Answer:
A:
Let x represent the price of a ticket and R represent the revenue.
Let r be the number of 1 $ reduction.
So,
=> .......(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;
Solving this we get;
B:
To maximize revenue, we have to find the derivative for
Setting this to zero:
=>
=>
x = 8.5
To maximize the revenue, the price should be $8.50.
C:
When R=0
=>
=> 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.
Plug in 3 for x
f(3)= 2(3)+8
f(3)=6+8
f(3)=14
g(3)=5
Then add the values
14+5=19
Final answer: 19
Answer:
7/2
Step-by-step explanation:
12x=4(x+2)+20
12x=4x+8+20
12x-4x=8+20
8x=28
x=28/8
simplify
x=7/2
Answer:
Step-by-step explanation: