/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Hotel Occupancy Rate The occupan... [FREE SOLUTION] | 91影视

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Chapter 0: Problem 68

Hotel Occupancy Rate The occupancy rate of the all-suite Wonderland Hotel, located near an amusement park, is given by the function $$ r(t)=\frac{10}{81} t^{3}-\frac{10}{3} t^{2}+\frac{200}{9} t+55 \quad 0 \leq t \leq 11 $$ where \(t\) is measured in months and \(t=0\) corresponds to the beginning of January. Management has estimated that the monthly revenue (in thousands of dollars) is approximated by the function $$ R(r)=-\frac{3}{5000} r^{3}+\frac{9}{50} r^{2} \quad 0 \leq r \leq 100 $$ where \(r\) (percent) is the occupancy rate. a. What is the hotel's occupancy rate at the beginning of January? At the beginning of July? b. What is the hotel's monthly revenue at the beginning of January? At the beginning of July?

Short Answer

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So, at the beginning of January (t=0), the occupancy rate is: $$ r(0) = \frac{10}{81}(0)^3 - \frac{10}{3}(0)^2 + \frac{200}{9}(0) + 55 = 55\% $$ #tag_title#Step 2: Find the occupancy rate at t=6 (Beginning of July)#tag_content#At the beginning of July, t=6. We need to plug in t=6 into the given formula for the occupancy rate: $$ r(6) = \frac{10}{81}(6)^3 - \frac{10}{3}(6)^2 + \frac{200}{9}(6) + 55 $$ Calculating the occupancy rate, we get: $$ r(6) \approx 89.84\% $$ #tag_title#Step 3: Find the hotel's monthly revenue at t=0 (Beginning of January)#tag_content#Now we need to find the hotel's monthly revenue at the beginning of January, using the given R(r) function: $$ R(r) = -\frac{3}{5000}r^3 + \frac{9}{50}r^2 $$ Using the occupancy rate at the beginning of January (r(0) = 55%), we can calculate the hotel's monthly revenue: $$ R(55) = -\frac{3}{5000}(55)^3 + \frac{9}{50}(55)^2 \approx 89.13 (\textrm{in thousands of dollars}) $$ #tag_title#Step 4: Find the hotel's monthly revenue at t=6 (Beginning of July)#tag_content#Finally, we need to find the hotel's monthly revenue at the beginning of July, using the given R(r) function: $$ R(r) = -\frac{3}{5000}r^3 + \frac{9}{50}r^2 $$ Using the occupancy rate at the beginning of July (r(6) 鈮� 89.84%), we can calculate the hotel's monthly revenue: $$ R(89.84) \approx 147.07 (\textrm{in thousands of dollars}) $$ In conclusion: At the beginning of January, the hotel's occupancy rate is 55%, and its monthly revenue is approximately $89.13 (\textrm{in thousands of dollars}). At the beginning of July, the hotel's occupancy rate is approximately 89.84%, and its monthly revenue is approximately $147.07 (\textrm{in thousands of dollars}).

Step by step solution

01

Find the occupancy rate at t=0 (Beginning of January)

At the beginning of January, t=0. We need to plug in t=0 into the given formula for the occupancy rate: $$ r(t) = \frac{10}{81} t^3 - \frac{10}{3} t^2 + \frac{200}{9} t + 55 $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Occupancy Rate Function
Understanding the occupancy rate function is crucial for managing a hotel's operations effectively. Occupancy rate is an expression of how many rooms are rented out compared to the total number of rooms available. The function provided, expressed as

\[\begin{equation}r(t)=\frac{10}{81} t^3 - \frac{10}{3} t^2 + \frac{200}{9} t + 55\end{equation}\]where \(t\) is the time in months, allows us to calculate the occupancy rate at any given time throughout the year. To find the occupancy rate at the start of January, for instance, we set \(t = 0\) which gives us an occupancy rate of 55%. This rate is a constant provided in the formula, representing the minimum occupancy rate when there is no increase due to passing months. As time progresses, the occupancy rate is impacted by the cubic, quadratic, and linear terms in the function, challenging the notion that occupancy changes linearly over time.
Revenue Calculation
Revenue calculation for a hotel is typically based on the occupancy rate because it directly relates to the number of rooms filled. The provided revenue function

\[\begin{equation}R(r) = -\frac{3}{5000} r^3 + \frac{9}{50} r^2\end{equation}\]
where \(r\) represents the occupancy rate percentage, shows a polynomial relationship between the occupancy rate and monthly revenue. To find the monthly revenue at the beginning of January, with an occupancy rate of 55%, we replace \(r\) in the function with 55 to calculate the corresponding revenue. This connection between occupancy rate and revenue empowers hotel management to use these models to predict financial outcomes and strategize accordingly. Applying this method also helps to estimate how fluctuations in the occupancy rate can impact revenue, dictating marketing and pricing strategies.
Polynomial Functions
Polynomial functions, like the ones used for calculating occupancy rates and revenue, are essential in business for modeling various phenomena. These functions are composed of variables and constants using the operation of addition, subtraction, multiplication, and non-negative integer exponents. The occupancy function we're dealing with is a cubic polynomial, as indicated by the highest exponent of 3 in the term \(\frac{10}{81} t^3\). Cubic polynomials have distinctive curves that can model changes more accurately than straight lines, which is beneficial when representing real-world business behavior that doesn't follow linear trends. The revenue function is also a polynomial, although it's a quadratic function, denoted by the highest exponent of 2 in the term \(\frac{9}{50} r^2\). Understanding how to manipulate and interpret these functions is vital for making informed business decisions.
Application of Calculus in Business
Calculus, particularly derivatives and integrals, plays a key role in understanding and optimizing business functions. In the context of hotel management, calculus can be used to find the rate of change of revenue with respect to occupancy and time鈥攁 core principle in business analysis. For example, finding the derivative of the occupancy rate function with respect to time (\[\begin{equation}r'(t)\end{equation}\]) will shed light on the rate at which occupancy is increasing or decreasing as the months pass. Similarly, by finding the derivative of the revenue function with respect to the occupancy rate (\[\begin{equation}R'(r)\end{equation}\]), we can analyze how small changes in occupancy can affect revenue. Moreover, calculus can be used for optimizing these functions, helping to determine the ideal pricing and marketing strategies to maximize revenue or achieve desired occupancy levels. By applying principles of calculus, hotel managers can refine their operations for better financial health.

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Most popular questions from this chapter

Overcrowding of Prisons The 1980 s saw a trend toward oldfashioned punitive deterrence of crime in contrast to the more liberal penal policies and community-based corrections that were popular in the 1960 s and early \(1970 \mathrm{~s}\). As a result, prisons became more crowded, and the gap between the number of people in prison and the prison capacity widened. The number of prisoners (in thousands) in federal and state prisons is approximated by the function $$ N(t)=3.5 t^{2}+26.7 t+436.2 \quad 0 \leq t \leq 10 $$ where \(t\) is measured in years, with \(t=0\) corresponding to 1983\. The number of inmates for which prisons were designed is given by $$ C(t)=24.3 t+365 \quad 0 \leq t \leq 10 $$ where \(C(t)\) is measured in thousands and \(t\) has the same meaning as before. a. Find an expression that shows the gap between the number of prisoners and the number of inmates for which the prisons were designed at any time \(t\) b. Find the gap at the beginning of 1983 and at the beginning of 1986 . Source: U.S. Department of Justice.

Suppose the slope of a line \(L\) is \(-\frac{1}{2}\) and \(P\) is a given point on \(L\). If \(Q\) is the point on \(L\) lying 4 units to the left of \(P\), then \(Q\) lies 2 units above \(P\).

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The line with equation \(A x+B y+C=0\), where \(B \neq 0\), and the line with equation \(a x+b y+c=0\), where \(b \neq 0\), are parallel if \(A b-a B=0\).

Find the point \((s)\) of intersection of the graphs of the functions. Express your answers accurate to five decimal places. $$ f(x)=0.3 x^{2}-1.7 x-3.2 ; \quad g(x)=-0.4 x^{2}+0.9 x+6.7 $$
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