Mastering Optimisation Problems | 优化问题的求解方法

📚 Mastering Optimisation Problems | 优化问题的求解方法

Optimisation is one of the most powerful applications of differentiation in A-Level Mathematics. It asks a simple question: among all possible choices, which one gives the largest or smallest value? Whether you are maximising profit, minimising material cost, or finding the fastest route, the underlying calculus is the same.

优化问题是A-Level数学中微分学最强大的应用之一。它提出一个简单的问题：在所有可能的选择中，哪一个能产生最大值或最小值？无论是最大化利润、最小化材料成本，还是寻找最快路线，其背后的微积分原理都是相同的。

1. What Is Optimisation? | 什么是优化问题

An optimisation problem involves finding the maximum or minimum value of a function, usually subject to a constraint. The function being optimised is called the objective function, and the constraint links the variables together. For example, you may need to maximise the volume of a box made from a fixed sheet of cardboard, or minimise the cost of fencing a rectangular field of given area.

优化问题涉及寻找函数的最大值或最小值，通常受某个约束条件的限制。被优化的函数称为目标函数，而约束条件将各个变量联系在一起。例如，你可能需要用一张固定大小的纸板制作一个盒子并最大化其体积，或者用给定面积的矩形围栏最小化围栏成本。

In mathematical terms, if the objective function is \(y = f(x)\), we seek values of \(x\) where \(f'(x) = 0\) or where the derivative does not exist, provided these lie within the feasible domain. These points are called critical points or stationary points.

用数学语言表达，如果目标函数为 \(y = f(x)\)，我们寻找使 \(f'(x) = 0\) 或导数不存在的 \(x\) 值，前提是这些值位于可行定义域内。这些点称为临界点或驻点。

2. Setting Up the Problem | 建立问题模型

The hardest part of optimisation is often translating a word problem into mathematics. A structured approach begins with three steps: draw a diagram, define variables clearly, and identify the quantity to be maximised or minimised.

优化问题中最困难的部分往往是将文字题转化为数学表达式。结构化的方法从三个步骤开始：画示意图、明确定义变量、识别需要最大化或最小化的量。

Suppose you are asked to find the maximum area of a rectangle with a fixed perimeter of 40 m. Let the length be \(x\) and the width be \(y\). The constraint is \(2x + 2y = 40\), so \(y = 20 – x\). The objective function (area) is \(A = xy = x(20 – x) = 20x – x²\).

假设要求找到周长为40米的矩形的最大面积。设长为 \(x\)，宽为 \(y\)。约束条件为 \(2x + 2y = 40\)，因此 \(y = 20 – x\)。目标函数（面积）为 \(A = xy = x(20 – x) = 20x – x²\)。

Once the objective is expressed as a single-variable function, the domain must be stated. Here, \(0 < x < 20\), since lengths must be positive. This domain will be important when checking endpoints.

一旦目标被表示为单变量函数，就必须明确定义域。此处 \(0 < x < 20\)，因为长度必须为正。这个定义域在检查端点时非常重要。

3. Finding Critical Points | 寻找临界点

To locate potential maxima and minima, we differentiate the objective function with respect to \(x\) and set the derivative equal to zero. These solutions are the stationary points.

为了找到可能的最大值和最小值，我们对目标函数关于 \(x\) 求导，并令导数等于零。这些解就是驻点。

For \(A = 20x – x²\), we have \(\frac{dA}{dx} = 20 – 2x\). Setting \(20 – 2x = 0\) gives \(x = 10\). This is the only critical point in the interval.

对于 \(A = 20x – x²\)，我们有 \(\frac{dA}{dx} = 20 – 2x\)。令 \(20 – 2x = 0\) 得 \(x = 10\)。这是区间内唯一的临界点。

It is essential to check whether the derivative exists everywhere in the domain. For polynomial and most rational functions encountered in A-Level, this is true. However, for functions involving absolute values or roots, extra care is needed.

必须检查导数在定义域内是否处处存在。对于A-Level中遇到的多项式和大多数有理函数，这通常是成立的。然而，对于涉及绝对值或根号的函数，需要格外小心。

4. Determining the Nature of Critical Points | 判断临界点的性质

Once a critical point is found, we must decide whether it corresponds to a maximum, a minimum, or a point of inflection. The first derivative test and the second derivative test are the two standard methods.

找到临界点后，我们必须判断它对应最大值、最小值还是拐点。一阶导数检验法和二阶导数检验法是两种标准方法。

First derivative test: Examine the sign of \(f'(x)\) just to the left and right of the critical point. If the sign changes from positive to negative, it is a local maximum; from negative to positive, it is a local minimum; if there is no sign change, it is a point of inflection.

一阶导数检验法：观察临界点左右两侧 \(f'(x)\) 的符号。如果符号从正变为负，则为局部最大值；从负变为正，则为局部最小值；如果没有符号变化，则为拐点。

Second derivative test: Compute \(f”(x)\) at the critical point \(x = c\). If \(f”(c) < 0\), the function is concave down, indicating a local maximum. If \(f''(c) > 0\), the function is concave up, indicating a local minimum. If \(f”(c) = 0\), the test is inconclusive.

二阶导数检验法：计算临界点 \(x = c\) 处的 \(f”(x)\)。如果 \(f”(c) < 0\)，函数为凹向下，表明是局部最大值。如果 \(f''(c) > 0\)，函数为凹向上，表明是局部最小值。如果 \(f”(c) = 0\)，则检验法无效。

For the area example, \(\frac{d²A}{dx²} = -2 < 0\), so \(x = 10\) gives a maximum. The maximum area is \(A = 10(20 - 10) = 100\) m².

对于面积示例，\(\frac{d²A}{dx²} = -2 < 0\)，因此 \(x = 10\) 给出最大值。最大面积为 \(A = 10(20 - 10) = 100\) 平方米。

5. Global Extrema on a Closed Interval | 闭区间上的全局最值

When the domain is a closed interval \([a, b]\), the global maximum or minimum may occur at a critical point inside the interval or at one of the endpoints. The Extreme Value Theorem guarantees that a continuous function on a closed interval attains both a maximum and a minimum.

当定义域为闭区间 \([a, b]\) 时，全局最大值或最小值可能出现在区间内部的临界点，也可能出现在端点处。极值定理保证闭区间上的连续函数一定能取得最大值和最小值。

The procedure is straightforward: list all critical points in \((a, b)\), evaluate the function at these points and at \(x = a\) and \(x = b\), and compare the values. The largest is the global maximum; the smallest is the global minimum.

求解步骤很直接：列出 \((a, b)\) 内所有临界点，计算这些点以及 \(x = a\) 和 \(x = b\) 处的函数值，然后进行比较。最大值为全局最大值，最小值为全局最小值。

For example, consider \(f(x) = x³ – 3x + 2\) on \([0, 2]\). The derivative \(f'(x) = 3x² – 3 = 3(x – 1)(x + 1)\) gives a critical point at \(x = 1\) (since \(x = -1\) is outside the interval). Evaluating: \(f(0) = 2\), \(f(1) = 0\), \(f(2) = 4\). Therefore, the global maximum is 4 at \(x = 2\), and the global minimum is 0 at \(x = 1\).

例如，考虑 \([0, 2]\) 上的函数 \(f(x) = x³ – 3x + 2\)。导数 \(f'(x) = 3x² – 3 = 3(x – 1)(x + 1)\) 给出临界点 \(x = 1\)（因为 \(x = -1\) 在区间外）。计算函数值：\(f(0) = 2\)，\(f(1) = 0\)，\(f(2) = 4\)。因此，全局最大值为 4（在 \(x = 2\) 处），全局最小值为 0（在 \(x = 1\) 处）。

6. Optimisation with Geometric Constraints | 带几何约束的优化问题

Many optimisation problems involve geometric shapes—cylinders, cones, spheres, or boxes—where the constraint is a fixed surface area or volume. The key is to use the constraint to eliminate one variable.

许多优化问题涉及几何形状——圆柱体、圆锥体、球体或盒子——此时约束条件是固定的表面积或体积。关键在于使用约束条件消去一个变量。

Worked example: A cylindrical tin has a fixed surface area of \(S = 600\pi\) cm². Find the radius and height that maximise the volume \(V = \pi r² h\).

例题：一个圆柱形罐头具有固定的表面积 \(S = 600\pi\) cm²。求使体积 \(V = \pi r² h\) 最大的半径和高度。

The surface area is the sum of the two circular ends and the curved surface: \(S = 2\pi r² + 2\pi r h = 600\pi\). Dividing by \(2\pi\): \(r² + rh = 300\), so \(h = \frac{300 – r²}{r}\).

表面积等于两个圆形底面与侧面的面积之和：\(S = 2\pi r² + 2\pi r h = 600\pi\)。两边除以 \(2\pi\)：\(r² + rh = 300\)，因此 \(h = \frac{300 – r²}{r}\)。

Substitute into the volume formula:

将 \(h\) 代入体积公式：

\(V(r) = \pi r² \cdot \frac{300 – r²}{r} = \pi(300r – r³)\)

Differentiate: \(\frac{dV}{dr} = \pi(300 – 3r²) = 3\pi(100 – r²)\). Setting this equal to zero gives \(r = 10\) cm (since \(r > 0\)). Then \(h = \frac{300 – 100}{10} = 20\) cm.

求导：\(\frac{dV}{dr} = \pi(300 – 3r²) = 3\pi(100 – r²)\)。令其等于零得 \(r = 10\) cm（因为 \(r > 0\)）。则 \(h = \frac{300 – 100}{10} = 20\) cm。

To confirm this is a maximum, compute the second derivative: \(\frac{d²V}{dr²} = -6\pi r\). At \(r = 10\), this is \(-60\pi < 0\), confirming a maximum. The maximum volume is \(V = \pi(300 \times 10 - 1000) = 2000\pi\) cm³.

为了确认这是最大值，计算二阶导数：\(\frac{d²V}{dr²} = -6\pi r\)。在 \(r = 10\) 处，其值为 \(-60\pi < 0\)，确认是最大值。最大体积为 \(V = \pi(300 \times 10 - 1000) = 2000\pi\) cm³。

7. Optimisation in Economics: Cost and Profit | 经济学中的优化：成本与利润

In economics, firms aim to minimise average cost or maximise profit. These problems follow the same calculus rules, but the functions often arise from given revenue and cost models.

在经济学中，企业追求平均成本最小化或利润最大化。这些问题遵循相同的微积分规则，但函数通常来自给定的收益和成本模型。

A typical model: total revenue is \(R(x) = p(x) \cdot x\), where \(p(x)\) is the price per unit and \(x\) is the number of units sold. Total cost \(C(x)\) includes fixed and variable costs. Profit is defined as \(P(x) = R(x) – C(x)\).

典型模型：总收入为 \(R(x) = p(x) \cdot x\)，其中 \(p(x)\) 是每单位价格，\(x\) 是销售数量。总成本 \(C(x)\) 包括固定成本和可变成本。利润定义为 \(P(x) = R(x) – C(x)\)。

To maximise profit, set \(P'(x) = 0\), which is equivalent to \(R'(x) = C'(x)\). This is the famous condition “marginal revenue equals marginal cost.” The second derivative \(P”(x) < 0\) ensures we have a maximum.

要使利润最大化，令 \(P'(x) = 0\)，这等价于 \(R'(x) = C'(x)\)。这就是著名的条件“边际收益等于边际成本”。二阶导数 \(P”(x) < 0\) 确保我们得到的是最大值。

Worked example: A company finds that the profit from producing \(x\) units is \(P(x) = -2x² + 80x – 500\). Find the production level that maximises profit.

例题：一家公司发现生产 \(x\) 件产品时的利润为 \(P(x) = -2x² + 80x – 500\)。求使利润最大的生产水平。

Differentiate: \(P'(x) = -4x + 80\). Setting to zero gives \(x = 20\). Since \(P”(x) = -4 < 0\), this is a maximum. The maximum profit is \(P(20) = -2(400) + 80(20) - 500 = 300\) units of currency.

求导：\(P'(x) = -4x + 80\)。令其等于零得 \(x = 20\)。由于 \(P”(x) = -4 < 0\)，这是最大值。最大利润为 \(P(20) = -2(400) + 80(20) - 500 = 300\) 货币单位。

8. Optimisation Involving Rates of Change | 涉及变化率的优化问题

Some optimisation problems require the use of related rates, especially when multiple quantities change over time. The key is to write the objective function in terms of a single variable before differentiating.

有些优化问题需要使用相关变化率，尤其是当多个量随时间变化时。关键在于先将要优化的函数写成单变量的形式，然后再求导。

A classic example involves a ladder sliding down a wall. If the bottom slides away at a constant rate, find the angle at which the area of the triangle formed by the ladder, wall, and ground is maximised.

经典例子是梯子沿墙滑落。如果梯子底部以恒定速率滑离墙壁，求梯子、墙壁和地面构成的三角形面积最大时的角度。

However, in exam questions, the time variable is usually eliminated before differentiation. For instance, if the ladder has length \(L\), let the bottom be \(x\) from the wall and the top be \(y\) above the ground. The constraint is \(x² + y² = L²\), and the area is \(A = \frac{1}{2}xy\). Using the constraint to express \(y = \sqrt{L² – x²}\), the area becomes \(A = \frac{1}{2}x\sqrt{L² – x²}\).

然而，在考试题中，通常会在求导之前消去时间变量。例如，如果梯子长度为 \(L\)，设底部距墙壁为 \(x\)，顶部距地面为 \(y\)。约束条件为 \(x² + y² = L²\)，面积为 \(A = \frac{1}{2}xy\)。利用约束条件将 \(y = \sqrt{L² – x²}\) 代入，面积变为 \(A = \frac{1}{2}x\sqrt{L² – x²}\)。

Rather than solving this directly, note that maximising \(A\) is equivalent to maximising \(A² = \frac{1}{4}x²(L² – x²)\), which is easier to differentiate. This “square trick” is very useful for functions involving square roots.

不过，与其直接求解，不如注意到最大化 \(A\) 等价于最大化 \(A² = \frac{1}{4}x²(L² – x²)\)，这样求导更容易。这种“平方技巧”对于涉及平方根的函数非常有用。

9. Common Mistakes and Pitfalls | 常见错误与易错点

Optimisation problems are notorious for small errors that lead to incorrect answers. Being aware of these common pitfalls can save valuable marks in the exam.

优化问题因小错误导致错误答案而闻名。了解这些常见陷阱可以在考试中保住宝贵的分数。

Forgetting the domain: Always state the feasible range of the variable. A critical point outside the domain is invalid, and endpoints may be the true extrema.

忘记定义域：始终明确变量的可行范围。定义域外的临界点是无效的，而端点可能是真正的最值。
Using the second derivative test when it fails: If \(f”(c) = 0\), the test is inconclusive. Use the first derivative test or examine the sign of \(f’\) around the point.

在二阶导数检验失效时仍使用它：如果 \(f”(c) = 0\)，检验法无效。应使用一阶导数检验法或观察 \(f’\) 在该点附近的符号。
Confusing local and global extrema: On a closed interval, always check endpoints. A local maximum is not necessarily the global maximum.

混淆局部最值和全局最值：在闭区间上，始终检查端点。局部最大值不一定是全局最大值。
Not simplifying before differentiating: Products and quotients can be simplified first to make differentiation easier and reduce errors.

求导前不化简：乘积和商可以先化简再进行求导，这样更容易且可以减少错误。
Neglecting units: In applied problems, always include units in the final answer, such as cm², m, or monetary units.

忽略单位：在应用题中，最终答案务必包含单位，如 cm²、m 或货币单位。

10. Step-by-Step Solution Strategy | 分步解题策略

When facing any optimisation problem, following a consistent strategy reduces errors and saves time. Use this checklist as a guide during the exam.

面对任何优化问题时，遵循一致的策略可以减少错误并节省时间。考试时可以将此检查清单作为指导。

Step / 步骤	Action / 行动
1	Read the problem and identify the quantity to be maximised or minimised. 阅读题目，确定要最大化或最小化的量。
2	Draw a diagram if possible; label variables clearly. 如果可能，画出示意图；清楚标注变量。
3	Write the constraint equation and use it to express the objective in terms of one variable. 写出约束方程，并用它把目标函数表示为单变量形式。
4	State the domain of the variable. 指明变量的定义域。
5	Differentiate the objective function; find critical points by setting \(f'(x) = 0\). 对目标函数求导；通过令 \(f'(x) = 0\) 寻找临界点。
6	Use the first or second derivative test to confirm whether each critical point is a max or min. 使用一阶或二阶导数检验法确认每个临界点是最大还是最小。
7	If the domain is closed, compare function values at critical points and endpoints. 如果定义域是闭区间，比较临界点和端点处的函数值。
8	Write the final answer in a sentence, including units and justifying that it is indeed the optimum. 用完整句子写出最终答案，包括单位，并说明它确实是最优值。

11. Worked Exam-Style Problem | 考试风格例题精讲

Let us apply the full strategy to a classic exam-style question: an open-top box is made by cutting equal squares of side \(x\) from each corner of a 30 cm by 20 cm rectangular sheet, then folding up the sides. Find the maximum volume.

让我们将完整策略应用于一道经典考试风格题目：从一块 30 cm × 20 cm 的矩形纸板的四个角各剪去边长为 \(x\) 的正方形，然后折起四周，制成一个无盖盒子。求最大体积。

Step 1 – Objective: The volume is \(V = \text{length} \times \text{width} \times \text{height}\). After cutting and folding, the dimensions are \((30 – 2x)\), \((20 – 2x)\), and \(x\). Thus:

步骤1——目标：体积为 \(V = \text{长} \times \text{宽} \times \text{高}\)。剪去并折叠后，尺寸为 \((30 – 2x)\)、\((20 – 2x)\) 和 \(x\)。因此：

\(V(x) = x(30 – 2x)(20 – 2x) = 4x³ – 100x² + 600x\)

Step 2 – Domain: The cut-out length must satisfy \(0 < x < 10\), because the smaller dimension is 20 cm, so \(20 - 2x > 0\).

步骤2——定义域：剪去的边长必须满足 \(0 < x < 10\)，因为较短边为 20 cm，所以 \(20 - 2x > 0\)。

Step 3 – Differentiate: \(\frac{dV}{dx} = 12x² – 200x + 600\). Set equal to zero: \(12x² – 200x + 600 = 0\), which simplifies to \(3x² – 50x + 150 = 0\).

步骤3——求导：\(\frac{dV}{dx} = 12x² – 200x + 600\)。令其等于零：\(12x² – 200x + 600 = 0\)，化简得 \(3x² – 50x + 150 = 0\)。

Solve using the quadratic formula:

使用二次公式求解：

\(x = \frac{50 \pm \sqrt{2500 – 1800}}{6} = \frac{50 \pm \sqrt{700}}{6} = \frac{50 \pm 10\sqrt{7}}{6}\)

This gives \(x \approx 12.74\) or \(x \approx 3.93\). Only \(x \approx 3.93\) lies in the domain \(0 < x < 10\).

得到 \(x \approx 12.74\) 或 \(x \approx 3.93\)。只有 \(x \approx 3.93\) 在定义域 \(0 < x < 10\) 内。

Step 4 – Confirm maximum: \(\frac{d²V}{dx²} = 24x – 200\). At \(x \approx 3.93\), \(\frac{d²V}{dx²} \approx 94.3 – 200 = -105.7 < 0\), confirming a maximum.

步骤4——确认最大值：\(\frac{d²V}{dx²} = 24x – 200\)。在 \(x \approx 3.93\) 处，\(\frac{d²V}{dx²} \approx 94.3 – 200 = -105.7 < 0\)，确认为最大值。

Step 5 – Final answer: \(V_{\text{max}} \approx 3.93(30 – 7.86)(20 – 7.86) \approx 3.93 \times 22.14 \times 12.14 \approx 1056\) cm³.

步骤5——最终答案：\(V_{\text{max}} \approx 3.93(30 – 7.86)(20 – 7.86) \approx 3.93 \times 22.14 \times 12.14 \approx 1056\) cm³。

12. Summary of Key Results | 核心结论总结

The table below summarises the essential techniques and their purposes:

下表总结了关键技术及其用途：

Technique / 技巧	Purpose / 用途
\(f'(x) = 0\)	Find critical points / 寻找临界点
Sign test of \(f'(x)\)	Classify maxima, minima, inflection / 分类最大值、最小值、拐点
\(f”(x) > 0\) or \(< 0\)	Confirm minimum or maximum / 确认最小值或最大值
Endpoint evaluation	Find global extrema on closed intervals / 在闭区间上寻找全局最值
Substitute constraint	Reduce to one variable / 化为一元函数
Squaring trick	Simplify square-root objectives / 化简含平方根的目标函数

Mastering optimisation requires practice. With a systematic approach—setting up the model, finding critical points, confirming their nature, and checking endpoints—you can confidently solve any optimisation problem in the A-Level syllabus.

掌握优化问题需要练习。通过系统化的方法——建立模型、寻找临界点、确认其性质并检查端点——你可以自信地解决A-Level考纲中的任何优化问题。

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Mastering Optimisation Problems | 优化问题的求解方法