Asked by michelle
Help pleaseeee
The Hotel Regal has 500 rooms. Currently the hotel is filled. The daily rental is $ 400 per room.
For every $4 increase in rent the demand for rooms decreases by 8 rooms.
Let x be the number of rooms that are being rented in the hotel.
What should x be so as to maximize the revenue of the hotel ?
What is the rent per room when the revenue is maximized?
What is the maximum revenue?
The Hotel Regal has 500 rooms. Currently the hotel is filled. The daily rental is $ 400 per room.
For every $4 increase in rent the demand for rooms decreases by 8 rooms.
Let x be the number of rooms that are being rented in the hotel.
What should x be so as to maximize the revenue of the hotel ?
What is the rent per room when the revenue is maximized?
What is the maximum revenue?
Answers
Answered by
Damon
x</= 500
room rent = r = 400 + i
x = 500 - (8/4)i
cash = [ 500 - (8/4)i ] [400 +i]
cash = c = (500-2i)(400+i)
c = 20,000 - 300 i - 2i^2
dc/di = 0 for max or min = -300 - 4 i
d^2c/di^2 = -4 so max
i = -300/4 = -75 for max
x = 500 -2(-75) = more rooms than we have
so we can not make more money by going up on the rent
rent = 400 per room
cash = 400 * 500 = $20,000
room rent = r = 400 + i
x = 500 - (8/4)i
cash = [ 500 - (8/4)i ] [400 +i]
cash = c = (500-2i)(400+i)
c = 20,000 - 300 i - 2i^2
dc/di = 0 for max or min = -300 - 4 i
d^2c/di^2 = -4 so max
i = -300/4 = -75 for max
x = 500 -2(-75) = more rooms than we have
so we can not make more money by going up on the rent
rent = 400 per room
cash = 400 * 500 = $20,000
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