A motel finds it can rent 180 rooms when the room rate is $ 60 per day. For each $ 2 increase in the rate, two fewer rooms are rented. Find the room rate that maximizes daily revenue. The rate that maximizes revenue is $ .

$7,200

Step-by-step explanation:

r=p × n

where

r=revenue

p = price per item

n = number of items sold

A change in price changes quantity ( number sold)

Variable x changes with price, then

Let

p+x=a dollar price increase

p-x= a dollar price decrease

n+x= an increase in quantity sold

n-x= a decrease in quantity sold

Therefore,

A $2 dollar increase in price=p+2x

2 decrease in room quantity sold= n-2x

Substituting in the above equation

r=(p+2x)*(n-2x)

P=$60

n=180

Then

r=(60+2x)*(180-2x)

r=10,800-120x+360x-4x^2

r=-4x^2-240x+10,800

Solve the quadratic equation

The maximum value of x in the vertex = - b/2*a

Where

b=240

a=-4

Value of x=-240/2*(-4)

X=30

r=-4x^2-240x+10,800

=-4(30)^2-240(30)+10,800

=-4(900)-7200+10,800

=3,600-7200+10800

=$7,200

The rate that maximises revenue=$7,200

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