Mathematical Modeling: Building and Applying Basic Models | 数学模型：基础模型的建立与应用

📚 Mathematical Modeling: Building and Applying Basic Models | 数学模型：基础模型的建立与应用

Mathematical modeling is the art of translating real-world problems into mathematical language, solving them, and then interpreting the results back into the real-world context. It is a core skill in modern mathematics education and appears in almost every examination board, from GCSE to A-Level.

数学模型是将现实世界的问题转化为数学语言、进行求解，再将结果回归到现实情境中的一门艺术。它是现代数学教育的核心技能，从GCSE到A-Level的几乎所有考试局中都会出现。

In this article, we will explore the fundamental principles of building mathematical models, examine common types of models, and work through practical examples that you may encounter in your exams. We will focus on linear models, quadratic models, and exponential models, as these form the backbone of basic mathematical modeling.

本文将探讨建立数学模型的基本原则，分析常见的模型类型，并通过例题来演练考试中可能遇到的实际问题。我们将聚焦于线性模型、二次模型和指数模型，因为它们构成了基础数学建模的骨干。

1. What Is a Mathematical Model? | 什么是数学模型？

A mathematical model is a simplified representation of a real-world situation using mathematical concepts, symbols, and equations. It helps us understand, predict, and control real-world phenomena. For example, the distance a car travels can be modeled as d = vt, where v is speed and t is time.

数学模型是使用数学概念、符号和方程对现实情境进行的简化表示。它帮助我们理解、预测和控制现实世界中的现象。例如，汽车行驶的距离可以建模为 d = vt，其中 v 是速度，t 是时间。

Models are not exact copies of reality — they are simplifications. We deliberately ignore certain details to focus on the most important relationships. A good model captures the essential features of a situation while remaining simple enough to analyze and solve.

模型不是现实世界的精确复制品——它们是简化。我们有意忽略某些细节，以聚焦于最重要的关系。一个好的模型能够捕捉情境的本质特征，同时保持足够简单以便分析和求解。

Key features of a good model: Accuracy, simplicity, and applicability.
好模型的关键特征： 准确性、简洁性和适用性。

2. The Modeling Process | 建模过程

The process of mathematical modeling typically follows several systematic steps. In your exams, you will often be asked to identify which step is being described, or to complete a specific step in the process.

数学建模的过程通常遵循几个系统性的步骤。在考试中，你常被要求识别描述的是哪一步，或完成过程中的某个特定步骤。

Step \| 步骤	Description \| 描述	Example \| 示例
1. Identify the problem	Clearly define what we want to find out.	How fast does a cup of coffee cool?
2. Identify variables	Determine which quantities change and which stay fixed.	Temperature (T), time (t), room temperature (T₀).
3. Make assumptions	Simplify the real world to make the problem tractable.	Cooling rate is proportional to temperature difference.
4. Formulate the model	Write equations that describe the relationships.	dT/dt = -k(T − T₀).
5. Solve the model	Use mathematical techniques to find a solution.	T(t) = T₀ + (T(0) − T₀)e^(−kt).
6. Interpret and validate	Check the solution against real data; refine if necessary.	Compare predicted and actual cooling curves.

The first step is the most important: if you do not identify the correct problem, your model will be irrelevant no matter how mathematically beautiful it is. Step 6 is equally critical — a model that fails to match observed data must be revised.

第一步是最重要的：如果你没有识别出正确的问题，无论模型在数学上多么优美，它都是无关紧要的。第6步同样关键——一个与观测数据不匹配的模型必须被修正。

3. Linear Models | 线性模型

Linear models are the simplest and most frequently tested type of model in secondary mathematics. They describe situations where there is a constant rate of change. The general form is y = mx + c, where m is the gradient (rate of change) and c is the y-intercept (initial value).

线性模型是中学数学中最简单、最常考的类型。它们描述的是变化率恒定的情况。一般形式为 y = mx + c，其中 m 是斜率（变化率），c 是截距（初始值）。

Linear models are appropriate when the relationship between two variables appears to be a straight line. For instance, if you earn a fixed hourly wage, your total pay is a linear function of the number of hours worked.

当两个变量之间的关系看起来是一条直线时，线性模型是合适的。例如，如果你的时薪固定，那么你的总工资就是工作小时数的线性函数。

P = 15h + 0, where P is total pay (in £) and h is hours worked.

P = 15h + 0，其中 P 是总工资（英镑），h 是工作小时数。

Gradient (m): The rate of change — how much y changes for each unit increase in x.
斜率 (m)： 变化率——x 每增加 1 个单位，y 变化多少。
Intercept (c): The value of y when x = 0 — the starting point.
截距 (c)： 当 x = 0 时 y 的值——起始点。

A classic exam question involves converting between temperature scales. The relationship between Celsius (C) and Fahrenheit (F) is a linear model:

一个经典的考试题涉及摄氏度（C）和华氏度（F）之间的转换。它们的关系是一个线性模型：

C = (5/9)(F − 32)

To build this model from two data points — the freezing point (0°C = 32°F) and boiling point (100°C = 212°F) — we use the two-point formula for a straight line. Try this in an exam: if you are given two (x, y) pairs, substitute both into y = mx + c and solve simultaneously.

为了从两个数据点——冰点（0°C = 32°F）和沸点（100°C = 212°F）——构建这个模型，我们使用直线的两点式公式。在考试中尝试这样做：如果你得到了两组 (x, y) 值，将它们代入 y = mx + c 并用联立方程求解。

4. Quadratic Models | 二次模型

Quadratic models are used when the relationship between variables involves acceleration, curvature, or a maximum/minimum point. The general form is y = ax² + bx + c, where a ≠ 0. Quadratic models appear frequently in projectile motion, area optimization, and profit maximization problems.

当变量之间的关系涉及加速度、曲率或最大值/最小值点时，使用二次模型。一般形式为 y = ax² + bx + c，其中 a ≠ 0。二次模型在抛体运动、面积优化和利润最大化问题中经常出现。

Consider the trajectory of a ball thrown upward. Its height h (in meters) after t seconds can be modeled as:

考虑一个向上抛出的球的轨迹。t 秒后其高度 h（米）可以建模为：

h(t) = −5t² + 20t + 2

Here, −5 comes from half the acceleration due to gravity (approximately 9.8 m/s², but conveniently rounded), 20 is the initial upward velocity (m/s), and 2 is the initial height (m).

这里，−5 来自重力加速度的一半（约为 9.8 m/s²，为简便取了近似值），20 是初始向上速度（m/s），2 是初始高度（m）。

Maximum height: Found by completing the square or using t = −b/(2a).
最大高度： 通过配方法或使用 t = −b/(2a) 求得。
Time when ball hits ground: Solve h(t) = 0 using the quadratic formula.
球落地时间： 解 h(t) = 0，使用求根公式。

t = −b ± √(b² − 4ac) / (2a)

For this example, a common exam question would be: “Find the maximum height of the ball.” The answer is found by computing t = −b/(2a) = −20/(2(−5)) = 2 seconds, then substituting back: h(2) = −5(2)² + 20(2) + 2 = 22 meters.

以本题为例，一个常见的考试问题是：”求球的最大高度。”答案通过计算 t = −b/(2a) = −20/(2(−5)) = 2 秒，然后代入：h(2) = −5(2)² + 20(2) + 2 = 22 米。

5. Exponential Models | 指数模型

Exponential models describe situations involving growth or decay where the rate of change is proportional to the current amount. The general form is y = A₀e^(kt) for continuous growth/decay, or y = A₀(1 ± r)^t for discrete growth/decay. They are used for population growth, radioactive decay, compound interest, and cooling of objects.

指数模型描述的是变化率与当前量成正比的情况，涉及增长或衰减。一般形式为 y = A₀e^(kt)（连续增长/衰减），或 y = A₀(1 ± r)^t（离散增长/衰减）。它们用于人口增长、放射性衰变、复利和物体冷却等问题。

A classic example is the depreciation of a car’s value. Suppose a car costs £20,000 and depreciates by 15% each year. The model is:

一个经典例子是汽车价值的折旧。假设一辆车价值 20,000 英镑，每年贬值 15%。模型为：

V(t) = 20000(0.85)^t

Where V(t) is the value after t years and 0.85 = 1 − 0.15 is the depreciation factor. After 3 years, the car’s value would be:

其中 V(t) 是 t 年后的价值，0.85 = 1 − 0.15 是折旧因子。3 年后，汽车的价值为：

V(3) = 20000 × (0.85)³ = 20000 × 0.614125 ≈ £12,282.50

In such problems, be careful with the interpretation of r. A depreciation of 15% per year means r = 0.15 is subtracted from 1. An appreciation (growth) of 10% per year would use 1.10 as the base.

在解决此类问题时，注意 r 的解释。每年 15% 的折旧意味着 r = 0.15 从 1 中减去。每年 10% 的增值（增长）则使用 1.10 作为底数。

6. Piecewise Models | 分段模型

Not all real-world situations can be described by a single simple equation. Sometimes, the relationship between variables changes after a certain point. In such cases, we use piecewise functions — different equations for different intervals of the independent variable.

并非所有现实情境都能用单一简单方程描述。有时，变量之间的关系会在某一点后发生变化。这种情况下，我们使用分段函数——在自变量的不同区间使用不同的方程。

A typical example is delivery charges. Suppose a courier charges £5 for deliveries under 2 kg, £8 for deliveries between 2 kg and 5 kg, and £10 plus £0.50 per kg above 5 kg. The cost function C(w) could be:

一个典型例子是快递费。假设一位快递员对 2 kg 以下的包裹收取 5 英镑，2 kg 到 5 kg 之间的包裹收取 8 英镑，超过 5 kg 的包裹收取 10 英镑加上每公斤 0.50 英镑。成本函数 C(w) 可以写成：

C(w) = { 5, if 0 < w < 2;
8, if 2 ≤ w < 5;
10 + 0.5(w − 5), if w ≥ 5 }

When solving piecewise model problems, always check the domain of each piece and ensure that the value of the variable falls into the correct interval before evaluating.

在解决分段模型问题时，始终要检查每段的定义域，并确保变量的值在计算前落入正确的区间。

7. Assumptions and Limitations | 假设与局限性

Every mathematical model rests on assumptions. These assumptions are simplifications of reality that make the mathematics tractable. However, they also impose limitations on the validity of the model.

每个数学模型都建立在假设之上。这些假设是对现实的简化，使数学处理变得可行。然而，它们也限制了模型的有效性。

For instance, in the linear pay model P = 15h, we assume: (1) the hourly wage stays constant regardless of hours worked; (2) there are no taxes or other deductions; (3) the worker is always productive. In reality, overtime may pay more, tax brackets apply, and productivity may decline.

例如，在线性工资模型 P = 15h 中，我们假设：(1) 无论工作多少小时，时薪保持不变；(2) 没有税收或其他扣款；(3) 工人始终保持高效。在现实中，加班可能支付更高报酬、存在税率档位、生产力也可能下降。

Assumptions in linear growth: Constant rate of change, no external factors.
线性增长中的假设： 变化率恒定，没有外部因素。
Assumptions in quadratic motion: No air resistance, constant gravity.
二次运动中的假设： 没有空气阻力，重力恒定。
Assumptions in exponential decay: Decay rate stays proportional, environment unchanged.
指数衰减中的假设： 衰减率始终成比例，环境不变。

In exam questions, you may be asked to critique a model. Common response points include: the model is too simplified, it does not account for random variation, or it is only valid within a certain range. Always mention that a model is an approximation, not an exact representation of reality.

在考试题目中，你可能会被要求评价一个模型。常见的回答要点包括：模型过于简化、没有考虑随机变化，或仅在某个范围内有效。始终要提到模型是近似，不是现实的精确表示。

8. Real-world Applications and Exam Examples | 实际应用与考试例题

Now let us work through a complete exam-style example to consolidate what we have learned.

现在让我们完整地做一道考试风格的例题来巩固所学知识。

Problem (Edexcel-style): A company sells widgets. The profit P (in £) from selling x hundred widgets is modeled by P(x) = −x² + 8x − 7, where x ≥ 0.

题目（Edexcel风格）： 一家公司销售小配件。卖出 x 百个配件获得的利润 P（英镑）由 P(x) = −x² + 8x − 7 建模，其中 x ≥ 0。

(a) Find the number of widgets that maximises profit.

(a) 求利润最大时的配件销售数量。

x = −b/(2a) = −8/(2(−1)) = 4

So 400 widgets. The maximum profit is P(4) = −(4)² + 8(4) − 7 = −16 + 32 − 7 = 9, that is £900.

所以是 400 个配件。最大利润为 P(4) = −(4)² + 8(4) − 7 = −16 + 32 − 7 = 9，即 900 英镑。

(b) Determine the break-even points (where profit is zero).

(b) 确定盈亏平衡点（利润为零的点）。

−x² + 8x − 7 = 0 → x² − 8x + 7 = 0 → (x − 1)(x − 7) = 0 → x = 1 or x = 7

Break-even occurs at 100 widgets and 700 widgets. In between, the company makes a profit; outside this range, it makes a loss.

盈亏平衡点出现在 100 个配件和 700 个配件时。在这之间，公司盈利；在这个范围之外，公司亏损。

(c) State one limitation of this model.

(c) 指出这个模型的一个局限性。

Acceptable answers include: the model assumes profit depends only on quantity sold, ignoring external factors such as advertising costs, competition, or seasonal demand; the quadratic shape implies profit will eventually become negative for very large x, but in reality a company might change its pricing structure.

可接受的答案包括：模型假设利润仅取决于销售数量，忽略了广告费用、竞争或季节性需求等外部因素；二次函数的形状意味着当 x 非常大时利润最终会变为负值，但实际上公司可能会改变其定价结构。

9. Tips for Building Models in Exams | 考试中构建模型的技巧

In exam situations, you will often be asked to build a model from a word problem. Here are proven strategies to help you succeed.

在考试中，你经常会被要求从文字题中构建模型。以下是被验证的成功策略。

Read carefully: Identify what quantity is to be maximised/minimised or found.
仔细阅读： 确定要求最大化/最小化或需要求解的量。
Name variables: Choose clear letters (x, y, t, P, etc.) and define their units.
命名变量： 选择清晰的字母（x, y, t, P 等）并定义它们的单位。
Look for keywords: “Constant rate” suggests linear; “maximum/minimum” suggests quadratic; “growth/decay percentage” suggests exponential.
寻找关键词： “恒定速率”暗示线性；”最大值/最小值”暗示二次；”增长/衰减百分比”暗示指数。
Check units: Ensure consistency — don’t mix hours and minutes, or meters and kilometers.
检查单位： 确保一致性——不要混用小时和分钟，或米和千米。
Test your model: Substitute a simple value to see if the result makes sense.
验证模型： 代入一个简单值，看结果是否符合常理。

Additionally, always state your assumptions clearly when constructing a model. In marking schemes, assumptions often carry marks. For example, “Assume the relationship is linear” might earn you a method mark even if your subsequent calculations contain a small error.

此外，在构建模型时始终清晰陈述你的假设。在评分标准中，假设往往带有分值。例如，”假设该关系是线性的”可能会让你获得方法分，即使后续计算有少量错误。

10. Validation and Refinement | 验证与优化

After building a model, it is essential to test it against real data or common sense. This is called validation. If the model fails to predict observed results, it needs refinement.

构建模型后，必须用真实数据或常识来检验它。这称为验证。如果模型无法预测已观察到的结果，就需要优化。

Suppose we model the height of a plant as h(t) = 2t centimeters, where t is weeks. If after two weeks the actual measurement is 5 cm instead of 4 cm, the model underestimates growth. The discrepancy could be due to variables we ignored, such as soil quality, sunlight, or water.

假设我们将植物的高度建模为 h(t) = 2t 厘米，其中 t 为周数。如果两周后实际测量值为 5 厘米而不是 4 厘米，则模型低估了生长。差异可能来自我们忽略的变量，如土壤质量、阳光或水分。

Refining a model might involve: adding more variables, changing the type of function, or adjusting parameter values. In this case, we could change the gradient to 2.5, giving h(t) = 2.5t, or we could add an intercept: h(t) = 2.5t + 0.5, reflecting that the plant was already 0.5 cm tall at t = 0.

优化模型可能涉及：添加更多变量、改变函数类型或调整参数值。在这种情况下，我们可以将斜率改为 2.5，得到 h(t) = 2.5t，或者添加一个截距：h(t) = 2.5t + 0.5，反映植物在 t = 0 时已经有 0.5 厘米高。

In exam questions, you may be asked to judge whether a model is fit for purpose. Consider both the residuals (differences between observed and predicted values) and the range of validity (whether the model works for all values of the domain or only a subset).

在考试题中，你可能会被要求判断一个模型是否适用。考虑残差（观测值与预测值之间的差异）以及有效性范围（模型是对定义域中的所有值都有效，还是只对部分值有效）。

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Mathematical Modeling: Building and Applying Basic Models | 数学模型：基础模型的建立与应用