Question

A hotel rents 210 rooms at a rate of $ 60 per day. For each $ 2 increase in the rate, three fewer rooms are rented. Find the room rate that maximizes daily revenue.

Answer 1

Answer:

r=$14,400

The hotel should charge $120

Step-by-step explanation:

Revenue (r)= p * n

where,

p = price per item

n = number of items sold

A change in price leads to a change in number sold

A variable to measure the change in p and n needs to be introduced

Let the variable=x

Such that

p + x means a one dollar price increase

p - x means a one dollar price decrease

n + x means a one item number-sold increase

n - x means a one item number-sold decrease

for each $2 price increase (p + 2x) there are 3 fewer rooms are rented (n-3x)

know that at $60 per room, the hotel rents 210 rooms

r = (60 + 2x) * (210 - 3x)

=12,600-180x+420x-6x^2

=12,600+240x-6x^2

r=2100+40x-x^2

= -x^2 +40x+2100=0

Solve the quadratic equation

x= -b +or- √b^2-4ac / 2a

a= -1

b=40

c=2100

x= -b +or- √b^2-4ac / 2a

= -40 +or- √(40)^2 - (4)(-1)(2100) / (2)(-1)

= -40 +or- √1600-(-8400) / -2

= -40 +or- √ 1600+8400 / -2

= -40 +or- √10,000 / -2

= -40 +or- 100 / -2

x= -40+100/-2 OR -40-100/-2

=60/-2 OR -140/-2

= -30 OR 70

x=70

The quadratic equation has a maximum at x=70

p+2x

=60+2(30)

=60+60

=$120

r= (60 + 2x) * (210 - 3x)

={60+2(30)}*{(210-3(30)}

r=(60+60)*(210-90)

=120*120

=$14,400