Theater Ticket Mix
A community theater has 220 seats. Adult tickets cost $12 and student tickets $7. The theater needs at least $1,800 in ticket revenue to cover production costs. To keep the show accessible, the board requires that the number of student tickets sold be at least twice the number of adult tickets. The box office manager proposes selling 60 adult tickets and 160 student tickets, arguing that this fills every seat and raises enough money. Before approving the plan, the board asks you to model the constraints, determine the full range of adult ticket numbers that can work when all 220 seats are sold, and evaluate the manager's proposal.
1. Write the inequalities for seats, revenue, and the student-to-adult ratio. Then test whether the manager's proposal of 60 adult and 160 student tickets satisfies all constraints.
Read the worked answer
Let a = adult tickets and s = student tickets. Constraints:
- Seats: a + s ≤ 220
- Revenue: 12a + 7s ≥ 1800
- Access: s ≥ 2a
- a, s are nonnegative integers.
Proposal (60, 160): seats 60 + 160 = 220 ≤ 220; revenue 12(60) + 7(160) = 720 + 1120 = 1840 ≥ 1800; access 160 ≥ 2(60) = 120. The proposal satisfies all three constraints.
2. Assuming all 220 seats are sold, find the minimum and maximum number of adult tickets that satisfy both the revenue and ratio constraints. Verify the endpoints.
Read the worked answer
With all seats sold, s = 220 − a. Revenue: 12a + 7(220 − a) = 1540 + 5a ≥ 1800, so 5a ≥ 260 and a ≥ 52. Access: 220 − a ≥ 2a, so 220 ≥ 3a and a ≤ 73.33; since a is an integer, a ≤ 73. Thus 52 ≤ a ≤ 73.
Check endpoints: a = 52 gives s = 168, revenue 624 + 1176 = 1800, and 168 ≥ 104. a = 73 gives s = 147, revenue 876 + 1029 = 1905, and 147 ≥ 146. a = 74 gives s = 146, but 146 < 148, so access fails.
3. Suppose only 200 tickets are sold. Using s = 200 − a, show whether any adult ticket number satisfies both the revenue and ratio requirements. If none does, find the smallest total number of tickets that would allow both constraints to be met.
Read the worked answer
With 200 tickets, s = 200 − a. Revenue: 12a + 7(200 − a) = 1400 + 5a ≥ 1800, so a ≥ 80. Access: 200 − a ≥ 2a, so a ≤ 66.67, or a ≤ 66. No integer a is both ≥ 80 and ≤ 66, so no 200-ticket mix meets both requirements.
Find smallest total N. Need a ≥ (1800 − 7N)/5 and a ≤ N/3. Require (1800 − 7N)/5 ≤ N/3: 5400 − 21N ≤ 5N, so 5400 ≤ 26N, N ≥ 207.69, so N ≥ 208. At N = 208, a ≥ 68.8 → 69 and a ≤ 69.33 → 69, so a = 69, s = 139 works: revenue 828 + 973 = 1801 and 139 ≥ 138.