Question

A hotel rents 220 rooms at a rate of $ 40 per day. For each $ 1 increase in the rate, two fewer rooms are rented. Find the room rate that maximizes daily revenue. The rate that maximizes revenue is $ .

Answer 1

The rooms should be rented at $75 per day for a maximum income of $11250 per day.

Explanation:

If the daily rental is increased by $ x

then

Rental: R (x )=( 40 + x ) dollars per room-day

Number of rooms rented: N ( x ) = ( 220 − 2 x ) and

Income: I ( x ) = ( 40 + x ) ( 220 − 2 x ) =8800+140x-2x² dollars/day

The maximum will be achieved when the derivative of I ( x ) is zero.

`(dI(x))/(dx) =140-4x=0`

x=35

so, ($40+$35)=75$per day

I ( x35) =8800+140(35)-2(35)²= 11250