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???

Question

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.

Answer 1

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.