Question

an average of people visit riverside park each day in the summer. the park charges $ for admission. consultants predict that for each $ increase in the entrance price, the park would lose an average of customers per day. (a) express the daily revenue from ticket sales, as a function of the number of $ price increases, . (b) what ticket price maximizes the revenue from ticket sales? $ (round to nearest cent)

Answer 1

The daily revenue from ticket sales, R, as a function of the number of $1.00 price increases, x is;

R = 1,080,000 - 40,500x - 40
`x^2`

The ticket price that maximizes the revenue from ticket sales is approximately $533.

How to solve the problem

To express the daily revenue from ticket sales, R, as a function of the number of $1.00 price increases, x, consider the relationship between the entrance price and the number of customers.

Given that the average number of customers decreases by 1,500 for each $1.00 increase in the entrance price, express the number of customers, N, as a function of x:

N = 40,000 - 1,500x

The revenue from ticket sales, R, is calculated by multiplying the entrance price by the number of customers:

R = (40,000 - 1,500x) * (27 + x)

Simplifying the expression:

R = (40,000 - 1,500x) * (27 + x)

R = 1,080,000 - 40,500x - 40
`x^2`

Now, we have expressed the daily revenue from ticket sales, R, as a function of the number of $1.00 price increases, x.

To find the ticket price that maximizes the revenue, we need to find the value of x that maximizes the function R(x).

One way to determine the maximum value is by finding the vertex of the quadratic function. The x-coordinate of the vertex can be calculated using the formula:

x = -b / (2a)

In this case, a = -40 and b = -40,500.

x = -(-40,500) / (2 * -40)

x = 40,500 / 80

x = 506.25

Since x represents the number of $1.00 price increases, it cannot be a fraction. Therefore, we round it to the nearest whole number:

x ≈ 506

To find the corresponding ticket price, we substitute x = 506 into the expression:

Price = 27 + x

Price ≈ 27 + 506

Price ≈ 533

Therefore, the ticket price that maximizes the revenue from ticket sales is approximately $533.

An average of 40,000 people visit Riverside Park each day in the summer. The park charges $27.00 for admission. Consultants predict that for each $1.00 increase in the entrance price, the park would lose an average of 1, 500 customers per day. Express the daily revenue from ticket sales, R as a function of the number of $1.00 price increases, x. R = f(x) = What ticket price maximizes the revenue from ticket sales? $ (round to nearest cent)