Break-Even With Free Seats
Tickets for a benefit concert cost $12 in advance and $18 at the door. The hall seats 300. Fixed costs are $1,850, and the hall charges $2.10 per person admitted - per person admitted, not per ticket sold, and complimentary admissions count. Advance sales have closed at 180 tickets. Twenty-five seats are reserved as comps for volunteers and performers, who pay nothing and are in addition to the 180 advance holders.
The organisers want the smallest number of door tickets that covers all costs, and they intend to buy no more than that many seats' worth of anything.
1. Compute the break-even number of door tickets, showing the revenue and cost expressions and verifying both sides of the threshold.
Read the worked answer
Let d be door tickets sold.
- Revenue: 180 x $12 + 18d = 2160 + 18d
- Attendees: 180 advance + 25 comps + d = 205 + d
- Cost: 1850 + 2.10(205 + d) = 1850 + 430.50 + 2.10d = 2280.50 + 2.10d
Setting revenue equal to cost:
2160 + 18d = 2280.50 + 2.10d 15.90d = 120.50 d = 7.58
Door tickets are whole, so the minimum is 8.
Check the boundary:
- d = 7: revenue 2160 + 126 = 2286.00; cost 2280.50 + 14.70 = 2295.20; short by $9.20.
- d = 8: revenue 2160 + 144 = 2304.00; cost 2280.50 + 16.80 = 2297.30; ahead by $6.70.
The answer must be the first whole number above the crossing point, not the nearest one - d = 7.58 rounds down to 8 only by luck of the decimal, and a value like 7.4 would still require 8.
2. Justify treating the 25 comps as incurring the $2.10 charge, and quantify the error if they are excluded.
Read the worked answer
They incur the charge. The hall bills per person admitted. A comp is waived revenue, not waived occupancy: the volunteer still occupies a seat, still passes through the door, and still counts in the hall's headcount. Charging rules follow occupancy; pricing rules follow payment. Nothing in the agreement ties the $2.10 to a ticket.
The error if excluded:
- Cost would read 1850 + 2.10(180 + d) = 2228 + 2.10d.
- Break-even: 2160 + 18d = 2228 + 2.10d, so 15.90d = 68, d = 4.28, giving 5 tickets.
That understates the requirement by 3 tickets, and at d = 5 the event actually loses money:
- Revenue 2160 + 90 = 2250
- True cost 1850 + 2.10 x 210 = 2291
- Shortfall $41
Five tickets look like a comfortable margin over a 'break-even' of 4.28, which is exactly why the mistake survives review. The comps are a fixed 25-person cost of 25 x $2.10 = $52.50 that the error drops entirely.
3. The hall adds a $300 surcharge if attendance exceeds 250. Does this change the break-even figure? Show why or why not.
Read the worked answer
It does not change it.
Attendance is 205 + d, which exceeds 250 only when d is at least 46.
Test the surcharge regime over its whole range: at d of 46 or more, break-even requires
2160 + 18d = 1850 + 300 + 2.10(205 + d) 2160 + 18d = 2580.50 + 2.10d 15.90d = 420.50 d = 26.4
So within the surcharge regime, any d of 46 or more is already past break-even - the condition is satisfied with room to spare. Check d = 46: revenue 2160 + 828 = 2988; cost with surcharge 1850 + 300 + 527.10 = 2677.10; ahead by $310.90.
The general point: a step cost that switches on only after you have already cleared the break-even point is irrelevant to break-even, even though it is very relevant to profit. The correct procedure is to solve within each regime and then check whether the solution actually lies inside the regime that produced it - the surcharge never binds because d = 8 lies in the no-surcharge regime.