The data below is from a survey of hotels which shows that room occupancy during the off-season is related to the price charged for a basic room.

1. What price per day will maximize the daily off-season revenue for a typical hotel in this group if it has 200 rooms available?
   Solution: The price per day that will maximize the daily off-season revenue for a typical hotel in this group is $104.

   Price per Day	Occupancy Rate, %	Total Rooms Available	Room Occupancy (Total Avail Rooms\% Occupancy)	Revenue (Price*Room Occupancy)
   $74	53%	200	106	$7,844
   $94	47%	200	94	$8,836
   $100	46%	200	92	$9,200
   $104	45%	200	90	$9,360
   $114	40%	200	80	$9,120
   $134	32%	200	64	$8,576

2. Suppose that for this typical hotel the daily cost is $4,592 plus $30 per occupied room. What price will maximize the profit for this hotel in the off-season?

Solution: In order to maximize the profit, we get the following table:

As it can be observed, the profit is maximized when the price is $114.

What price per day will maximize the off-season profit for this typical hotel applies to this group of hotels. To find the room price per day that will maximize the daily revenue and the room price per day that will maximize profit for this hotel (and thus the group of hotels) in the off-season, complete the following.

1. Multiply each occupancy rate by 200 to get the hypothetical room Occupancy. Create the revenue data points that compare the price with the revenue, R, which is equal to price times

   Price per Day	Occupancy Rate, %	Total Rooms Available	Room Occupancy (Total Avail Rooms % Occupancy)	Revenue (Price*Room Occupancy)
   $74	53%	200	106	$7,844
   $94	47%	200	94	$8,836
   $100	46%	200	92	$9,200
   $104	45%	200	90	$9,360
   $114	40%	200	80	$9,120
   $134	32%	200	64	$8,576

2. Use technology to create an equation that models the revenue, R, as a function of the price per day, x.
   Solution: The type of model to be used depends on the visual trend exhibited by the data. Based on the quality of fit, we’ll use a cubic model.
   Using Excel we get the following results:

   SUMMARY OUTPUT
   Regression Statistics
   Multiple R	0.981863451
   R Square	0.964055837
   Adjusted R Square	0.910139592
   Standard Error	166.2735947
   Observations	6
   ANOVA
   	df	SS	MS	F	Significance F
   Regression	3	1483031.517	494343.8389	17.88061919	0.053428809
   Residual	2	55293.81658	27646.90829
   Total	5	1538325.333
   	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
   Intercept	-3388.876564	15562.43698	-0.217760018	0.847813989	-70348.68514	63570.93
   Price	220.6306895	472.8966529	0.466551599	0.686706646	-1814.080801	2255.342
   Price^2	-0.873085032	4.670931935	-0.186918808	0.868968008	-20.97049706	19.22433
   Price^3	-0.000804091	0.015012202	-0.053562499	0.962152729	-0.065396427	0.063788

   
   The model is
   Revenue = -3,388.877 + 220.6307*Price – 0.8731*Price ² – 0.000804*Price ³
3. Use maximization techniques to find the price that these hotels should charge to maximize the daily revenue.
   Solution: We have the function
   \[R\left( x \right)\text{ }=-\text{3},\text{388}.\text{877}+\text{22}0.\text{63}0\text{7}x-0.\text{8731}{{x}^{\text{2}}}-0.000\text{8}0\text{4}{{x}^{\text{3}}}\]
   where x represents price. Differentiating, we get
   \[R'\left( x \right)=220.6037-1.7462x-0.00241{{x}^{2}}\]
   Setting the derivative equal to zero, we get the following results:
   \[220.6037-1.7462x-0.00241{{x}^{2}}=0\]
   \[\Rightarrow \,\,\,\,x=-833.59\text{ or }x=109.72\]
   The only point that makes sense in this context is x* = $109.72. In fact this point is a maximum, because \(R''\left( 109.72 \right)=-2.27552<0\).
   The maximum profit is R* = $9,246.01.
   Graphically:
4. Use technology to get the occupancy as a function of the price, and use the occupancy function to create a daily cost function.
   Solution: Using a quadratic model, we get:

   SUMMARY OUTPUT
   Regression Statistics
   Multiple R	0.997424
   R Square	0.994855
   Adjusted R Square	0.991425
   Standard Error	0.006611
   Observations	6
   ANOVA
   	df	SS	MS	F	Significance F
   Regression	2	0.02535223	0.012676113	290.0547741	0.000369026
   Residual	3	0.00013111	4.37025E-05
   Total	5	0.02548333
   	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
   Intercept	0.584855	0.07273874	8.040494667	0.004017469	0.353368095	0.816342792
   Price	0.000777	0.00141387	0.549669864	0.620820182	-0.003722413	0.005276739
   Price^2	-2.1E-05	6.7399E-06	-3.055829483	0.055173927	-4.20451E-05	8.53399E-07

   
   The model is
   Occupancy Rate = 0.584855 + 0.000777*Price – 0.000021*Price ²
   Graphically,
   The cost function is therefore:
   \[C\left( x \right)=8101.13266+4.662977x-0.12358{{x}^{2}}\]
   This is obtained multiplying by 200 to obtain the number of rooms that are occupied, then multiplying by the variable cost (which is $30) and then adding the fixed cost.
5. Form the profit function.
   Solution: The profit function is
   \[P\left( x \right)=R\left( x \right)-\text{C}\left( x \right)\text{ }=-\text{3},\text{388}.\text{877}+\text{22}0.\text{63}0\text{7}x-0.\text{8731}{{x}^{\text{2}}}-0.000\text{8}0\text{4}{{x}^{\text{3}}}-\left( 8101.13266+4.662977x-0.12358{{x}^{2}} \right)\]
   which means that
   \[P\left( x \right)=-11,490+215.968x-0.749505{{x}^{2}}-0.000804091{{x}^{3}}\]
6. Use maximization techniques to find the price that will maximize the profit .

Solution: Differentiating the profit function and setting the derivative equal to zero:

which means that the price that maximizes the profit is P* = $120.649. The optimal daily profit is $2,244.21.

Graphically, we have:

Price: $10.9

Solution: The downloadable solution consists of 6 pages, 490 words and 3 charts.
Deliverable: Word Document