A 300 room hotel in Guatemala is filled to capacity every ni

A 300 room hotel in Guatemala is filled to capacity every night at a rate of $80 per room. For each $1 increase in the nightly rate. 3 fewer rooms are rented. If each rented room costs $10 a day to service, how much should the management charge per room in order to maximize gross profit? What is this maximum profit? The price-demand equation for a fancy watch is given by p(x) = 1200e^-0.01x. where x represents the monthly demand and p is the price in dollars. Find the production level, x. and price per unit, p. that produce the maximum revenue. What is this maximum revenue?

Solution

Solution : (3)

a )

Let x be the amount over $80 for a room.

When x = 0, the hotel rents 300 rooms.

When x = 1, the hotel rents 300 – 3 = 297 rooms

When x = 2, the hotel rents 297 – 3 = 294 rooms

Number of rooms rented = N(x) = 300 – 3x

Profit per room = (rent earned) – (cleaning costs)

R(x) = (80 + x) – 10 = 70 + x

Total Profit = P(x) = N(x) R(x) = (300 – 3x)(70 + x)

P(x) = -3x² + 90x + 21000

P(x) = -3x2 + 90x + 21000

P’(x) = -6x + 90

Set P’(x) = 0

to find x = 15.

The room charge should be $95.

Number of rooms rented

   = 300 – 3(15) = 265

Total profit P(15) = $21675.