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Question 159109: The question states "Maximizing Revenue- A large hotel is considering the following group discount on room rates. The regular price for a room is $120, but for each room rented the price decreases by $2 per room. ie. one room costs 118, 2 rooms cost $116 X 2=232 and so on.
a)What is the maximum revenue?
b) What is the number of rooms that should be rented?
This questions comes from "College Algebra with Modeling and Visualization" Gary Rockswold, 3rd edition, section 3.1
Thank you,
Jessica
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! The question states "Maximizing Revenue- A large hotel is considering the following group discount on room rates. The regular price for a room is $120, but for each room rented the price decreases by $2 per room. ie. one room costs 118, 2 rooms cost $116 X 2=232 and so on.
a)What is the maximum revenue?
b) What is the number of rooms that should be rented?
.
Let r = number of rooms rented at the group discount
.
Revenue = n(120 - 2n)
Revenue = 120n - 2n^2
Revenue = -2n^2 + 120n
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Because the 'a' coefficient of the equation is negative, we KNOW that the parabola opens downward (upside down U).
.
To find the vertex, rearrange the equation into the "vertex form":
y= a(x-h)^2+k
.
Basically, we're manipulating our equation into the "vertex form":
Revenue = -2n^2 + 120n
Revenue = -2(n^2 - 120n)
Completing the square:
Revenue = -2(n^2 - 60n - 900) + 1800
Revenue = -2(n^2 - 30)^2 + 1800
.
Now, we know that (h,k) is at (30, 1800)
This vertex represents the peak of the parabola -- answering your questions:
a)What is the maximum revenue? $1800
b) What is the number of rooms that should be rented? 30
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