Interesting Mathematics Introduction to Mathematical Modeling 3 Two Examples of Introduction to Mathematical Modeling Using Calculus to Solve

1. Getting Started Example 1

1. Problem description

A hotel has 150 rooms. After a period of operation, the hotel manager got some data: if each room is priced at 200 yuan, the occupancy rate is 55%; if the price is 180 yuan, the occupancy rate is 65%; if the price is 160 yuan , the occupancy rate is 75%; the price is 140 yuan, and the occupancy rate is 85%.

The manager wants to maximize daily revenue and asks what the price of each room should be.

2. Model assumptions

Assumption 1: The maximum price per room is 200 yuan.

Hypothesis 2: According to the data provided in the question, it can be assumed that as house prices decrease, the occupancy rate increases linearly.

Assumption 3: Each hotel room is priced equally.

3. Model establishment

Suppose $and$ represents the total revenue of the hotel for one day. Compared with 200 yuan, the reduced room price per room is $x$ yuan. From hypothesis 2, it can be seen that for every 1 yuan decrease in house price, the occupancy rate will increase $\frac{0.1}{20} = 0.005$ .

Therefore, 150 rooms, the maximum room price is 200, and the occupancy rate of 1 yuan is 0.005, the following formula can be used:

$y=150\times (200-x)\times (0.55+0.005x)$

From $0.55+0.005x\leq 1$ , it can be seen that $0\leq x\leq 90$ . So the problem is transformed into finding what is the maximum value of total income when $0\leq x\leq 90$ ? $and$

4. Model solution

Let’s sort out the above equation.

$y = 150 \times (200\times 0.55 + 200 \times 0.005x -0.55x - 0.005 \times x^2)$

$y = 150 \times (110 - 0.45x - 0.005x^2)$

Outline being made by using the one-variable function to differentiate ${y}' =150\times (0.45-0.01x) =0$

It can be appropriate $x=45$ , that is, when the house price is set at 155 yuan, the maximum income can be obtained at 18018.75 yuan. At this time, the corresponding occupancy rate is 77.5%.

2. Getting Started Example 2

1. Problem description

Demographers have found that the city center of each city has the highest population density, and the farther away from the city center, the sparser the population and the smaller the density. The most common population density model is $f = ce^{-ar^2}$ (population per square kilometer), where $a$ and $c$ are constants greater than 0. , $r$ is the distance from the city center. How to find the total population of a city?

According to relevant data: the population density of a certain city’s city center is: $f=10^5$ .

The population density when 10km away from the city center is: $f = \frac{10^5}{e^2}$ .

The city is a circular area with a radius of 30km.

2. Problem analysis

In order to determine the interval, assume that the city center is located at the origin of the coordinates, so $r = \sqrt{x^2 + y ^2}$ , and the population density function is $f = ce^{-a(x^2 + y ^2)}$ .

3. Model solution

First determine the constants a and c in the population density.

Yu $r=0$ , $f=10^5$ ; $r=10$ , $f = \frac{10^5}{e^2}$ , Possible a> $a=\frac{1}{50}$ , $c=10^5$ ,

So the population density function is: $f = 10^5 \cdot e ^ {- \frac{x^2 + y ^2}{50}}$

Therefore, the total population of the city is the integral of the population density function, where the integral area D is $0\leq r\leq 30$ , $0 \leq \theta \leq 2\pi$ , that is,