Talent Show Ticket Sales: System of Inequalities
The student council is selling tickets to a talent show in an auditorium that seats at most 250 people. Student tickets cost $5 each and adult tickets cost $8 each. The council must sell at least 60 adult tickets to satisfy a sponsor, and it wants ticket revenue of at least $1,600. Let x be the number of student tickets and y the number of adult tickets. A council member proposes selling 150 student tickets and 90 adult tickets and claims this meets every condition. Another member wants to fill all 250 seats and still meet the revenue goal, and asks what mix of tickets would allow the greatest possible number of student tickets. The council also needs to explain to the sponsor why some proposed ticket counts are not feasible, using the inequalities rather than guesswork.
1. Write a system of inequalities that represents the ticket-sale conditions. Define each variable and explain what each inequality means in context.
Read the worked answer
Let (x) be the number of student tickets and (y) the number of adult tickets. Both must be whole numbers, so (x\ge 0) and (y\ge 0).
The auditorium seats at most 250 people:
[x+y\le 250.]
The sponsor requires at least 60 adult tickets:
[y\ge 60.]
Revenue must be at least $1,600. Student tickets bring in (5x) dollars and adult tickets bring in (8y) dollars:
[5x+8y\ge 1600.]
So the system is
[\begin{cases}x+y\le 250\ y\ge 60\ 5x+8y\ge 1600\ x\ge 0,\ y\ge 0\end{cases}]
with (x) and (y) integers. A feasible ticket plan must satisfy all of these at once.
2. Test the proposal \(x=150\), \(y=90\). Does it meet every condition? If not, explain which condition fails and whether simply adding more student tickets could fix it while keeping \(y=90\).
Read the worked answer
Check each condition.
Seating: (150+90=240\le 250), so the seating limit is met.
Adult tickets: (90\ge 60), so the sponsor condition is met.
Revenue: (5(150)+8(90)=750+720=1470). This is less than 1600, so the revenue condition fails by $130.
Keeping (y=90), the revenue inequality becomes (5x+720\ge 1600), so (5x\ge 880) and (x\ge 176). But (176+90=266>250), which exceeds the seating limit. Therefore, adding student tickets alone cannot fix the proposal while keeping 90 adult tickets; the council would need more adult tickets or a different combination.
3. If all 250 seats are sold, find the minimum number of adult tickets needed to meet the $1,600 revenue goal, and the corresponding number of student tickets. Show the algebra and check the result.
Read the worked answer
If all 250 seats are sold, then (x+y=250), so (x=250-y). Substitute into the revenue inequality:
[5(250-y)+8y\ge 1600.]
Simplify:
[1250-5y+8y\ge 1600] [1250+3y\ge 1600] [3y\ge 350] [y\ge 116.\overline{6}.]
Since (y) must be a whole number, the minimum number of adult tickets is 117. Then (x=250-117=133) student tickets.
Check: revenue is (5(133)+8(117)=665+936=1601), which is at least 1600, and the sponsor condition (117\ge 60) is met. If (y=116), then (x=134) and revenue is (5(134)+8(116)=670+928=1598), which is short of 1600. So the minimum adult count is 117, with 133 student tickets, and that gives the greatest possible number of student tickets when all 250 seats are sold.