SOLUTION: The revenue at the Assembly Center depends on the number of seats sold for the Willie Williams and the Wranglers concert. At $10 per ticket, they will fill all 8000 seats. The mana

Algebra ->  Functions -> SOLUTION: The revenue at the Assembly Center depends on the number of seats sold for the Willie Williams and the Wranglers concert. At $10 per ticket, they will fill all 8000 seats. The mana      Log On


   

Question 1147418: The revenue at the Assembly Center depends on the number of seats sold for the Willie Williams and the Wranglers concert. At $10 per ticket, they will fill all 8000 seats. The manager knows that for every $1 increase in the price, 500 tickets will go unsold. If the revenue in dollars, 𝑅(𝑝), is given by
𝑅(𝑝) = −500𝑝^2 + 13000𝑝, where 𝑝 is the price per ticket sold.
What ticket price will produce a maximum revenue? What is the maximum
revenue? You must show this algebraically.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +k+ = the number of $1 increases in price/ticket
+R%28p%29+=+p%2A%28+8000+-+500k+%29+
+-500p%5E2+%2B+13000p+=+p%2A%28+8000+-+500k+%29+
+-500p+%2B+13000+=+8000+-+500k+
+-500p+=+-500k+-+5000+
+p+=+k+%2B+10+
This shows the equation works
--------------------------------------
+R%28p%29+=+-500p%5E2+%2B+13000p+
The maximum is at +p+=+-b%2F2a+ when the equation has the form
+R%28p%29+=+a%2Ap%5E2+%2B+b%2Ap+
+p%5Bmax%5D is at +-13000%2F%28+2%2A%28-500%29%29+
+p%5Bmax%5D+=+13+
The maximum revenue is at a ticket price of $13/ticket
which is after three $1 increases in price/ticket +k=3+
Here's the plot:
+graph%28+400%2C+400%2C+-5%2C+30%2C+-10000%2C+100000%2C+-500x%5E2+%2B+13000x%29+