📚 Modelling with Differentiation | 用微分法建模
Modelling with differentiation is one of the most practical parts of Edexcel A-Level Mathematics. It turns real-world problems about maximum volume, minimum cost, fastest rate or changing quantities into equations, then uses derivatives to find optimal or instantaneous values.
用微分法建模是 Edexcel A-Level 数学中最实用的内容之一。它将现实世界中最大体积、最小成本、最快变化率或变化中的量转化为方程，再利用导数求出最优值或瞬时值。
In the exam, questions often begin with a written description, ask you to form a function, and then require you to differentiate, solve and interpret. This article covers the full modelling cycle and the most common techniques.
在考试中，题目通常先给出一段文字描述，要求你建立函数，然后求导、求解并解释结果。本文涵盖完整的建模流程和最常见的方法。
1. What Is Modelling with Differentiation? | 什么是微分建模？
Differentiation modelling means using a function to represent a real quantity and applying dy/dx or f'(x) to locate optimum values or to connect rates of change.
微分建模是指用一个函数表示实际量，并应用 dy/dx 或 f'(x) 来寻找最优值或联系变化率。
Typical Edexcel contexts include maximising the volume of a box made from a cut sheet, minimising the surface area of a can for a fixed volume, or finding the speed of a moving shadow.
Edexcel 的典型情境包括：用裁剪的纸板制作盒子并使其体积最大、在固定体积下使罐体表面积最小，或求移动影子的速度。
Mathematically, we write quantity Q = f(x), define a realistic domain, and solve f'(x) = 0. We then use f''(x) to identify a maximum or minimum.
从数学上，我们写出量 Q = f(x)，确定符合实际的定义域，并解 f'(x) = 0。然后用 f''(x) 判断最大值或最小值。
2. The Core Modelling Cycle | 建模核心流程
The modelling cycle has four stages: define variables, build an expression, differentiate, and solve or interpret.
建模循环有四个阶段：定义变量、建立表达式、求导、求解或解释。
- Define each variable and state its units. | 定义每个变量并注明单位。
- Write the objective function in one variable if possible. | 尽可能将目标函数写成单变量函数。
- Differentiate and find stationary points. | 求导并找到驻点。
- Classify the stationary points and answer in context. | 判定驻点性质并在题目情境中作答。
Most marks in Edexcel modelling questions are method marks, so a clear labelled method is more important than a quick answer.
Edexcel 建模题中的大多数分数是方法分，因此清晰标注步骤比快速得出答案更重要。
3. Translating Words into Variables | 将文字转化为变量
Read the question and label unknowns such as x, h, r or t. Always note what the question asks you to maximise or minimise.
阅读题目并用 x、h、r 或 t 等标记未知量。始终注意题目要求最大化或最小化什么。
Worked translation: a farmer uses 200 m of fencing for three sides of a rectangular pen built against a wall. Let the equal widths be x and the side parallel to the wall be y.
转化示例：一位农民用 200 m 围栏围一个靠墙的矩形围栏的三边。设两个等宽为 x，与墙平行的边为 y。
onstraint is 2x + y = 200, so y = 200 − 2x. The area is A = xy = x(200 − 2x) = 200x − 2x².
约束为 2x + y = 200，因此 y = 200 − 2x。面积为 A = xy = x(200 − 2x) = 200x − 2x²。
This is the translation step: every extra condition must become an equation linking the variables.
这就是转化步骤：每一个附加条件都必须变成联系变量的方程。
4. Constructing the Objective Function | 构造目标函数
Most optimisation problems have a constraint linking two variables. Use the constraint to eliminate one variable so the quantity to optimise is expressed in terms of one variable only.
大多数优化问题都有一个联系两个变量的约束。利用约束消去一个变量，使待优化量只用一个变量表示。
V = πr²h gives h = V₀/(πr²). The surface area S = 2πr² + 2πrh becomes S(r) = 2πr² + 2V₀/r.
对于固定体积 V₀ 的密闭圆柱体，V = πr²h 得 h = V₀/(πr²)。表面积 S = 2πr² + 2πrh 变为 S(r) = 2πr² + 2V₀/r。
This is the key step: optimisation only works cleanly when the function is in one variable. Do not differentiate before substituting the constraint.
这是关键步骤：只有将函数化为单变量，优化才能顺利进行。不要在代入约束条件之前就求导。
5. Stationary Points and Optimisation | 驻点与最优化
the maximum or minimum of y = f(x), solve dy/dx = 0. The solutions are stationary points.
要求 y = f(x) 的最大值或最小值，解 dy/dx = 0。所得解为驻点。
V = 2x² − 12x + 10, dV/dx = 4x − 12, dV/dx = 0 ⇒ x = 3
n the farmer example, A = 200x − 2x², so dA/dx = 200 − 4x. Setting dA/dx = 0 gives x = 50.
在农民围栏示例中，A = 200x − 2x²，所以 dA/dx = 200 − 4x。令 dA/dx = 0 得 x = 50。
Always check the domain: x may be limited by physical constraints, such as 0 < x < 100 for the fencing problem.
始终检查定义域：x 可能受物理条件限制，例如围栏问题中 0 < x < 100。
6. Classifying Maxima and Minima | 判定极大值与极小值
Stationary points can be local maxima, local minima or points of inflection. Use the second derivative test when possible.
驻点可能是局部最大值、局部最小值或拐点。尽可能使用二阶导数判定。
| Condition | Conclusion |
|---|---|
| d²y/dx² < 0 | Local maximum | 局部最大值 |
| d²y/dx² > 0 | Local minimum | 局部最小值 |
| d²y/dx² = 0 | Use first derivative test | 使用一阶导数判定 |
For A = 200x − 2x², d²A/dx² = −4 < 0, so x = 50 gives a maximum area.
对于 A = 200x − 2x²，d²A/dx² = −4 < 0，所以 x = 50 处取得最大面积。
In a closed interval, also compare the function values at the endpoints before concluding.
在闭区间上，得出结论前还要比较端点的函数值。
7. Common Geometrical Models | 常见几何模型
Edexcel questions often use rectangles, boxes, cylinders, sectors and triangles. Learn the standard area and volume formulas so you can build models quickly.
Edexcel 考题常使用矩形、长方体、圆柱、扇形和三角形。熟练掌握标准面积和体积公式能帮助你快速建模。
- Box: V = lwh, surface area S = 2lw + 2lh + 2wh | 盒子：V = lwh，表面积 S = 2lw + 2lh + 2wh
- Closed cylinder: V = πr²h, surface area S = 2πr² + 2πrh | 密闭圆柱：V = πr²h，表面积 S = 2πr² + 2πrh
- Open-top box from a rectangle: after cutting squares of side x, V = x(L − 2x)(W − 2x) | 从矩形纸板四角剪去边长为 x 的正方形后，V = x(L − 2x)(W − 2x)
If a question gives a fixed area or volume, use that information as the constraint before differentiating.
如果题目给出固定面积或体积，在求导前应先将该信息作为约束条件使用。
8. Related Rates of Change | 相关变化率
Differentiation can also link rates of change using the chain rule. This is useful when two quantities change over time.
微分还可以通过链式法则把变化率联系起来。当两个量随时间变化时，这非常有用。
dy/dt = (dy/dx) × (dx/dt)
ple: a spherical balloon is inflated so that dV/dt = 20 cm³ s⁻¹. Find dr/dt when r = 5 cm.
例如：一个球形气球被充气，dV/dt = 20 cm³ s⁻¹。求 r = 5 cm 时的 dr/dt。
Since V = (4/3)πr³, dV/dr = 4πr². By the chain rule, dr/dt = (dV/dt)/(dV/dr).
因为 V = (4/3)πr³，dV/dr = 4πr²。由链式法则，dr/dt = (dV/dt)/(dV/dr)。
At r = 5, dV/dr = 4π(5)² = 100π, so dr/dt = 20/(100π) = 1/(5π) cm s⁻¹.
当 r = 5 时，dV/dr = 4π(5)² = 100π，所以 dr/dt = 20/(100π) = 1/(5π) cm s⁻¹。
9. Interpreting and Validating Results | 结果解释与检验
Always return to the original question. State the value found, its units, and whether it is a maximum or minimum. Reject any solution outside the physical domain.
一定要回到原题。指出所求得的值、单位以及它是最大值还是最小值。剔除所有超出实际定义域的解。
ivative gives x = −2 for a length, it must be rejected because lengths cannot be negative.
如果导数给出 x = −2 作为长度，必须舍去，因为长度不能为负。
Check that the answer makes sense: a maximum volume should not be infinite, and a minimum cost should not be negative.
检查答案是否合理：最大体积不应为无穷大，最小成本不应为负。
roblem, x = 50 gives y = 200 − 2(50) = 100, so the maximum area is A = 50 × 100 = 5000 m².
对于农民围栏问题，x = 50 得 y = 200 − 2(50) = 100，所以最大面积为 A = 50 × 100 = 5000 m²。
10. Exam Technique and Common Pitfalls | 考试技巧与常见易错点
ur variables, write dQ/dx, solve dQ/dx = 0, classify with d²Q/dx², and write a conclusion.
采用清晰的方法：写出变量，给出 dQ/dx，解 dQ/dx = 0，用 d²Q/dx² 判定，最后写出结论。
- Do not forget units in the final answer. | 最终答案不要忘记单位。
- Do not differentiate before substituting the constraint. | 不要在代入约束条件之前就求导。
- Check endpoints for closed intervals. | 在闭区间上要检查端点。
- Show complete working because Edexcel awards method marks. | 展示完整步骤，因为 Edexcel 会给方法分。
- Use exact values such as π or √2 unless the question asks for decimals. | 除非题目要求小数，否则使用 π 或 √2 等精确值。
Practice past-paper questions that require proving a formula, then optimising it, then interpreting the result. This is the standard pattern in Edexcel A-Level Mathematics.
练习历年真题中需要先证明公式、再优化公式、最后解释结果的题型。这是 Edexcel A-Level 数学的标准模式。
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