Question 810255: An average of 100,000 people visit Riverside Park each day in the summer. The park charges $21.00 for admission. Consultants predict that for each $1.00 increase in the entrance price, the park would lose an average of 3,500 customers a day
A) Express the daily revenue from ticket sales, R as a function of the number of $100 price increases, x.
R=f(x)=
B) What ticket price maximizes the revenue from ticket sales? (round to the nearest cent)
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! An average of 100,000 people visit Riverside Park each day in the summer.
The park charges $21.00 for admission.
Consultants predict that for each $1.00 increase in the entrance price, the park would lose an average of 3,500 customers a day
:
A) Express the daily revenue from ticket sales, R as a function of the number of $1.00 price increases, x.
R=f(x)= (100000-3500x)(21 + x)
FOIL
f(x) = 2100000 + 100000x - 73500x - 3500x^2
f(x) = 2100000 + 26500x - 3500x^2
usually written
f(x) = -3500x^2 + 26500x + 2100000
:
B) What ticket price maximizes the revenue from ticket sales? (round to the nearest cent)
Simplify equation, divide by 500
f(x) = -7x^2 + 53x + 4200
Maximum occurs at the axis of symmetry which is: x = -b/(2a)
x =
x = +3.79 * $1 = $3.79 increase
A price of $24.79 would give max revenue
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