📚 Exercise 14D: Optimisation Problems | 练习14D：最优化问题
Optimisation is a central theme in IB Mathematics: Analysis and Approaches, and Exercise 14D is designed to help students master the technique of finding maximum and minimum values of functions derived from real-world or geometric contexts. This article explores the concepts, strategies, and common pitfalls associated with such problems, providing a thorough revision guide for tackling Exercise 14D with confidence.
最优化是IB数学：分析与方法的核心主题，练习14D旨在帮助学生掌握从现实或几何情境中推导函数并求其最大值和最小值的技术。本文探讨相关概念、策略和常见错误，为自信应对练习14D提供全面的复习指南。
1. Understanding Optimisation Problems | 理解最优化问题
An optimisation problem asks for the 'best' possible outcome under given constraints. In calculus, this usually means finding the maximum or minimum value of a function, such as maximising profit, minimising surface area for a fixed volume, or finding the dimensions that give the greatest area. The function to be optimised is called the objective function.
最优化问题要求在给定约束下寻找“最佳”可能结果。在微积分中，这通常意味着求函数的最大值或最小值，例如最大化利润、在固定体积下最小化表面积，或找出使面积最大的尺寸。被优化的函数称为目标函数。
2. Step 1: Define Variables and Constraints | 步骤1：定义变量与约束
Begin by clearly defining the variables involved, usually with a diagram. Identify the quantity to be maximised or minimised (e.g., area A, volume V, cost C). Then express any constraints as equations linking the variables. For instance, if a wire of fixed length is bent into a shape, the perimeter is fixed; if a container has a given volume, the dimensions are related by that volume.
首先要清晰地定义所涉及的变量，通常借助示意图。确定待最大化或最小化的量（如面积A、体积V、成本C）。然后将任何约束条件表示为变量间的方程。例如，若一根固定长度的铁丝弯成某个形状，周长是固定的；若容器有给定的容积，各尺寸便由该体积相互关联。
3. Step 2: Construct the Objective Function | 步骤2：构建目标函数
n A(x) = x((P/2) - x). This reduction to one variable is essential before differentiation.
利用约束方程消去一个变量，使目标函数仅用单一自变量表达。例如，若周长为 P = 2x + 2y，且要最大化面积 A = xy，代入 y = (P/2) - x 得 A(x) = x((P/2) - x)。在求导之前将问题化为一元函数至关重要。
4. Step 3: Use Calculus to Find Critical Points | 步骤3：用微积分求驻点
Differentiate the objective function with respect to the chosen variable, set the first derivative equal to zero, and solve for the variable. These solutions are critical points where a maximum or minimum may occur. Always confirm that the critical point lies within the practical domain of the problem (positive lengths, positive time, etc.).
对所选变量求导，令一阶导数为零并解出变量。这些解就是可能发生极大值或极小值的驻点。务必确认该驻点落在问题的实际定义域内（例如正的长度、正的时间等）。
dA/dx = 0
5. Step 4: Verify Nature of Stationary Points | 步骤4：验证驻点性质
After finding a critical point, determine whether it yields a maximum or a minimum. Two common methods are the second derivative test and the first derivative test. For the second derivative test: if f''(x) < 0 at the point, it is a local maximum; if f''(x) > 0, it is a local minimum. A sign change of the first derivative around the critical point also indicates its nature.
找到驻点后，需判断其对应极大值还是极小值。两种常用方法是二阶导数检验和一阶导数检验。对于二阶导数检验：若该点处 f''(x) < 0，则为局部极大值；若 f''(x) > 0，则为局部极小值。一阶导数在驻点两侧的符号变化同样可以确定极值类型。
6. Handling Domain Restrictions | 处理定义域限制
In optimisation problems, the independent variable is often limited by physical reality. For example, lengths cannot be negative, and the dimensions derived must satisfy all constraints. Always define the domain explicitly, and check the endpoints if the function is continuous on a closed interval. The global maximum or minimum might occur at an endpoint rather than at a stationary point.
在最优化问题中，自变量常受现实条件限制。例如长度不能为负，推导出的尺寸必须满足所有约束。务必明确定义域，若函数在闭区间上连续，还需检查端点。全局最大值或最小值可能出现在端点而非驻点。
7. The First Derivative Test vs Second Derivative Test | 一阶导数检验与二阶导数检验
The first derivative test examines the sign of f'(x) on either side of a critical point. If f' bust when the second derivative is difficult to compute or equals zero. The second derivative test is quicker but inconclusive when f''(x) = 0 or does not exist.
一阶导数检验观察驻点两侧 f'(x) 的符号。若 f'(x) 由正变负，驻点为局部极大值；由负变正则为局部极小值。当二阶导数难以计算或为零时，此方法更可靠。二阶导数检验更为快捷，但当 f''(x) = 0 或不存在时则无法判断。
8. Typical Optimisation Models in Exercise 14D | 练习14D中的典型优化模型
Exercise 14D features several classic contexts. These include: maximising the area of a rectangle inscribed in a curve; minimising the surface area of a cylinder or cuboid for a fixed volume; determining the dimensions of a box made by cutting squares from corners; shortest paths and minimising distance; and economic models such as minimising average cost or maximising revenue. Recognising these patterns speeds up problem-solving.
练习14D涉及几种经典情境。包括：最大化内接于曲线的矩形面积；在固定体积下最小化圆柱或长方体的表面积；通过切去四角制作盒子以最大化容积；最短路径及最小化距离问题；以及经济学模型，如最小化平均成本或最大化收益。识别这些模式有助于快速解题。
| Context | Objective | Typical Constraint |
|---|---|---|
| Rectangle in semicircle | Max area A | Equation of circle |
| Cylinder with fixed volume | Min surface area S | V = πr²h |
| Box from a sheet | Max volume V | Dimensions of original sheet |
| Cost model | Min total cost C | Production level x > 0 |
9. Worked Example: Maximising Volume of a Box | 演练示例：使盒子体积最大化
An open box is made from a 30 cm by 20 cm rectangular sheet by cutting squares of side x cm from each corner and folding up the sides. Find the value of x that maximises the volume, and verify it is a maximum.
用一块30 cm × 20 cm的矩形纸板制作无盖盒，方法是从每个角切去边长为 x cm 的正方形，然后折起边框。求使体积最大的 x 值，并验证其为极大值。
(x) = x(30 - 2x)(20 - 2x) = 4x³ - 100x² + 600x. Domain: 0 < x < 10. Differentiate: V'(x) = 12x² - 200x + 600. Set V'(x) = 0 → 12x² - 200x + 600 = 0 → divide by 4: 3x² - 50x + 150 = 0.
体积函数：V(x) = x(30 - 2x)(20 - 2x) = 4x³ - 100x² + 600x。定义域：0 < x < 10。求导：V'(x) = 12x² - 200x + 600。令 V'(x) = 0 得 12x² - 200x + 600 = 0，除以 4：3x² - 50x + 150 = 0。
Solve using quadratic formula: x = [50 ± √2500 - 1800] / 6 = [50 ± √700] / 6 = [50 ± 10√7 ) and x ≈ (50 - 26.46)/6 ≈ 3.92 cm. Thus x ≈ 3.92 cm is the only critical point in domain.
用二次公式求解：x = [50 ± √2500 - 1800] / 6 = [50 ± √700] / 6 = [50 ± 10√7] / 6。近似值：x ≈ (50 + 26.46)/6 ≈ 12.74（舍去，因为 >10）以及 x ≈ (50 - 26.46)/6 ≈ 3.92 cm。故 x ≈ 3.92 cm 是定义域内唯一的驻点。
Second derivative: V''(x) = 24x - 200. At x ≈ 3.92, V'' he maximum volume is approximately V(3.92) = 4(3.92)³ - 100(3.92)² + 600(3.92) ≈ 1050 cm³.
二阶导数：V''(x) = 24x - 200。当 x ≈ 3.92 时，V''(3.92) = 24(3.92) - 200 = 94.08 - 200 = -105.92 < 0，确认是最大值。最大体积约为 V(3.92) = 4(3.92)³ - 100(3.92)² + 600(3.92) ≈ 1050 cm³。
10. Common Errors and How to Avoid Them | 常见错误及避免方法
Many students forget to express the objective function in one variable before differentiating, or they fail to check the domain. Another frequent mistake is solving the derivative equation but not verifying the nature of the critical point. In the box problem above, both solutions of the quadratic are mathematically valid, but only one lies inside the practical domain. Always interpret results in context.
许多学生忘记在求导前将目标函数化为一元表达式，或未检查定义域。另一个常见错误是求解了导函数方程却没有验证驻点性质。在上述盒子问题中，二次方程的两个解在数学上均有效，但只有一个落在实际定义域内。务必在情境中解读结果。
- Write a clear diagram with variables labelled.
- State the constraint and the objective function explicitly.
- Simplify to one variable before differentiating.
- Solve f'(x) = 0 and check domain endpoints.
- Confirm max/min using first or second derivative test.
- 画出清晰示意图并标注变量。
- 明确写出约束条件和目标函数。
- 求导前化简为一元函数。
- 解 f'(x) = 0 并检查定义域端点。
- 用一阶或二阶导数检验确认极大/极小值。
11. IB Exam Tips for Optimisation Questions | IB考试最优化问题提示
Optimisation problems often appear in Paper 1 (without calculator) and Paper 2 (with calculator) in both SL and HL. In Paper 1, answers may include surds such as √7; in Paper 2, you may be asked to give decimal approximations. Marks are awarded for setting up the function correctly (method marks), differentiating accurately, solving, and interpreting the result. Always justify your conclusion with a sign diagram or second derivative test.
最优化问题常在SL和HL的试卷1（无计算器）和试卷2（可用计算器）中出现。在试卷1中，答案可能包含根式如√7；在试卷2中，可能要求十进制近似值。分数通常分配给正确建立函数（方法分）、准确求导、求解及解读结果。务必用符号图或二阶导数检验来论证你的结论。
12. Conclusion: Practising Exercise 14D Effectively | 结论：有效练习练习14D
Mastering Exercise 14D means internalising a structured approach: define, reduce, differentiate, solve, and verify. Working through each problem step by step builds fluency not only in calculus but also in connecting abstract mathematics to tangible situations. As you complete the exercise, challenge yourself by varying the given numbers and creating your own similar problems. This deepens understanding and prepares you for any optimisation question on the IB exam.
掌握练习14D意味着内化一套结构化的方法：定义、化简、求导、求解和验证。逐步完成每一道题不仅能提高微积分熟练度，还能锻炼将抽象数学与现实情境联系起来的思维。在完成练习的过程中，尝试更换题中数字并自编相似题目，这能加深理解，让你从容应对IB考试中的任何最优化问题。
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