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
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?
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