Answer :
To maximize revenue, the ticket price should be set at $7.40.
To find the demand function, we need to determine the relationship between the ticket price and the number of spectators. We are given two data points:
At a ticket price of $10, the average attendance is 24,000.
At a ticket price of $9, the average attendance is 29,000.
Let's assume that the demand function is linear, which means it can be represented by the equation p(x) = mx + b, where p(x) is the ticket price, x is the number of spectators, m is the slope, and b is the y-intercept.
Using the given data points, we can create two equations to solve for m and b:
When p(x) = 10 and x = 24,000:
10 = 24,000m + b
When p(x) = 9 and x = 29,000:
9 = 29,000m + b
We can now solve this system of equations to find the values of m and b.
Subtracting the second equation from the first equation, we get:
10 - 9 = 24,000m + b - (29,000m + b)
1 = -5,000m
Dividing both sides by -5,000:
m = -1/5,000
Now, substituting the value of m into one of the original equations, let's use the first equation:
10 = 24,000(-1/5,000) + b
Simplifying:
10 = -4.8 + b
b = 14.8
Therefore, the demand function p(x) is:
p(x) = (-1/5,000)x + 14.8
To maximize revenue, we need to find the ticket price that will generate the maximum revenue. Revenue can be calculated by multiplying the ticket price by the number of spectators.
Let's denote the revenue as R(x), where x is the number of spectators. Using the demand function p(x) = (-1/5,000)x + 14.8, we can express R(x) as:
R(x) = p(x) * x
R(x) = [(-1/5,000)x + 14.8] * x
R(x) = (-1/5,000)x² + 14.8x
To find the maximum revenue, we can take the derivative of R(x) with respect to x and set it equal to zero:
dR(x)/dx = (-2/5,000)x + 14.8
dR(x)/dx = 0
Solving for x:
(-2/5,000)x = -14.8
x = (5,000/2) * 14.8
x = 37,000
The maximum revenue occurs when the number of spectators is 37,000. To find the corresponding ticket price, we can substitute x = 37,000 into the demand function:
p(x) = (-1/5,000) * 37,000 + 14.8
p(x) = -7.4 + 14.8
p(x) = 7.4
Therefore, to maximize revenue, the ticket price should be set at $7.40.
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