At a price of $7.5 per ticket, a musical theater group can fill every seat in their 1500 seat performance hall. For every additional dollar charged for admission, the number of tickets sold drops by 95.
a) What ticket price maximizes revenue? Round your answer to the nearest cent.
price = $
b) How many seats are sold at that price? Round your answer to the nearest whole number.
number of seats sold =
To find the ticket price that maximizes revenue and the number of seats sold at that price, we need to analyze the relationship between ticket price, number of tickets sold, and revenue.
Let's break down the information given step by step:
1) At a price of $7.5 per ticket, the theater group can fill every seat in their 1500 seat performance hall. This means that if they charge $7.5 per ticket, they will sell all 1500 tickets.
2) For every additional dollar charged for admission, the number of tickets sold drops by 95. This means that as the ticket price increases by $1, the number of tickets sold decreases by 95.
To start solving the problem, we can create equations to represent the relationship between ticket price, number of tickets sold, and revenue. Let's define the variables:
- P: Ticket price in dollars
- N: Number of tickets sold
- R: Revenue
Equations:
1) N = 1500 - 95*(P - 7.5)
- This equation shows the relationship between the ticket price and the number of tickets sold. Starting with a full capacity of 1500, for each dollar increase in price, 95 tickets are deducted from the total.
2) R = P * N
- This equation calculates the revenue by multiplying the ticket price (P) by the number of tickets sold (N).
Now, let's find the ticket price that maximizes revenue (a) and the number of seats sold at that price (b).
a) Maximizing Revenue:
To find the ticket price that maximizes revenue, we need to find the value of P that maximizes R. We can do this by finding the critical points of the revenue equation.
Taking the derivative of the revenue equation with respect to P:
dR/dP = N + P * dN/dP
- dN/dP represents the rate at which the number of tickets sold changes with respect to the ticket price.
Setting dR/dP = 0 and solving for P:
N + P * dN/dP = 0
1500 - 95*(P - 7.5) + P * (-95) = 0
1500 - 95P + 712.5 - 95P = 0
225P = 2187.5
P = 9.72
Therefore, the ticket price that maximizes revenue is $9.72.
b) Number of Seats Sold at that Price:
To find the number of seats sold at the price of $9.72, we substitute this value into the equation for N:
N = 1500 - 95*(P - 7.5)
N = 1500 - 95*(9.72 - 7.5)
N = 1500 - 95*(2.22)
N = 1500 - 209.9
N ≈ 1290.1
Rounding to the nearest whole number, the number of seats sold at the price of $9.72 is 1290.
Therefore, the answers to the questions are:
a) The ticket price that maximizes revenue is $9.72.
b) The number of seats sold at that price is approximately 1290.