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 $ .

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

Step-by-step 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.

x=35

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

 I ( x35) =8800+140(35)-2(35)²= 11250

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5

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