Respuesta :
Answer:
a) R(x) =  51000 x - 3000 x²
b) When the Revenue is maximum, The price is $ 8.5
c) When No revenue is generated, The price of the ticket is $17
Explanation:
Given - A baseball team plays in a stadium that holds 54,000 spectators. With the ticket price at $10, the average attendance at recent games has been 21,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000.
To find - (a) Find a function that models the revenue in terms of ticket price. Â
       (b) Find the price that maximizes revenue from ticket sales. Â
       (c) What ticket price is so high that no revenue is generated?
Proof -
Let us assume that,
The price of a ticket = x
The revenue = R
Now,
a)
Total number of tickets sold , N = 21000 + 3000(10 - x)
                           = 21000 + 30000 - 3000 x
                           = 51000 - 3000 x
⇒Total number of tickets sold , N = 51000 - 3000 x
Now,
Revenue , R(x) = x * N
            = x [ 51000 - 3000 x ]
            = 51000 x - 3000 x²
⇒R(x) =  51000 x - 3000 x²
b)
For maximum Revenue,
Put [tex]\frac{dR(x)}{dx} = 0[/tex]
Now,
[tex]\frac{dR(x)}{dx}[/tex] = 51000 - 6000 x
⇒51000 - 6000 x = 0
⇒6000 x = 51000
⇒6x = 51
⇒x = [tex]\frac{51}{6}[/tex] = 8.5
∴ we get
When the Revenue is maximum, The price = $ 8.5
c)
If No Revenue is generated
⇒R(x) = 0
⇒51000 x - 3000 x² = 0
⇒1000 x ( 51 - 3 x ) = 0
⇒1000 x = 0, 51 - 3 x = 0
⇒x = 0, 3 x = 51
⇒x = 0, x = [tex]\frac{51}{3}[/tex]
⇒x = 0, x = 17
∴ we get
When No revenue is generated, The price of the ticket is $17
