  ### Home > CALC > Chapter 5 > Lesson 5.3.1 > Problem5-115

5-115. Ms. Platinum has a hit on her hands. A survey she commissioned showed that she could sell $100$ seats to her smashing new musical for $40$ each, and for each $1$ she drops the price, she could sell $10$ more seats. Too bad the theater she rented only seats $200$! How much should she charge for tickets to maximize her revenue?

Before we even start, notice that there is boundaries to the domain of this story. There are only $200$ seats to rent! We will return to that at the end of the problem.

So, we have not answered the question. Sometimes, when a function has a restricted domain, the maximums is NOT where the derivative equals $0$, but at an endpoint (where the derivative does not exist). Go back and determine how much Ms. Platinum will need to charge to sell $200$ tickets.

Let $x =$ number of times she decreases the price. Write a Revenue function, $R(x)$.
Revenue $=$ (seats sold)(price per seat)
$R(x) =$ ($100 + 10x$)($40$ minus: $1x$)

Maximize the revenue.
$R^\prime(x) = 10(40 − x) − 1(100 + 10x)$ Product Rule
$= 300 − 20x$
$0 = 300 − 20x$
$x = 15$

What does this mean?
She should charge $40 −15$ dollars, or $25$ and she will sell $100 + 10(15)$ tickets $= 450$ tickets
But this is impossible because the theater only has $200$ seats.