Question

The data in the table, from a survey of resort hotels with comparable rates on Hilton Head Island, show that room occupancy during the off-season (November through February) is related to the price charged for a basic room.

Price per Day $	Occupancy Rate %
104	53
134	47
143	46
149	45
164	40
194	32

• First make a linear equation using linear regression on your calculator where x = price and y = occupancy rate.
• Convert occupancy rate to quantity of rooms in a 200-room hotel.
• Write down a revenue function for a 200-room hotel.
• What price per day will maximize the daily off-season revenue for a typical 200-room hotel? Use Calculus to determine the maximum.
• If this 200-room hotel has daily operating costs of $5510 plus $30 per occupied room. What price will maximize the daily profit during the off-season? Again use calculus to determine the maximum

More detailed instructions are given on page 690 of the textbook (12th edition).

Answer 1

Solution

Back-up Theory

The linear regression model: Y = β₀ + β₁X + ε, …………………………………......……..(1)

where ε is the error term, which is assumed to be Normally distributed with mean 0 and variance σ².

Estimated Regression of Y on X is given by: Y_cap = β_0cap + β_1capX, …………………….(2)

where β_1cap = Sxy/Sxx and β_0cap = Ybar – β_1cap.Xbar..…………………………..….…..(3)

Mean X = Xbar = (1/n)sum of x_i ………………………………………….………............….(4)

Mean Y = Ybar = (1/n)sum of y_i ………………………………………….……..........….….(5)

Sxx = sum of (x_i – Xbar)² ……………………………………….........………..…………....(6)

Syy = sum of (y_i – Ybar)² …………………………………………..........…..………………(7)

Sxy = sum of {(x_i – Xbar)(y_i – Ybar)} ……………………………............………………….(8)

All above sums are over i = 1, 2, …., n,n = sample size ………......……………………..(9)

Now to work out the solution,

Given y = Occupancy Rate % and x = Price per Day $

Part (a)

Linear regression equation: Y_cap = 78.5627 – 0.2347x Answer 1

Details of calculations

n	6
Xbar	148.00
ybar	43.8333333
Sxx	4530
Syy	254.833333
Sxy	-1063
β1cap	-0.2346578
β0cap	78.5626932

Part (b)

Conversion of occupancy rate to quantity of rooms in a 200-room hotel

Multiplying the Y_cap obtained in Part (a) by (200/100), the quantity of rooms in a 200-room hotel,

Q = 157.1254 – 0.4694x Answer 2

Part (c)

Revenue function for a 200-room hotel.

Revenue function, R(x) = quantity of rooms x Price per Day

= Qx

= 157.1254x – 0.4694x²[vide answer of Part (b)] Answer 3

Part (d)

Revenue maximizing price per day for the off-season for a typical 200-room hotel

dR/dx = 157.1254 – 0.9388x

Equating (dR/dx) to zero and solving for x,

x = 167.37

Thus, revenue maximizing price per day = $167.37 Answer 4

Part (e)

Profit maximizing price per day for the off-season for a typical 200-room hotel with daily operating costs of $5510 plus $30 per occupied room.

Profit function, P(x) = R(x) – C(x), where, C(x) = cost function = 5510 + 30x [given]

So, P(x) = 127.1254x – 0.4694x² – 5510

dP/dx = 127.1254 – 0. 9388x

Equating (dP/dx) to zero and solving for x,

x = 135.41

Thus, profit maximizing price per day = $135.41 Answer   5

DONE