Help please >.<
The Hotel Lakewell has 450 rooms. Currently the hotel is filled . The daily rental is $ 750 per room.
For every $ 4 increase in rent the demand for rooms decreases by 7 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? $
number of rooms rented = 450 - 7x
price per room = 750 + 4x
revenue = (450-7x)(750+4x)
expand, then either find the vertex of the resultiong quadratic function, or
differentiate, set equal to zero, and solve for x
revenue is demand*price
r(x) = (450-7x)(750+4x)
Hmmm. I get max revenue when x = -61.
Sure those figures are right?
Sure my function looks good?
To maximize the revenue of the hotel, we need to find the value of x that maximizes the revenue function.
Step 1: Calculate the initial demand for rooms.
Since we know that the hotel is currently filled, the initial demand for rooms is 450.
Step 2: Determine the demand function.
We are given that for every $4 increase in rent, the demand for rooms decreases by 7. This means that the demand function can be expressed as:
Demand = 450 - 7x,
where x represents the number of $4 increases in rent.
Step 3: Calculate the daily revenue.
The daily revenue is given by multiplying the rent per room by the demand for rooms. In this case, the rent per room is $750. Therefore, the revenue function can be expressed as:
Revenue = (450 - 7x) * $750.
Step 4: Simplify the revenue function.
Multiply the rent by 750:
Revenue = 337,500 - 5,250x.
To maximize the revenue, we need to find the value of x that corresponds to the maximum point of the revenue function.
Step 5: Find the maximum point.
The revenue function is in the form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is -5,250 and the y-intercept is 337,500.
The formula to find the x-coordinate of the maximum point of a linear equation is x = -b/2m.
Therefore, the x-coordinate of the maximum point is:
x = -337,500 / (2 * (-5,250))
x = 32.14
Step 6: Round the value of x to the nearest whole number.
Since we can only have a whole number of $4 increases, we need to round x to the nearest whole number, which is 32.
Step 7: Calculate the rent per room when revenue is maximized.
To find the rent per room when the revenue is maximized, substitute the value of x into the demand function:
Demand = 450 - 7 * 32
Demand = 450 - 224
Demand = 226
Step 8: Calculate the maximum revenue.
To calculate the maximum revenue, substitute the value of x into the revenue function:
Revenue = (450 - 7 * 32) * $750
Revenue = 226 * $750
Revenue = $169,500
Therefore, the value of x that maximizes the revenue of the hotel is 32. The rent per room when the revenue is maximized is $750. The maximum revenue is $169,500.