Question

At a price of $8.5 per ticket, a musical theater group can fill every seat in their 1800 seat performance hall. For every additional dollar charged for admission, the number of tickets sold drops by 65. a) What ticket price maximizes revenue? Round your answer to the nearest cent.

Answer 1

The ticket price that maximizes revenue for the musical theater group is approximately $18.10. This is found by establishing a revenue function based on a downward-sloping demand curve and using calculus to find the price at which the derivative of the revenue function is zero.

Step-by-step explanation:

To calculate the ticket price that maximizes revenue for a musical theater group, we need to use the demand function that relates the price and quantity sold. We start with the information that at a price of $8.5, all 1800 seats are filled. For each additional dollar charged, 65 fewer tickets are sold. The revenue function, R(p), can be expressed as the product of the number of tickets sold and the price per ticket.

The number of tickets sold can be represented as 1800 - 65(x - 8.5), where x is the new ticket price. We use x - 8.5 since the decline in ticket sales starts after an increase from the initial $8.5 price.

The revenue function is then: R(x) = x * (1800 - 65(x - 8.5)). Expanding this, we get: R(x) = x * (1800 - 65x + 552.5) = 1800x - 65x2 + 552.5x. Now, let's calculate the derivative of R(x) and set it to zero to find the maximum.

The derivative of R(x) is: R'(x) = 1800 - 130x + 552.5. Setting this to zero, we get 130x = 2352.5, resulting in x = 2352.5 / 130 ≈ $18.10. Therefore, the ticket price that maximizes revenue is approximately $18.10.

Raising the ticket price beyond $8.5 decreasing the audience size due to the downward-sloping demand curve, but up to a certain point, the increase in ticket price compensates for the lower number of tickets sold, thus maximizing revenue.

Answer 2

The ticket price that maximizes revenue is $18.10

Step-by-step explanation:

Hi, first we need to construct the revenue equation in terms of the additional dollar charge (that would be X). That is:

`Revenue=Price*Quantity`

`Revenue=(8.5-X)(1,800-65X)`

So we expand it:

`Revenue=15,300-552.5X+1,800X-65X^(2)`

`Revenue=-65X^(2) +1247.5X+15,300`

This is a parabola, and we need to find its vertex, which in our case that would be the maximum additional dollar charge in order to obtain the highest revenue possible, to find the vertex, we need to consider that:

`Y(X)=AX^(2)+ BX+C`

And to find the X-coordenate we have to use the following equation.

`Vertex(X)=(-B)/(2A)`

In our case, A= -65; B= 1,247.5, so, all should look like this:

`Vertex (X)=(-(1247.50)/(2(-65)) =9.6`

That means, we need to make 9.6 increments of $1 in order to obtain the max revenue possible, therefore, the price would be

Price = $8.50 + $1(9.6)= $8.50 + $9.6 =$18.10

Best of luck.