Ticket Pricing at the School Play
A drama club sells adult tickets for $12 and student tickets for $7. The auditorium seats 240 people, and the club needs at least $2,000 in ticket revenue to cover production costs. On opening night they sold 150 tickets total, of which 60 were student tickets. The advisor wants to know how many more adult tickets than student tickets they can sell across the remaining capacity while still meeting the revenue goal, assuming the same price structure and that every seat need not be filled.
1. Write a system of inequalities for a full-house scenario with a adult and s student tickets, and state the revenue inequality.
Read the worked answer
Capacity: $a + s \le 240$ with $a \ge 0, s \ge 0$. Revenue: $12a + 7s \ge 2000$. Solving the boundary $12a + 7s = 2000$ together with $a + s = 240$: from $s = 240 - a$, $12a + 7(240 - a) = 2000 \Rightarrow 5a + 1680 = 2000 \Rightarrow a = 64$, so $s = 176$. The feasible region is the segment of the capacity line with $a \ge 64$.
2. Opening night sold 150 tickets (60 student). What is the minimum number of additional adult-only tickets needed to reach $2,000?
Read the worked answer
Opening revenue: $12(90) + 7(60) = 1080 + 420 = 1500$. Shortfall: $2000 - 1500 = 500$. Each additional adult ticket adds $12$: $500/12 \approx 41.67$, so 42 adult tickets are needed. Check: $1500 + 12(42) = 2004 \ge 2000$. Also confirm capacity: $150 + 42 = 192 \le 240$, so it fits.
3. If they instead sell only student tickets for the remaining seats, can they still reach $2,000? Justify with an inequality.
Read the worked answer
Remaining seats: $240 - 150 = 90$. Max student-only revenue: $1500 + 7(90) = 1500 + 630 = 2130 \ge 2000$, so yes it is possible. But the per-ticket margin is lower, so the number needed is $500/7 \approx 71.4 \Rightarrow 72$ student tickets. Since $72 \le 90$, the goal is achievable with student tickets alone, though it consumes more of the remaining capacity.