📚 Mastering Optimization Problems (IB Math Exercise 19C) | 掌握优化问题 (IB 数学练习19C)
Optimization is one of the most powerful applications of differentiation in the IB Mathematics Analysis and Approaches course. In Exercise 19C, you are asked to find maximum and minimum values of quantities such as area, volume, cost, or distance. These real‑world problems may seem challenging at first, but they become straightforward once you master the method: translate a word problem into a single‑variable function, then use derivatives to locate and verify the optimum.
优化是IB数学分析与方法课程中微分应用中最强大的工具之一。在练习19C中，你需要求解面积、体积、成本或距离等量的最大值和最小值。这些实际问题初看起来可能很棘手，但一旦你掌握了方法——把文字题转化为单变量函数，然后用导数定位并验证最优值——它们就会变得非常简单。
1. What Is Optimization? | 什么是优化？
Optimization is the process of finding the best possible solution to a problem under given constraints. In calculus, we achieve this by modelling the quantity we want to maximise or minimise as a function of one independent variable, then applying differentiation to locate the point where the derivative is zero or undefined. The fundamental idea comes from Fermat's theorem: if a differentiable function has a local extremum at an interior point, the derivative at that point must be zero.
优化就是在给定约束条件下，找到问题的最佳可能解的过程。在微积分中，我们通过将想要最大化或最小化的量建模成一个自变量的函数，然后应用微分来找到导数为零或未定义的点，从而实现这一点。其基本思想来自费马定理：如果一个可微函数在某个内点有局部极值，那么该点处的导数必定为零。
2. Key Concepts: Derivatives and Critical Points | 关键概念：导数和临界点
rval. An interior point where f′(x)=0 is often called a stationary point or turning point.
一阶导数 f′(x) 告诉我们切线的斜率。对于区间上的连续函数优化问题，绝对最大值或最小值的候选点是临界点（即 f′(x)=0 或 f′(x) 不存在的点）以及区间的端点。使得 f′(x)=0 的内点通常称为驻点或转向点。
ext: a negative length or a radius that breaks the model must be excluded, so the domain is crucial.
在IB课程中，要求通过解方程 f′(x)=0 来寻找驻点。务必检查实际背景：负的长度或破坏模型的半径必须排除，因此定义域至关重要。
3. The First Derivative Test | 一阶导数检验
the first derivative test determines whether it is a local maximum, a local minimum, or neither. Examine the sign of f′(x) just to the left and just to the right of c. If f′(x) changes from positive to negative, f has a local maximum at c. If it changes from negative to positive, f has a local minimum. If the sign does not change, the point is a stationary inflection.
找到驻点 x=c 后，一阶导数检验可以确定它是局部最大值、局部最小值抑或两者都不是。检查 c 点左侧和右侧 f′(x) 的符号。如果 f′(x) 由正变负，f 在 c 处取得局部最大值。如果由负变正，则取得局部最小值。若符号不变，该点是一个驻拐点。
This test is especially helpful when the second derivative is difficult to compute or zero. In optimisation word problems, you usually only have one physical stationary point; the first derivative test can quickly confirm it is the desired maximum or minimum.
当二阶导数难以计算或为零时，该检验尤其有用。在优化文字题中，通常只有一个符合实际意义的驻点；一阶导数检验可快速确认它就是所需的最大值或最小值。
4. The Second Derivative Test and Concavity | 二阶导数检验和凹凸性
′(c) = 0, the test is inconclusive, and you should fall back on the first derivative test.
二阶导数 f′′(x) 衡量斜率的变化率。若 f′′(c) < 0，函数在 c 处为凹向下，f(c) 为局部最大值；若 f′′(c) > 0，函数为凹向上，f(c) 为局部最小值。若 f′′(c) = 0，该检验无法得出结论，应转而使用一阶导数检验。
For many simple polynomial and power‑law models, the second derivative test is faster. Just plug the stationary x‑value into f′′(x) and check the sign.
对于许多简单的多项式模型和幂函数模型，二阶导数检验更快。只需将驻点的 x 值代入 f′′(x) 并检查符号。
| Test | Condition | Conclusion |
|---|---|---|
| First derivative | f′(x) changes + → − | Local maximum |
| f′(x) changes − → + | Local minimum | |
| Second derivative | f′′(c) < 0 | Local maximum |
| f′′(c) > 0 | Local minimum |
上表总结了两种常用检验方法。在实际考试中，只要清晰展示你的推理，两种方法均可满分。
5. Five Steps to Solve an Optimisation Problem | 解决优化问题的五个步骤
Every optimisation question in Exercise 19C can be handled with a consistent strategy:
练习19C中的每个优化问题都可以通过一套固定的策略来解决：
Step 1: Identify variables and the quantity to optimise. Give clear definitions, drawing a diagram if possible. Express the quantity Q you wish to maximise or minimise in terms of one variable, using any given constraint to eliminate other letters.
第1步：确定变量和优化量。给出清晰的变量定义，并尽可能画图。利用所给约束条件消去其他字母，将目标量 Q 用单一变量表示。
Step 2: State the domain. The domain is determined by physical or geometric restrictions. For example, lengths must be positive, and products must remain non‑negative.
第2步：声明定义域。定义域由物理或几何限制决定。例如，长度必须为正，乘积必须保持非负。
Step 3: Differentiate. Compute dQ/dx and set it equal to zero to find stationary points. Simplify the derivative before solving.
第3步：求导。计算 dQ/dx 并令其为零，求出驻点。在求解前先将导数表达式化简。
Step 4: Test the stationary point(s). Apply either the first or second derivative test to confirm the nature. If there is only one stationary point in the domain and the context clearly implies a maximum or minimum exists, you may justify with a brief sign table or concavity argument.
第4步：检验驻点。用一阶或二阶导数检验确认极值类型。若定义域内仅有一个驻点且实际背景明确存在极值，可以用简单的符号表或凹凸性来论证。
Step 5: Answer the question. Substitute the optimal x back into the original expression for Q and any other required dimensions. Write a concluding sentence with appropriate units.
第5步：回答问题。将最优的 x 代回 Q 的原始表达式以及任何其他需要求出的尺寸。写一句包含合适单位的结论。
6. Worked Example 1: Maximising a Rectangular Area | 实例1：矩形面积最大化
Problem: A farmer has 200 m of fencing and wants to enclose a rectangular field bordering a straight river. No fence is needed along the river. Find the dimensions that maximise the enclosed area.
问题：一位农民有200米围栏，想沿着一条笔直的河流围出一个矩形田地。靠河一边无需围栏。求使得围成面积最大化的尺寸。
Solution: e only three sides are fenced. The area A = x(200 − 2x) = 200x − 2x². Domain: 0 < x < 100.
解：设垂直于河流的边长为 x 米。则平行于河流的边长为 (200 − 2x) 米，因为只围三边。面积 A = x(200 − 2x) = 200x − 2x²。定义域：0 < x < 100。
Differentiate: dA/dx = 200 − 4x. Set dA/dx = 0 → 200 − 4x = 0 → x = 50.
求导：dA/dx = 200 − 4x。令 dA/dx = 0 → 200 − 4x = 0 → x = 50。
the domain, it yields the absolute maximum area. The parallel side is 200 − 2(50) = 100 m.
二阶导数：d²A/dx² = −4 < 0，故函数凹向下，x=50 处为局部最大值。因为它是定义域内唯一的驻点，所以得到绝对最大面积。平行边长为 200 − 2(50) = 100 米。
Maximum area = 50 × 100 = 5000 m².
最大面积 = 50 × 100 = 5000 平方米。
7. Worked Example 2: Minimising Surface Area of a Cylinder | 实例2：圆柱体表面积最小化
Problem: A cylindrical can is to hold 500 cm³ of soup. The material for the top and bottom costs twice as much per cm² as the material for the side. Find the radius r and height h that minimize the cost. (Ignore the thickness.)
问题：一个圆柱形罐头需容纳500 cm³汤。顶面和底面材料每平方厘米的成本是侧面材料的两倍。求半径 r 和高 h 使成本最小。（忽略材料厚度。）
Solution: h + 4πr². Substitute for h: C(r) = 2πr[500/(πr²)] + 4πr² = (1000/r) + 4πr². Domain: r > 0.
解：体积：V = πr²h = 500 → h = 500/(πr²)。成本 C 正比于：(侧面积)×1 + (顶底面积)×2 = (2πrh)×1 + (2πr²)×2 = 2πrh + 4πr²。代入 h：C(r) = 2πr[500/(πr²)] + 4πr² = (1000/r) + 4πr²。定义域：r > 0。
/r² + 8πr. Set C′(r)=0: 8πr = 1000/r² → 8πr³ = 1000 → r³ = 1000/(8π) = 125/π → r = (125/π)1/3. Numeric approximation: r ≈ 3.41 cm.
求导：C′(r) = −1000/r² + 8πr。令 C′(r)=0：8πr = 1000/r² → 8πr³ = 1000 → r³ = 1000/(8π) = 125/π → r = (125/π)1/3。数值约为 r ≈ 3.41 cm。
gives a local (and absolute) minimum. Then h = 500/(πr²). Using the exact r, h = 2×(125/π)1/3 = 2r. So the cost‑minimising can has h = 2r.
二阶导数：C″(r) = 2000/r³ + 8π > 0 对所有 r>0 成立，故 C 凹向上，驻点为局部（也是绝对）最小值。然后 h = 500/(πr²)。带入精确的 r，得 h = 2×(125/π)1/3 = 2r。所以成本最小的罐头满足 h = 2r。
Optimal radius r ≈ 3.41 cm, height h ≈ 6.82 cm.
最优半径 r ≈ 3.41 cm，高 h ≈ 6.82 cm。
8. Common Pitfalls and How to Avoid Them | 常见陷阱及避免方法
Forgetting the domain: A derivative test only works inside the domain. Always list the physically acceptable interval and check endpoints if appropriate. In many IB problems, the domain is an open interval (e.g. x>0) and the single stationary point is automatically the optimum, but you must state this justification.
忘记定义域：导数检验仅在定义域内有效。务必列出物理上允许的区间，并视情况检查端点。在许多IB问题中，定义域是开区间（如 x>0），唯一的驻点自然就是最优点，但你必须陈述这一理由。
Not eliminating a variable correctly: If the constraint equation contains a square or root, isolate one variable before substitution. Algebraic slips often occur when squaring or simplifying fractions, so double‑check your expression before differentiating.
变量消去错误：若约束方程含有平方或根号，先分离出一个变量再代入。涉及平方或分式化简时极易发生代数错误，求导前务必检查表达式。
Confusing maximum and minimum: Always confirm with a sign line or second derivative. A quick sign diagram of the first derivative is often the safest method when the context is not obvious.
混淆最大值和最小值：务必用符号线或二阶导数确认。当问题背景不明显时，画出一阶导数的符号图通常是最稳妥的方法。
Forgetting units or a concluding statement: IB mark schemes award the final answer mark for presenting the optimal dimensions with correct units and a clear conclusion. Even a perfect derivation can lose a mark if you omit the concluding sentence.
忘记单位或结论语句：IB评分标准对给出带有正确单位的最优尺寸和清晰结论授予最后答案分。即使推导完美，若遗漏结论句也可能失分。
9. Examination Tips for Optimization Questions | 优化题的考试技巧
l number is wrong, marks are given for a correct derivative and stationary point equation.
优化题常出现在试卷一（无计算器）或试卷二（有计算器）中。在试卷一，数字通常整洁且可用根式表示；在试卷二，可能需要解三次方程或使用方程求解器。一定要展示导数和方程 f′(x)=0。即使最终数值错误，正确的导数和驻点方程也能得分。
When using a GDC, sketch the graph of the quantity you want to optimise to confirm that the stationary point matches the graph’s turning point. This visual check can catch sign errors early.
使用图形计算器时，画出优化量的函数图像，确认驻点与图像的转折点吻合。这种视觉检查能及早发现符号错误。
Finally, always relate your answer back to the original problem. If the question says “find the minimum cost”, give the cost, not just the dimensions. If it asks for “dimensions”, list them clearly with units.
最后，务必将答案与原始问题联系起来。如果问题要求“求最小成本”，就要给出成本，而不只是尺寸。如果要求“尺寸”，就要以带单位的形式清晰列出。
10. Practice Problem | 练习题
Problem: A piece of wire of length 80 cm is cut into two pieces. One piece is bent into a square, and the other into a circle. How should the wire be cut so that the total area enclosed is (a) a maximum, (b) a minimum?
问题：一根长80厘米的金属丝被剪成两段。一段弯成一个正方形，另一段弯成一个圆。应如何剪断，使得封闭的总面积（a）最大，（b）最小？
Hint: ck endpoints, and consider the second derivative. Pay attention to the closed interval; the absolute maximum and minimum may occur at a stationary point or at an endpoint.
提示：设用于正方形的长度为 x 厘米。用 x 表示正方形边长和圆的半径，然后写出总面积 A(x)。定义域为 0 ≤ x ≤ 80。计算 A′(x)=0，检查端点并考虑二阶导数。注意这是个闭区间，绝对最大值和最小值可能出现在驻点或端点。
Try this exercise yourself and then check the solution:
试着自己完成这个练习，然后对照答案：
ea; maximum area occurs at one endpoint (all wire used for circle, x=0, area ≈ 509.3 cm²).
正方形周长 x = 80π/(4+π) ≈ 44.6 cm 时总面积最小；最大面积出现在一个端点（全部金属丝围圆，x=0，面积 ≈ 509.3 cm²）。
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