Question 1189985: 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.
Found 2 solutions by Solver92311, ikleyn: Answer by Solver92311(821) (Show Source): Answer by ikleyn(52778) (Show Source):
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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.
ANSWER. Under given conditions, the revenue function is this quadratic function of the ticket price
R(p) = -3000p*2 + 57000p.
CHECK. 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.
Solved.
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