Asked by Clara
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? $
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? $
Answers
Answered by
Reiny
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
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
Answered by
Steve
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?
r(x) = (450-7x)(750+4x)
Hmmm. I get max revenue when x = -61.
Sure those figures are right?
Sure my function looks good?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.