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?
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
To maximize the revenue of the hotel, we need to determine the optimal number, x, of rooms to rent.
To find this, we can first derive an equation for the revenue based on the number of rooms rented.
Let's assume the initial rent per room is $400. Then, for every $4 increase in rent, the demand for rooms decreases by 8. So, the rent per room can be given as:
R(x) = 400 + (x/8)(-4)
Where x represents the number of rooms being rented.
The revenue, denoted by Rev(x), is given by:
Rev(x) = R(x) * x
Rev(x) = (400 + (x/8)(-4)) * x
Now, to maximize the revenue, we need to find the value of x that makes Rev(x) maximum.
To find this, we can take the derivative of Rev(x) with respect to x and set it equal to zero.
dRev(x)/dx = 0
Solving this equation will give us the value of x at the maximum revenue. We can then substitute that value of x into the revenue equation to find the maximum revenue.
Let's find the derivative of Rev(x):
dRev(x)/dx = (400 + (x/8)(-4)) + (x/8)(-4)
= 400 + (x/8)(-4) - (x/8)(4)
Setting this derivative equal to zero:
400 + (x/8)(-4) - (x/8)(4) = 0
Simplifying:
400 - (x/8)(8) = 0
400 - x = 0
400 = x
Therefore, x = 400 is the optimal number of rooms to rent to maximize revenue.
To find the rent per room when the revenue is maximized, we substitute this value of x into the rent equation:
R(x) = 400 + (400/8)(-4)
= 400 - 200
= 200
Thus, the rent per room when the revenue is maximized is $200.
To find the maximum revenue, we substitute the value of x into the revenue equation:
Rev(x) = (400 + (400/8)(-4)) * 400
= (400 - 200) * 400
= 200 * 400
= 80,000
Therefore, the maximum revenue is $80,000.