Question

A 300 room hotel can rent every one of its rooms at $80 per room. For every $1 increase in rent, 3 fewer rooms are rented.
How much should the hotel charge for each room to maximize its daily revenue? What is the maximum daily revenue?

Answer 1

Answer:

The hotel charge for each room to maximize its daily revenue and the maximum daily revenue is explained below in details.

Step-by-step explanation:

So for every $1 hike in price, the number of rooms goes underneath by 1.

So the earnings can be displayed by R(x) = (300-x)(80+x)

From that, you require to subtract 22(300-x) since it takes $22 per occupied room.

AND there is a framed cost of $1000 per day.

So your profit function is P(x) = (300-x)(80+x) - 22(300-x) - 1000

Simplified, this is P(x) = -x2 +242x + 16400.

P'(x) = -2x + 242  Set it = to 0

x = 121 which displayed the number of unoccupied rooms.

P(121) = -14641 + 29282 + 16400 = $31041

You can verify by computing P(120) and P(122) which both even $31040.

Hope this helps.