Answer
a. $f(x)=-3000x^2+57000x$
b. The maximum revenue is $270750$ and it is when $x=9.50$ ticket is sold.
c. the ticket price of $x=19$ is so high that no revenue is generated.
Work Step by Step
a. A function that models the revenue in terms of the ticket price is $f(x)=-3000x^2+57000x$
b. To find the maximum revenue from a ticket price we can turn an equation from a standard form into a vertex form by completing the square.
thus, to complete the square,
$f(x)=-3000(x^2-19x)$,
$f(x)=-3000(x^2-19x+(\frac{19}{2})^2)$, To balance an equation to equal the previous one just as we added $-3000(\frac{19}{2})$ into the equation we need to subtract $-3000(\frac{19}{2})$ from the equation,
thus,
$f(x)=-3000(x^2-19x+(\frac{19}{2})^2)-(-3000(\frac{19}{2}))$,
$f(x)=-3000(x-\frac{19}{2})^2+270750$,
Therefore, we can see from the equation that The maximum revenue is $270750$ and It is when $x=\frac{19}{2}$ ticket is sold.
c. $f(x)=0$,
$f(x)=x(-3000x+57000)=0$,
either $x=0$ when no ticket is sold or $-3000x+57000=0, x=19$.
thus, the ticket price of $x=19$ is so high that no revenue is generated.
