Answer to Question #172682 in Calculus for Todd Phillips

Question

The number of people attending a concert is represented by 9000(75-x), where x is the cost of the ticket in dollars. What ticket price will maximize the total revenue of the concert?

Answer 1

Given, Price of ticket is\ $x$ and number of people attending the concert are $\ 9000(75-x)$ .

So ,total revenue generated will be $R=x\times 9000(75-x)=9000x(75-x)$

To maximize $\ R,\frac{dR}{dx}=0$ and $\ \frac{d^2R}{dx^2}<0$

So, $\frac{dR}{dx}=\frac{d}{dx}(9000x(75-x))=9000[x(-1)+(1)(75-x)]=9000[75-2x]$

And $\frac{dR}{dx}=0\implies 9000[75-2x]=0\implies x=\frac{75}{2}=37.5$

Also ,$\frac{d^2R}{dx^2}=9000(-2)=-18000<0$

So, for $x= 37.5 ,$ R is maximum.

Hence,  ticket price will maximize the total revenue of the concert is $37.5$ dollars.