Answer
The ticket price to maximize revenue would be $11.50 per ticket, with 11,500 spectators, 132,250 dollars made.
Work Step by Step
Revenue is equal to ticket cost times spectators at the game (amount of tickets bought), we want to maximize revenue.
Let $R$ be revenue, $T$ being tickets bought, and $S$ being spectators.
$R=T*S$
The price of tickets is inversely proportional to the number of spectators, with every dollar increase yielding 1,000 less spectators.
Let $X$ be the discount of tickets in dollars.
$T=12-x$
$S=11,000+1,000x$
$R=(12-x)*(11,000+1,000x)$
$R=132,000+1,000x-1,000x^{2}$
Maximum possible price of tickets would be 24 (zero spectators), minimum would be 8 (maximum amount of spectators)
$8\leq T\leq24$
$-4\leq X\leq12$
$R′(x)=1,000-2,000x$
Find when $R′(x)=0$
$R′(x)=1,000=2,000x$
$R′(x)=\frac{1,000}{2,000x}$
$R′(x)=\frac{1}{2}x$
When $-4\leq x \lt \frac{1}{2}$, $R′(x)>0$
When $12 \geq x \gt \frac{1}{2} $, $R′(x)<0$
First Derivative Test gives $R(\frac{1}{2})$ as an absolute maximum.
Maximized Revenue can be found via plugging in $\frac{1}{2}$ into $R$
$R=132,000+1,000(\frac{1}{2})-1,000(\frac{1}{2})^{2}=132,250$
Plug $X$ back in to $T$ and $S$ to find the ticket price and spectators, respectively
$T=12-\frac{1}{2}-11.5$
$S=11,000+1,000(\frac{1}{2})=11,500$
Revenue is maximized when tickets are $11.50, where 11,500 spectators attend the game.
