A theatre company allows a group of 10 people to buy theatre tickets at a price of $28 per person. For each person in excess of 10, the price is decreased by $2 per person for everyone down to a minimum of $10 per person. What number of people will produce the maximum revenue for the theatre company???
PLEASE HELP? I WOULD BE EXTREMELY THANKFUL.
I tried to solve this problem myself, but I don't really understand how to write out the quadratic equation to do so.
I did this:
Let n be he number of tickets sold over 10.
revenue= (price)(quantity)
Revenue=(28x2n)(n)
I know here is a problem with the quadratic equation that I came up with, but I just don't understand how to write out the quadratic equation.
Please help me.
Let n be the number of persons in excess of 10. Then the number of persons is (10+n) and the ticket price is $(28-2n)$ per person.
So, the revenue is given by: R = (28-2n)(10+n)
To find the number of people that would produce the maximum revenue, we need to find when this quadratic function is at its maximum point.
Expanding the equation, we get:
R = 280 - 20n + 2n^2
The maximum (or minimum) point of a quadratic equation is given by the formula:
x = -b/(2a), where a is the coefficient of the squared term and b is the coefficient of the linear term. In our case, a = 2 and b = -20.
So, x = -(-20) / (2*2) = 20/4 = 5
Since x is the number of persons in excess of 10, the total number of people that would produce the maximum revenue is 10+5 = 15 people.
To find the number of people that will produce the maximum revenue for the theatre company, we need to write out the revenue equation and then find its maximum value.
Let's break down the problem:
For the first 10 people, the price per person is $28, so the revenue for these 10 people is:
Revenue1 = 10 * 28 = 280
For each additional person beyond the initial 10, the price decreases by $2 per person until it reaches a minimum of $10 per person.
If we let n be the number of people beyond the initial 10, then the price per person is given by:
Price per person = 28 - 2n
However, the price per person cannot go below $10, so we need to take the maximum of either 10 or (28 - 2n). We can express this as:
Price per person = max(10, 28 - 2n)
The total number of people, including the initial 10, will be:
Total number of people = 10 + n
Now, we can calculate the revenue for the additional n people using the price per person:
Revenue2 = (max(10, 28 - 2n)) * n
To find the total revenue, we add up the revenue from the first 10 people and the revenue from the additional n people:
Total Revenue = Revenue1 + Revenue2 = 280 + (max(10, 28 - 2n)) * n
To find the number of people that will produce the maximum revenue, we need to find the value of n that maximizes the Total Revenue equation.
To do that, we can differentiate the Total Revenue equation with respect to n and set it equal to zero:
d(Total Revenue)/dn = (max(10, 28 - 2n)) * (1) + n * (-2) = 0
Simplifying this equation:
max(10, 28 - 2n) - 2n = 0
Now, we can solve this equation to find the value of n that maximizes the Total Revenue.
To find the number of people that will produce the maximum revenue for the theatre company, you can start by analyzing the problem step by step.
Let's break down the pricing structure:
For the first 10 people, each ticket costs $28.
For each additional person beyond the initial 10, the price decreases by $2 for everyone.
However, the price cannot drop below $10 per person.
Based on this, we need to determine the optimal number of additional people to maximize revenue.
Let's denote the number of additional people as x.
We can calculate the price per person for these additional people using the given pricing structure:
Price per person = $28 - $2x (as the price decreases by $2 for each additional person)
However, this price cannot go below $10.
So the actual price per person becomes: min($28 - $2x, $10)
Now, we need to calculate the revenue generated by these additional people.
Revenue = (number of additional people) * (price per person)
Revenue = x * min($28 - $2x, $10)
To maximize the revenue, we need to find the value of x that maximizes the above equation.
One way to approach this is to plot a graph of revenue as a function of x, and find the peak of the graph. However, since you mentioned you're looking for a quadratic equation, we can rewrite the revenue equation as a quadratic function.
Consider the revenue equation:
Revenue = x * min($28 - $2x, $10)
To simplify, let's consider two cases:
Case 1: When $28 - $2x < $10:
In this case, revenue equation becomes: Revenue = x * ($28 - $2x)
Case 2: When $28 - $2x >= $10:
In this case, revenue equation becomes: Revenue = x * $10
We can combine these cases using the "max" function:
Revenue = x * max($10, $28 - $2x)
Now, we have the revenue equation in terms of x. This equation represents a quadratic function, which we can further simplify:
Revenue = x * ($28 - $2x) = 28x - 2x^2 for Case 1.
Revenue = x * $10 = 10x for Case 2.
Now we can compare the two cases to find which case produces a higher revenue. The case with the higher revenue represents the answer to the question.
To solve this quadratic equation, we can set the derivative of the equation equal to zero to find the maximum value. The derivative of the revenue equation is:
d(Revenue)/dx = 28 - 4x
Setting the derivative equal to zero and solving for x gives us:
28 - 4x = 0
4x = 28
x = 7
So, in this case, the derivative gives us x = 7, which represents the number of additional people, beyond 10, that will maximize the revenue for the theatre company.
Therefore, the total number of people that will produce the maximum revenue for the theatre company is:
Number of people = 10 + (number of additional people) = 10 + 7 = 17.
Thus, a group of 17 people will produce the maximum revenue for the theatre company.