📚 10E Optimization | 优化问题
Optimisation lies at the heart of applied calculus. In IB Mathematics, topic 10E challenges you to take a real-world scenario, model it with a function, and then use derivatives to find the maximum or minimum value of that function. Whether you are minimising the material for a box or maximising the profit from selling an item, the strategy remains consistent: express the quantity to be optimised as a single‑variable function, differentiate, and interpret the critical points. This article walks you through the entire process, from problem formulation to exam‑ready worked examples.
优化是应用微积分的核心。在IB数学中，10E专题要求你将现实情境转化为函数模型，然后利用导数找出该函数的最大值或最小值。无论是让盒子用料最少，还是让商品利润最大，策略都是一致的：把要优化的量表示成单变量函数，求导，再解读临界点。本文将带你走完整过程，从问题构建到考试实用的例题。
1. Understanding Optimisation | 理解优化
Optimisation means finding the 'best' possible outcome under given conditions. In calculus, 'best' usually refers to a maximum or minimum value of a function. These extreme values occur where the derivative is zero (stationary points) or at the boundaries of the domain. IB problems often ask you to optimise a quantity such as area, volume, cost, or profit.
优化意味着在给定条件下找出“最佳”结果。在微积分中，“最佳”通常指函数的最大值或最小值。这些极值出现在导数为零的点（驻点）或定义域的边界上。IB题目常要求你优化面积、体积、成本或利润等量。
Goal: Max or Min of f(x) for x ∈ [a, b]
目标：在区间 [a, b] 上求 f(x) 的最大值或最小值
2. Problem Formulation | 问题的建立
The first, and often most difficult, step is translating the word problem into mathematical language. Identify the quantity to be maximised or minimised, and name it (e.g. A for area, V for volume, P for profit). Then list any constraints that link the variables. Use the constraint to eliminate all but one variable, so that the quantity to be optimised becomes a function of a single variable.
第一步，往往也是最难的一步，就是把文字题翻译成数学语言。先确定要最大化或最小化的量，并给它命名（如面积 A、体积 V、利润 P）。然后列出联系变量的所有约束条件。利用约束消去多余的变量，使优化目标变成单一变量的函数。
- Read carefully: what exactly is to be optimised? 仔细阅读：究竟要优化什么？
- Draw and label a diagram if geometry is involved. 涉及几何时画出并标注图形。
- Write an equation for the constraint (e.g. perimeter = 100). 写出约束方程（如周长=100）。
- Express the objective in terms of one variable. 将目标表示成单一变量的表达式。
3. Objective Functions and Constraints | 目标函数与约束
In any optimisation problem there are two key ingredients: the objective function – the formula you want to maximise or minimise – and the constraint equation – the constraint for y gives y = 40 − x, so A(x) = x(40 − x). Now A is a function of only x.
任何优化问题都有两个关键要素：目标函数——你想最大化或最小化的式子，以及约束方程——变量之间必须满足的关系。例如，将一段长80 cm的铁丝弯成长方形，约束为2x + 2y = 80。面积A = xy 是目标。利用约束解出 y = 40 − x，于是 A(x) = x(40 − x)。此时 A 变成了只含 x 的函数。
Always check the feasible domain. In the rectangle example, x must be positive and less than 40, so x ∈ (0, 40). A clearly defined domain is essential for endpoint tests later.
务必检查可行域。在长方形例子中，x 必须为正且小于40，因此 x ∈ (0,40)。明确定义域对后面的端点检验至关重要。
4. Differentiation to Find Extrema | 求导以寻找极值
Once the objective function is written as a single‑variable expression f(x), differentiate it and set f' points. These are candidates for local maxima or minima. IB questions normally ask for the global maximum or minimum within the practical domain, but local extrema help you locate it.
一旦目标函数写成单变量表达式 f(x)，就对其求导并令 f'(x) = 0，找到驻点。这些点是局部最大或最小的候选点。IB题目通常要求实际定义域内的全局最大值或最小值，但局部极值能帮你定位。
f'(x) = 0 → stationary points
f'(x) = 0 → 驻点
If f(x) is a quadratic like A(x) = −x² + 40x, then A' point. For more complicated functions, you may need the product, quotient or chain rules.
如果 f(x) 是二次函数，比如 A(x) = −x² + 40x，那么 A'(x) = −2x + 40，唯一驻点为 x = 20。对于更复杂的函数，可能需要用积、商或链式法则。
5. First Derivative Test | 一阶导数检验
The first derivative test examines the sign of f'(x) on either side of a stationary point. If f' changes from positive to negative, the point is a local maximum. If it changes from negative to positive, it is a local minimum. If the sign does not change, the stationary point is a point of inflection.
一阶导数检验考察驻点两侧 f'(x) 的符号。若 f' 由正变负，则该点是局部极大值点；由负变正，则是局部极小值点；若符号不变，则为拐点。
e critical point you found really does give the desired maximum or minimum. For example, A'(x) = −2x + 40. For x < 20, A' > 0; for x > 20, A' < 0. Hence x = 20 yields a maximum area.
在优化情境中，你可以快速确认所找到的临界点确实给出了想要的最大或最小值。例如，A'(x) = −2x + 40。当 x < 20 时，A' > 0；当 x > 20 时，A' < 0。因此 x = 20 给出最大面积。
6. Second Derivative Test | 二阶导数检验
d, if the second derivative is easy to calculate, is the second derivative test. Compute f''(x) at the stationary point. If f''(x) < 0, the function is concave down and the point is a local maximum. If f''(x) > 0, the function is concave up and the point is a local minimum. If f''(x) = 0, the test is inconclusive and you must use the first derivative test.
如果二阶导数容易计算，更高效的方法是二阶导数检验。计算驻点处的 f''(x)。若 f''(x) < 0，函数下凹，该点为局部极大值点；若 f''(x) > 0，函数上凹，该点为局部极小值点；若 f''(x) = 0，检验失效，必须改为一阶导数检验。
Continuing the rectangle example, A'' nfirming a maximum. Most IB questions explicitly expect either the first or second derivative test as justification that a stationary point gives the required optimum.
继续长方形的例子，A''(x) = −2，恒为负，确认是最大值。大多数IB题目明确要求用一阶或二阶导数检验，作为驻点给出所求最优值的理由。
7. Endpoints and Global Extrema | 端点与全局极值
A stationary point found by f' he global maximum or minimum on a closed interval could occur at an endpoint. Therefore, after finding stationary points, always evaluate the objective function at the endpoints of the practical domain and compare values.
由 f'(x) = 0 得到的驻点可能是局部极值，但闭区间上的全局最大值或最小值可能出现在端点。因此，找到驻点后，务必在实用定义域的端点处计算目标函数，并比较数值。
In the wire‑ a is 400 cm². If the stationary point had been outside the domain, the optimum would lie at one of the endpoints.
在铁丝长方形问题中，A(0) = 0，A(40) = 0，而 A(20) = 400。最大面积是400 cm²。如果驻点落在定义域外，最优值就会出现在某个端点。
8. Worked Example: Geometric Optimisation | 例题：几何优化
Problem: A farmer has 60 m of fencing and wants to build a rectangular enclosure using a barn as one side. Find the dimensions that maximise the enclosed area.
问题：一位农场主有60 m围栏，想靠着一间谷仓建一个长方形围场。求使围场面积最大的尺寸。
Solution: y = 60 − 2x. Area: A = xy = x(60 − 2x) = 60x − 2x². Domain: x > 0 and y > 0 → 0 < x < 30.
解答：设垂直谷仓的两边宽为 x，平行谷仓的一边长为 y。约束：2x + y = 60 → y = 60 − 2x。面积：A = x(60 − 2x) = 60x − 2x²。定义域：x > 0 且 y > 0 → 0 < x < 30。
A'(x) = 60 − 4x = 0 → x = 15
A'(x) = 60 − 4x = 0 → x = 15
Second derivative: A'' m area = 15 × 30 = 450 m². Since A(0) = A(30) = 0, the global maximum is indeed at x = 15.
二阶导数：A''(x) = −4 < 0，确认最大。y = 60 − 2×15 = 30。最大面积 = 15×30 = 450 m²。因 A(0) = A(30) = 0，全局最大确实在 x = 15 处。
9. Worked Example: Economic Optimisation | 例题：经济优化
Problem: The cost of producing q units of a product is C(q) = 2q² + 100q + 5000. The items sell for $300 each. Find the production level that maximises profit.
问题：生产 q 件产品的成本为 C(q) = 2q² + 100q + 5000，每件售价300美元。求使利润最大化的产量。
R(q) − C(q) = 300q − (2q² + 100q + 5000) = −2q² + 200q − 5000. Domain: q ≥ 0 (practical).
收益：R(q) = 300q。利润：P(q) = −2q² + 200q − 5000。定义域：q ≥ 0。
P'(q) = −4q + 200 = 0 → q = 50
P'(q) = −4q + 200 = 0 → q = 50
P'' 0, so q = 50 gives a maximum. Maximum profit = P(50) = −2(2500) + 200(50) − 5000 = 0. Wait — profit is zero? Let' at is the best possible; producing more or less would result in a loss. Thus q = 50 units.
P''(q) = −4 < 0，故 q = 50 给出最大值。最大利润 = P(50) = −5000 + 10000 − 5000 = 0。稍等——利润为零？检验一下：P(50)确实是0，收入恰好抵消成本。这就是可能的最好情况；多产或少产都会亏损。因此最佳产量是50件。
10. IB Exam Tips and Summary | IB考试技巧与总结
In IB Mathematics exams, optimisation questions carry significant marks and follow a predictable pattern. Always show the constraint equation, the objective function in one variable, the derivative, the test for maximum/minimum, and the final answer with units. Marks are awarded for setting up the problem correctly, for calculus steps, and for interpretation. If a question asks you to 'justify that this gives the maximum', you must use the first or second derivative test – stating the derivative sign or concavity is enough. Never forget to check whether the value found lies within the feasible domain, and compare it with endpoint values when the domain is closed. Time management is crucial: practice drawing a quick sketch and writing concisely. The more you practise translating words into equations, the faster you become.
在IB数学考试中，优化题占分高且模式固定。一定要写出约束方程、单变量目标函数、导数、最大/最小值的检验以及最终答案（带单位）。正确设立问题、微积分步骤和结果解释都能得分。如果题目要求“证明这是最大值”，你必须使用一阶或二阶导数检验——指出导数的符号变化或凹凸性就足够了。永远不要忘记检查求出的值是否落在可行域内，并且在定义域为闭区间时与端点值比较。时间管理很关键：练习快速画草图和简明书写。你越多地练习将文字转化为方程，速度就会越快。
| Optimisation Workflow | 优化流程 |
| 1. Identify variables and constraint | 1. 确定变量与约束 |
| 2. Express objective in one variable | 2. 将目标表示为单变量函数 |
| 3. State the feasible domain | 3. 说明可行定义域 |
| 4. Differentiate and find f'(x)=0 | 4. 求导并解 f'(x)=0 |
| 5. Test using 1st or 2nd derivative | 5. 用一阶或二阶导数检验 |
| 6. Check endpoints if domain is closed | 6. 若定义域闭合则检查端点 |
| 7. State conclusion with units | 7. 给出结论并注明单位 |
Mastering optimisation equips you with a powerful tool that reaches far beyond the classroom – from designing efficient containers to planning production lines. Stick to the system, practise a variety of contexts, and you will find these problems both logical and rewarding.
掌握优化问题，能使你拥有一项超越课堂的强大工具——从设计高效容器到规划生产线。按照流程来，练习多样的情境，你会发现这些问题既富有逻辑又令人成就感十足。
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