a 6,000 seat theater has tickets for sale at $26 and $40 . How many tickets must be sold at each price for a sell out performance to generate a revenue of $178400?
Well, I have to say, this sounds like quite the "theater-ical" problem! Let's break it down and see what we can come up with.
Let's say we sell x tickets at $26 and y tickets at $40. Since there are 6,000 seats in total, we must have:
x + y = 6000
Now, given that the revenue is $178,400, we can set up an equation for that as well. The revenue from the $26 tickets would be 26x, and the revenue from the $40 tickets would be 40y. So, we have:
26x + 40y = 178400
Now, we can use these two equations to solve for x and y. I'd be happy to assist you further once you have the values of x and y. Trust me, it's going to be a "theater-rific" solution!
Let's assume x represents the number of tickets sold at $26 and y represents the number of tickets sold at $40.
We know that the total number of tickets sold must add up to 6,000, so we have the equation:
x + y = 6,000 (Equation 1)
We also know that the revenue from selling each ticket at $26 is $26x, and the revenue from selling each ticket at $40 is $40y. So, the total revenue is:
26x + 40y = 178,400 (Equation 2)
Now, we can solve this system of equations to find the values of x and y.
From Equation 1, we can isolate x:
x = 6,000 - y
Substituting this value of x into Equation 2, we have:
26(6,000 - y) + 40y = 178,400
Simplifying the equation:
156,000 - 26y + 40y = 178,400
14y = 22,400
y = 1,600
Substituting the value of y back into Equation 1, we find:
x + 1,600 = 6,000
x = 6,000 - 1,600
x = 4,400
Therefore, to generate a revenue of $178,400 for a sold-out performance, 4,400 tickets must be sold at $26 and 1,600 tickets must be sold at $40.
To find out how many tickets must be sold at each price, let's assign variables to represent the number of tickets sold at $26 and $40.
Let's say 'x' represents the number of tickets sold at $26, and 'y' represents the number of tickets sold at $40.
According to the given information, the total number of tickets sold should be sufficient to fill all 6,000 seats. Therefore, the equation for the total number of tickets sold is:
x + y = 6000 ----(1)
Now, let's calculate the revenue generated from each ticket price. The revenue from the $26 tickets would be 26x, and the revenue from the $40 tickets would be 40y. Since the total revenue required is $178,400, we can write:
26x + 40y = 178400 ----(2)
To solve this system of equations, we can use different methods such as substitution or elimination. Let's use the substitution method:
From equation (1), we can isolate the value of 'x' in terms of 'y':
x = 6000 - y
Now, substitute the value of 'x' in equation (2):
26(6000 - y) + 40y = 178400
Simplify and solve for 'y':
156000 - 26y + 40y = 178400
14y = 22400
y = 1600
Now substitute the value of 'y' back into equation (1) to find 'x':
x + 1600 = 6000
x = 6000 - 1600
x = 4400
Therefore, to generate a revenue of $178400, 4400 tickets must be sold at $26 each and 1600 tickets must be sold at $40 each.
work with seats:
If there are n seats at $26, then there are 6000-n seats at $40, if the place is sold out.
work with the money -- add up how much money is made for each ticket price.
26n + 40(6000-n) = 178400
26n + 240000 - 40n = 178400
14n = 61600
n = 4400
So, there were 4400 $26 seats and 1600 $40 seats