Theatre Pricing: Revenue Maximum versus Profit Maximum
A community theatre has 200 seats. At a ticket price of $80, every seat is sold. A market survey suggests that for each $5 increase in the ticket price, 10 fewer tickets will be sold. The theatre's fixed cost for each performance is $6,000, and each occupied seat costs the theatre an additional $20 in programme, cleaning and staffing costs. The theatre manager wants to choose a ticket price in $5 steps to maximise profit, but a volunteer has calculated the price that maximises revenue and proposes that price instead. The manager is also aware that the number of tickets sold cannot be negative and that the theatre cannot sell more than 200 tickets. Ticket prices must be whole multiples of $5 and attendance must be a whole number of tickets.
1. Let n be the number of $5 increases. Express the revenue and the profit in terms of n, stating the valid values of n.
Read the worked answer
Price = 80+5n; attendance = 200-10n.
For non-negative attendance, 200-10n ≥ 0, so n ≤ 20. Also n ≥ 0 and n is an integer, so n = 0,1,...,20.
Revenue: R(n)=(80+5n)(200-10n)=16000-800n+1000n-50n²=16000+200n-50n².
Cost: C(n)=fixed 6000 + variable 20(200-10n)=6000+4000-200n=10000-200n.
Profit: P(n)=R(n)-C(n)=(16000+200n-50n²)-(10000-200n)=6000+400n-50n².
2. Find the value of n that maximises revenue and the value that maximises profit. Explain why the volunteer's proposal is not necessarily the manager's best choice.
Read the worked answer
R(n)=16000+200n-50n² is a concave quadratic; its vertex is at n=-200/(2×-50)=2. Since 2 is allowed, revenue is maximised at n=2: price $90, attendance 180, R=16200.
P(n)=6000+400n-50n²; its vertex is at n=-400/(2×-50)=4, which is allowed. Profit is maximised at n=4: price $100, attendance 160, P(4)=6000+1600-800=6800.
At n=2, P(2)=6000+800-200=6600. Revenue is higher at n=2 (16200 vs 16000 at n=4), but profit is lower because the extra attendance at the lower price brings extra variable cost. The volunteer maximised revenue, not profit, so the proposal is not necessarily best for the theatre's profit.
3. Check the endpoints n=0 and n=20, and confirm the profit maximum is global on the allowed domain.
Read the worked answer
At n=0: P=6000+0-0=6000. At n=20: attendance=0; P=6000+8000-50(400)=6000+8000-20000=-6000.
The profit function is a concave quadratic (coefficient -50), so on the closed interval [0,20] its maximum is at the vertex if the vertex lies inside. The vertex n=4 lies inside, and P(4)=6800, which exceeds both endpoint values. Hence n=4 is the global maximum on the allowed domain. The theatre should set the price at $80+5×4=$100 and expect 160 tickets sold, giving maximum profit $6,800.