For an outdoor concert, a ticket price of $30 typically attracts 5000 people. For each $1 increase in the ticket price, 100 fewer people will attend. The revenue, R, is the product of the number of people attending and the price per ticket.
a) Let x represent the number of $1 price increases. Find an equation expressing the total revenue in terms of x.
b) State any restrictions on x. Can x be a negative number? Explain.
c) Find the ticket price that maximizes
revenue.
I have already got a and b. I only need help in c).
I hope your function looks something like this:
R(x) = (5000 - 100x)(30 + x)
= 150000+ 5000x - 3000x - 100x^2
= -100x^2 + 2000x + 150000
this is a standard parabola opening dowwards , so it will have a maximum
the x of the vertex is -b/(2a) = -2000/-200 = 10
So there should be 10 increases of $1, each ticket costing $40.
number of tickets sold = 4000
cost of ticket = 40
revenue = 40(4000) = 160000
thx!
c) To find the ticket price that maximizes revenue, we need to find the value of x that corresponds to the maximum revenue.
Let's start by determining the revenue equation in terms of x. From part a), we found that for each $1 increase in price, 100 fewer people will attend. Therefore, the number of people attending can be represented as:
P(x) = 5000 - 100x
The price per ticket can be represented as: $30 + $1x = 30 + x
The revenue equation is the product of the number of people attending and the price per ticket:
R(x) = P(x) * (30 + x)
= (5000 - 100x) * (30 + x)
To find the ticket price that maximizes revenue, we need to find the value of x that makes R(x) reach its maximum value. We can do this by analyzing the behavior of the revenue equation.
Taking the derivative of R(x) with respect to x:
R'(x) = 5000 - 100x - 100x + 5000 - 100x <- Distributed and simplified
= -200x + 10000
Setting R'(x) equal to zero and solving for x:
-200x + 10000 = 0
-200x = -10000
x = -10000 / -200
x = 50
Therefore, x = 50 corresponds to the maximum revenue.
Now that we have the value of x, we can find the ticket price by substituting x = 50 into the price per ticket equation:
Price per ticket = 30 + x
= 30 + 50
= $80
So, the ticket price that maximizes revenue is $80.
To find the ticket price that maximizes revenue, we need to determine the value of x that will result in the highest revenue.
Let's start by setting up the equation for revenue, R, in terms of x. We know that each $1 increase in ticket price leads to a decrease of 100 people attending, so the total number of people attending can be represented as 5000 - 100x.
The ticket price, P, is given as $30 + $1 per increase, so we can represent it as P = 30 + x.
Now, to find the total revenue, we multiply the number of people attending by the ticket price: R = (5000 - 100x)(30 + x).
Next, we want to find the value of x that maximizes revenue. We can do this by finding the x-value that corresponds to the vertex of the quadratic equation R = (5000 - 100x)(30 + x).
The x-coordinate of the vertex can be found using the formula: x = -b/2a, where a and b are the coefficients of the quadratic equation in standard form.
For our equation R = (5000 - 100x)(30 + x), the coefficients are a = -3000 and b = 5000.
Plugging these values into the formula, we get: x = -5000/(2*(-3000)) = -5000/(-6000) = 5/6.
However, we need to consider the restrictions on x. From part b, we know that the number of $1 price increases, x, cannot be negative. Therefore, x = 5/6 is not a valid solution.
In this case, there is no maximum ticket price that will maximize the revenue. The highest revenue would be obtained at the last valid x-value, which is x = 0 (no price increase).
To find the ticket price that maximizes revenue, we need to first express the revenue function in terms of x.
Let's start by understanding the relationship between the ticket price and the number of attendees.
We know that for each $1 increase in the ticket price, 100 fewer people attend. So, if the original ticket price is $30 and we increase it by x dollars, the corresponding number of attendees can be expressed as (5000 - 100x).
Now, let's express the revenue function in terms of x. Revenue (R) is the product of the number of attendees and the price per ticket, which is:
R = (Price per ticket) * (Number of attendees)
R = (30 + x) * (5000 - 100x)
Now, to find the ticket price that maximizes revenue, we need to find the value of x that maximizes the revenue function R = (30 + x) * (5000 - 100x). We can do this by finding the vertex of the quadratic function.
The vertex of a quadratic function in the form of f(x) = ax^2 + bx + c can be found using the formula: x = -b / (2a). In our case, a = -100, b = 3500.
x = -3500 / (2 * -100)
x = 35
So, x = 35 represents the number of $1 increases in the ticket price that maximizes revenue.
Now, let's find the corresponding ticket price. The original ticket price is $30, and for each $1 increase, the ticket price increases. Therefore, the ticket price that maximizes revenue is:
Ticket price = $30 + (35 * $1)
Ticket price = $65
Therefore, the ticket price that maximizes revenue is $65.