Answer:
A:
Let x represent the price of a ticket and R represent the revenue.
Let r be the number of 1 $ reduction.
So, ![x=10-r](https://tex.z-dn.net/?f=x%3D10-r)
=>
.......(1)
As given that for every dollar the ticket price is lowered, attendance increases by 3000.
So, if we reduce the ticket price by r dollars, the attendance will increase by 3000r.
Revenue = price of ticket X number of ticket sold
R =
.....(2)
Substituting the value of r from (1) in (2) we get;
![x\times(21000+3000(10-x))](https://tex.z-dn.net/?f=x%5Ctimes%2821000%2B3000%2810-x%29%29)
Solving this we get;
![R= 51000x- 3000x^{2}](https://tex.z-dn.net/?f=R%3D%2051000x-%203000x%5E%7B2%7D)
B:
To maximize revenue, we have to find the derivative for ![R= 51000x- 3000x^{2}](https://tex.z-dn.net/?f=R%3D%2051000x-%203000x%5E%7B2%7D)
![dR/dp=51000-6000x](https://tex.z-dn.net/?f=dR%2Fdp%3D51000-6000x)
Setting this to zero:
![51000-6000x=0](https://tex.z-dn.net/?f=51000-6000x%3D0)
=> ![6000x=51000](https://tex.z-dn.net/?f=6000x%3D51000)
=> ![x = 51000/6000](https://tex.z-dn.net/?f=x%20%3D%2051000%2F6000)
x = 8.5
To maximize the revenue, the price should be $8.50.
C:
When R=0
![0=51000x-3000x^{2}](https://tex.z-dn.net/?f=0%3D51000x-3000x%5E%7B2%7D)
=>![0=3000x(17-x)](https://tex.z-dn.net/?f=0%3D3000x%2817-x%29)
=> Either x=0 or x=17
x = $0 means there is no price for ticket, that is not possible, so we will neglect it.
And x = $17 means that the ticket price is so high, and no revenue will be generated.