The greatest revenue? A theatre seats 2000 people and charges $10 for a ticket. At this price all the tickets can be sold. A survey indicates that if the ticket price is increased, the number sold will decrease by 100 for every dollar of increase. What ticket price would result in the greatest revenue?
 A: Hint: If we have $x$ dollars of increase, we get: $$(2000-100x)(10+x)$$
We are attempting to maximize this value.
A: Hint. Let the price be increased by $x$ dollars. Then the number of people would be $2000-100x$, and the price per ticket would be $10+x$. What would be the revenue as a function of $x$? 
A: To solve this problem, you need to use some simple calculus. Let $x$ be the price of the ticket. When $x=10$, we know that $2000$ seats are sold and that $100$ fewer seats are sold for every dollar increase in price. This nets us the equation:
seats sold $= 2000-100(x-10)$
If we multiply this by the price of the ticket, we can find the total revenue $R$:
$R=x(2000-100(x-10))$
$=2000x-100x(x-10)$
$=2000x-100x^{2}+1000x$
$=3000x-100x^2$
Since we're trying to find a maximum for this expression, we can run this function for revenue through any monotonically increasing analytic function and their maximum will occur at the same value of $x$. Therefore, we can divide the equation by $100$ to find:
$30x-x^{2}$
Taking the derivative with respect to $x$, we have:
$30-2x$
... and setting this to zero, we have:
$0=30-2x$
$2x=30$
$x=15$
Lastly, we can verify that this is a maximum and not a minimum because the second derivative of our revenue function is negative.
Therefore, you'll want to price your tickets at $15.
