Solving for Break-Even Ticket Price
A school club is selling tickets to a fundraiser. It must pay a flat $240 for the venue and $3 per attendee for refreshments. The club charges a single ticket price p dollars. The club treasurer writes the profit as P = p·n − (240 + 3n), where n is the number of attendees. The club expects 80 attendees. The principal requires the club to raise at least $600 in profit, and the ticket price must be a whole-dollar amount.
1. Write and solve an inequality to find the minimum whole-dollar ticket price that meets the $600 goal at 80 attendees.
Read the worked answer
Set P ≥ 600 with n = 80: p·80 − (240 + 3·80) ≥ 600 80p − (240 + 240) ≥ 600 80p − 480 ≥ 600 80p ≥ 1080 p ≥ 13.5 Since price must be a whole dollar, round up to p = $14. Check: 14·80 − 480 = 1120 − 480 = $640 ≥ $600 ✓. At $13: 13·80 − 480 = $560, short of $600.
2. The treasurer instead wants to know the break-even price (profit exactly $0) at 80 attendees, and how many attendees are needed to break even if the price is set at $8.
Read the worked answer
Break-even means P = 0. At n = 80: 80p − 480 = 0 → 80p = 480 → p = $6.00. At p = $8: 8n − (240 + 3n) = 0 → 5n − 240 = 0 → 5n = 240 → n = 48 attendees. Interpretation: each attendee contributes p − 3 = $5 of margin toward the $240 fixed cost, so 240 ÷ 5 = 48 attendees covers the venue. This shows the break-even price falls as attendance rises, which is why the club should not confuse break-even with the profit goal.