Questions: A theatre rents a film for five showings. The rental fee is five hundred dollars. The theatre also has $100 in labor costs for each showing. The theatre is setting a ticket price of $7.00 and keeps all revenue from the ticket sale. The theatre has a concession stand. Half of the customers buy a treat and half do not. Each customer at the concession stand spends $9.00. The theatre keeps 70% of the concession purchase. (The rest covers the cost of the purchase over the already counted labor costs.)
How many ticket sales would result in this theatre breaking even?
How many ticket sales would lead the theatre to break even if it had no concession stand?
Answer: Total costs are $1,000 ($500 for the rental fee and 5 X $100 for the labor costs.)
The revenue received from the showing of the movies is
($7.00 X T) + (0.5 X $9.00 X 0.70 X T).
This simplifies to $10.15 X T.
The profit function is Profit = 10.15 X T -1000
Setting profit equal to 0 we find the break-even point
T=(1000/10.15) = 98.5
Since there is no such thing as half a customer the theater needs 99 people to go to the movie in order to break even.
An average of 20 views per viewing (there are 5 viewing) give us 100 customers, one to spare.
Without the concession stand the profit function is
Profit = 7T-1000
Setting profits equal to 0 we get a break even point of
Basically the theater needs 143 customers to break even without a concession stand.
Comment: In this example the theatre paid a fixed fee for the movie and was allowed to keep 100% of the movie ticket revenue. Most contracts between theaters and movie companies give the theater only a small take on each movie sale; hence, in the typical situation the reliance on concession revenue is much larger than depicted in this problem.
This is my second post on theaters, popcorn and soda. The first post mentioned the economic literature on the topic and presented a simple revenue function.
First post on movie tickets popcorn and soda:
Tomorrow I will create a post where the number of treat sales is variable and uncertain. It is a good problem for students of probability and statistics.