Ticket Pricing: Revenue Maximum Is Not Profit Maximum
A community theater seats 1,200 people. At a ticket price of $20, the box office sells 800 tickets for a typical Friday show. Past seasons show a consistent pattern: for each $1 increase in price, attendance falls by 40 people, and the pattern holds for prices from $15 to $40. Revenue equals price times tickets sold. The theater's fixed cost for a Friday show is $9,000, and in addition the theater pays a ticketing service $1.50 for every ticket actually sold. The board has asked the director to raise ticket prices, arguing that a higher price must mean more money. The director wants numbers before agreeing, and prices are set in whole dollars only.
1. Let x be the number of $1 price increases above $20. Write revenue R as a function of x and find the price that maximizes revenue.
Read the worked answer
Price: p = 20 + x. Attendance: 800 − 40x.
R(x) = (20 + x)(800 − 40x) = 16,000 − 800x + 800x − 40x² = 16,000 − 40x²
The linear terms cancel, so R is a downward parabola with its vertex at x = 0. Revenue is maximized at the current price of $20, with R = $16,000.
Any increase at all reduces revenue: each $1 raise costs 40 attendees × $20 = $800 in lost base revenue while adding only $1 × 760 = $760. The claim that a higher price means more money is false here.
2. Compare revenue at $22 and at $25. How much money does the theater lose by choosing $25 instead of $22?
Read the worked answer
At $22: x = 2, attendance = 800 − 80 = 720, revenue = 22 × 720 = $15,840. At $25: x = 5, attendance = 800 − 200 = 600, revenue = 25 × 600 = $15,000.
Difference: 15,840 − 15,000 = $840 lost by pricing at $25 rather than $22. (Relative to the $20 optimum, $25 gives up $1,000.)
Note the useful shortcut from the formula: R(2) = 16,000 − 40(4) = 15,840 and R(5) = 16,000 − 40(25) = 15,000. Both checks agree.
3. Profit P equals revenue minus the $9,000 fixed cost minus $1.50 per ticket sold. Using whole-dollar prices, which price maximizes profit, and why is it not the revenue-maximizing price?
Read the worked answer
P(x) = (16,000 − 40x²) − 9,000 − 1.5(800 − 40x) = 16,000 − 40x² − 9,000 − 1,200 + 60x = 5,800 + 60x − 40x²
Vertex at x = −60 ÷ (2 × −40) = 0.75, so the continuous optimum sits between $20 and $21. Whole-dollar check:
- x = 0 ($20): 5,800
- x = 1 ($21): 5,800 + 60 − 40 = 5,820
- x = 2 ($22): 5,800 + 120 − 160 = 5,760
$21 maximizes profit at $5,820.
Why not $20? Raising the price by $1 loses 40 sales, but each lost sale also saves $1.50 in ticketing fees, so the profit penalty per lost sale is only $18.50, not $20. Revenue falls by $40, yet fees fall by $60, so profit rises by $20. Profit and revenue peak at different prices because costs depend on the number of tickets.