Question 1189985
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A baseball team plays in a stadium that
holds 55,000 spectators. With the ticket price at $10, the
average attendance at recent games has been 27,000. A mar-
ket survey indicates that for every dollar the ticket price is
lowered, attendance increases by 3000.
(a) Find a function that models the revenue in terms of ticket
price.
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The problem tells us that if the ticket price is $10, then the attendance is 27000

and that ticket price change of $1 produces the attendance change of 3000 in opposite direction

    (when the price goes down, the attendance goes up, 
     and vice versa: when the price goes up, the attendance goes down).     (*)


In addition, it says that the attendance is a linear function of price.


It means that the slope of the plot is -3000, so we can write the attendance 
as a linear function of the ticket price in this form

        A(p) = 27000 - 3000*(p-10).


Indeed, this function is linear and satisfies the pointed properties (*).


Now, the revenue is the product of the ticket price by the attendance

    R(p) = p*A(p) = p*(27000 - 3000*(p-10)) = p*(27000 - 3000p + 30000) = p*(57000 - 3000p) = -3000p*2 + 57000p.


<U>ANSWER</U>.  Under given conditions, the revenue function is this quadratic function of the ticket price

                R(p) = -3000p*2 + 57000p.


<U>CHECK</U>.   At the ticket price of $10,  the revenue is  R(10) = -3000*10^2 + 57000*10 = -300000 + 570000 = 270000 dollars.

         Compare it with 27000*10 = 270000: these numbers coincide.


The given formula is valid until the attendance is not greater than the maximum capacity of the stadium of 55000 spectators.
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Solved.