The Hotel Thomasina has 500 rooms. Currently the hotel is filled . The daily rental is $ 1000 per room.
For every $ 4 increase in rent the demand for rooms decreases by 8 rooms.
Let x = the number of $ 4 increases that can be made.
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?
revenue
r(x) = (1000+4x)(500-8x)
Just find the vertex of that parabola to get the x desired, then figure the other items based on that.
To determine the value of x that maximizes the hotel's revenue, we need to analyze the relationship between the rent per room, the demand for rooms, and the total revenue.
Let's start by determining the initial rent per room and the initial revenue. The daily rental is $1000 per room, so initially, the rent per room is $1000. The number of rooms in the hotel is 500, so the initial revenue is 500 * $1000 = $500,000.
Next, let's consider the impact of increasing the rent by $4 on the demand for rooms. For every $4 increase in rent, the demand decreases by 8 rooms. This means that for each $4 increase, the hotel will lose 8 potential customers.
To find the total revenue after x increases in rent, we need to determine the new rent per room and the new number of rooms rented. Since each increase of $4 reduces the demand by 8 rooms, the new number of rooms rented will be 500 - 8x, assuming we can still fill all the remaining rooms. Therefore, the new rent per room will be $1000 + $4x.
Now, we can calculate the new revenue based on the new rent per room and the new number of rooms rented. The new revenue can be calculated as follows: (500 - 8x) * ($1000 + $4x).
To maximize the revenue, we need to find the value of x that results in the maximum revenue. We can do this by finding the derivative of the revenue function with respect to x and setting it equal to 0. We will then solve for x to find the critical point.
Let's calculate the derivative of the revenue function and set it equal to zero:
d(revenue)/dx = (500 - 8x) * 4 + (1000 + 4x) * (-8) = 0
Now, solve for x:
2000 - 32x - 8000 - 8x = 0
-40x - 6000 = 0
-40x = 6000
x = -6000 / -40
x = 150
Therefore, x should be 150 in order to maximize the hotel's revenue.
To find the rent per room when the revenue is maximized, we substitute the value of x into the expression for the new rent per room:
Rent per room = $1000 + $4(150) = $1600
Therefore, the rent per room when the revenue is maximized is $1600.
Finally, let's calculate the maximum revenue by plugging in the value of x into the expression for the new revenue:
Maximum Revenue = (500 - 8(150)) * (1000 + 4(150))
= (500 - 1200) * (1000 + 600)
= (-700) * (1600)
= -$1,120,000
Hence, the maximum revenue for the hotel is -$1,120,000. Note that the negative value indicates a loss, which implies that increasing the rent beyond a certain point actually results in a decrease in revenue.