5x, so R = (9 + 0.5x)(1080 − 40x) = 9720 − 360x + 540x − 20x² = 9720 + 180x − 20x².
This is a downward parabola (coefficient of x² negative), so the maximum is at the vertex: x = −b/(2a) = −180/(2 × −20) = 4.5.
Price = 9 + 0.5(4.5) = £11.25, attendance = 1080 − 180 = 900, R = 11.25 × 900 = £10,125.
Check with the formula: 9720 + 810 − 405 = £10,125. ✓
x = 4.5 is not a whole number of 50p steps. If prices must be in 50p steps, x = 4 and x = 5 both give R = £10,120 (11.00 × 920 and 11.50 × 880). The owner is effectively indifferent between £11.00 and £11.50; the continuous optimum is not attainable.
2. Find the price that maximises weekly profit and explain why it differs from the revenue-maximising price.
Read the worked answer
Profit = revenue − fixed costs − platform fees P = 9720 + 180x − 20x² − 6800 − 1.2(1080 − 40x) = 9720 + 180x − 20x² − 6800 − 1296 + 48x = 1624 + 228x − 20x².
Vertex: x = 228/40 = 5.7, so price = 9 + 2.85 = £11.85, attendance = 1080 − 228 = 852. P(5.7) = 1624 + 1299.6 − 649.8 = £2,274 per week.
Check directly: revenue = 11.85 × 852 = £10,096.20; costs = 6800 + 1.2 × 852 = £7,822.40; profit = £2,273.80. ✓
The prices differ because each extra ticket costs £1.20. Profit therefore penalises attendance, while revenue treats every ticket as free to serve. That pushes the profit optimum to a higher price and 48 fewer tickets per week than the revenue optimum.
3. Compare the two attendance models at a £12.00 ticket and advise which the owner should rely on near the top of the surveyed range.
Read the worked answer
£12.00 means x = 6.
Linear: n = 1080 − 240 = 840, revenue = 12 × 840 = £10,080. Exponential: n = 1080 × 0.97^6 = 1080 × 0.83297 = 899.6, revenue = 12 × 899.6 = £10,795.
Difference ≈ £715, about 7% of revenue - enough to change a pricing decision.
The exponential model loses 3% of a shrinking attendance, so its absolute losses fall as attendance drops: it stays optimistic at high prices. The linear model loses a constant 40 per step. Note also that the linear model predicts zero attendance at x = 27 (£22.50), far outside the surveyed £15 ceiling, so neither model can be extrapolated.
Advice: within the surveyed range up to £15 the linear model is the one supported by the survey, so use it, but quote the exponential figure as a sensitivity check. Above £15 the owner has no data and should not price on either model.