SOLUTION: The park had initially planned to charge $8 for admission and expected to have 2400 visitors a day. Allison and Hannah were assigned the task of analyzing the parks admission reven

Question 559038: The park had initially planned to charge $8 for admission and expected to have 2400 visitors a day. Allison and Hannah were assigned the task of analyzing the parks admission revenues.
a) How much revenue would the park have for one day at the current price?
b) Market research shows for every $0.50 the admission price is raised, the park will have 80 fewer visitors. How much would you expect the park revenue to be if the park raised their admission price by $1?
After a few calculations, Allison and Hannah realize the park will make more money if they raise the price of admission. However, they also understand that there must be a limit to how much the park can charge.
As a result, they model the situation with the equation, R = (2400 - 80x)(8 + 0.5x), where R represents the revenue from sales and x represents the number of price increases.
Use this Factored form Quadratic equation to solve for the zeroes, and then use the zeroes to calculate the number of price increases that will generate the maximum Revenue.
Now demonstrate a second strategy for solving such a problem. This process will ultimately double check your results from above...
a) Express this same Quadratic R = (2400 - 80x)(8 + 0.5x), in Standard form: R = ax2 + bx + c.
b) Complete the Square to again determine the number of price increases that will generate the maximum Revenue.
c) What is the Maximum Revenue that can be generated? (from just looking at your results!)
I would be
GRateful for any help you give me

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The park had initially planned to charge $8 for admission and expected to have 2400 visitors a day. Allison and Hannah were assigned the task of analyzing the parks admission revenues.
a) How much revenue would the park have for one day at the current price?
2400*8 = $19200
-------------------------
b) Market research shows for every $0.50 the admission price is raised, the park will have 80 fewer visitors. How much would you expect the park revenue to be if the park raised their admission price by $1?
(8+2*0.50)*(2400-2*80) = $20160
------------------------------------------
After a few calculations, Allison and Hannah realize the park will make more money if they raise the price of admission. However, they also understand that there must be a limit to how much the park can charge.
As a result, they model the situation with the equation, R = (2400 - 80x)(8 + 0.5x), where R represents the revenue from sales and x represents the number of price increases.
Use this Factored form Quadratic equation to solve for the zeroes, and then use the zeroes to calculate the number of price increases that will generate the maximum Revenue.
Revenue = (2400 - 80x)(8 + 0.5x)
Zeroes:
2400-80x = 0 when x = 30
8+0.5x = 0 when x = -16
Mid-value = c = (30-16)/2 = 7
-----------------------------------
Maximum Revenue = ?
R(x) = (2400 - 80x)(8 + 0.5x)
R(x) = -40x^2+1200x-640x+8*2400
R(x) = -40x^2+560x+19200
Max occurs when x = -b/(2a) = -560/(80) = 7
R(7) = (2400-560)(8+3.5) = $21,160
----------------------------------------------------
Now demonstrate a second strategy for solving such a problem. This process will ultimately double check your results from above...
a) Express this same Quadratic R = (2400 - 80x)(8 + 0.5x), in Standard form: R = ax2 + bx + c.
R = -40x^2+560x+19200
b) Complete the Square to again determine the number of price increases that will generate the maximum Revenue.
R = -40(x^2-14x+49)+19200+40*49
R = -40(x-7)^2 + 21160
-------------------------------
c) What is the Maximum Revenue that can be generated? (from just looking at your results!)
Max = $21160
==================
Cheers,
Stan H.
======================

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