📚 Exercise 6F.3 | 练习 6F.3
Exercise 6F.3 in the IB Mathematics course represents a pivotal step in mastering applied calculus, specifically targeting optimization problems that connect abstract derivatives to tangible real-world scenarios. Whether you are following the Analysis and Approaches or Applications and Interpretation syllabus, this exercise set challenges you to model situations, identify constraints, and use first and second derivatives to find maximum or minimum values. The problems range from geometric puzzles about area and volume to economic contexts involving cost, revenue, and profit, all designed to solidify your understanding of stationary points and their nature.
IB 数学课程中的练习 6F.3 是掌握应用微积分的关键一步，专门针对优化问题，将抽象的导数与具体的现实场景联系起来。无论你是在遵循分析与方法还是应用与解释大纲，这一练习集都要求你建立情境模型、识别约束条件，并运用一阶和二阶导数找到最大值或最小值。题目涵盖从关于面积和体积的几何谜题到涉及成本、收入和利润的经济情境，所有设计都是为了巩固你对驻点及其性质的深刻理解。
1. The Big Picture of Exercise 6F.3 | 练习 6F.3 的全局视角
Exercise 6F.3 is not merely about performing differentiation mechanically; it is about interpreting the derivative as a rate of change and leveraging it to make optimal decisions. The section bridges the gap between the technology of finding derivatives and the art of problem-solving, often requiring you to express one variable in terms of another before undertaking any calculus. By completing these exercises, you will learn to translate a written optimization problem into a function of a single variable, a skill that is assessed frequently in Paper 1 and Paper 2.
练习 6F.3 并不仅仅是机械地进行求导运算；它要求你将导数解释为变化率，并利用它做出最优决策。这一部分弥合了求导技术与解题艺术之间的鸿沟，常常需要你在进行任何微积分运算之前用一个变量表示另一个变量。通过完成这些练习，你将学会将文字描述的优化问题转化为单变量函数，这项技能在试卷一和试卷二中经常被考查。
In this article, we break down the core techniques and typical question styles found in Exercise 6F.3. Each section pairs an English explanation with its Chinese counterpart, ensuring that dual-language learners can grasp both the linguistic and mathematical nuances. We will walk through setting up a function, finding critical points, verifying maxima or minima, and wrapping up with a clear final answer in context.
在本文中，我们分解了练习 6F.3 中的核心技巧和典型题型。每一小节都配对了英文讲解和中文讲解，确保双语学习者能同时掌握语言和数学的细节。我们将逐步讲解如何建立函数、寻找临界点、验证极大值或极小值，最后结合情境写出清晰的最终答案。
2. Essential Tool: The First Derivative and Critical Points | 必备工具：一阶导数与临界点
At the heart of Exercise 6F.3 lies the first derivative f'(x). To locate where a function might achieve a maximum or minimum, we solve f'(x) = 0. These solutions are called critical points or stationary points. However, f' ses often first ask you to find these x-values as a stepping stone to the full optimization process.
练习 6F.3 的核心在于一阶导数 f'(x)。为了定位函数可能取得最大值或最小值的位置，我们求解 f'(x) = 0。这些解被称为临界点或驻点。然而，f'(x) = 0 是必要条件而非充分条件；导数为零的点也可能是水平拐点。这些练习通常会先要求你求出这些 x 值，作为完整优化过程的垫脚石。
It is vital to check the domain of your function, which in Exercise 6F.3 is often restricted by practical constraints (e.g., lengths must be positive, volume must be non-negative). The global maximum or minimum might occur at a critical point inside the interval or at an endpoint. A table of signs for f'(x) around each critical point quickly reveals whether the function is increasing or decreasing, paving the way for the first derivative test.
检查函数定义域至关重要，在练习 6F.3 中，定义域常常受到实际条件的限制（例如，长度必须为正，体积必须非负）。全局最大值或最小值可能出现在区间内的临界点或端点处。用一个符号表标记每个临界点附近的 f'(x) 符号，可以快速揭示函数是在递增还是递减，为应用一阶导数检验法铺平道路。
- Solve f'(x) = 0 and also note where f'(x) does not exist.
- 解决 f'(x) = 0 并注意 f'(x) 不存在的点。
- Create a sign diagram: pick test points in each interval determined by critical points.
- 绘制符号示意图：在由临界点划分的每个区间内选取测试点。
- If f'(x) changes from positive to negative at x = c, then f(c) is a local maximum.
- 若 f'(x) 在 x = c 处由正变负，则 f(c) 为局部最大值。
- If f'(x) changes from negative to positive, f(c) is a local minimum.
- 若 f'(x) 由负变正，f(c) 为局部最小值。
3. Confirming with the Second Derivative Test | 用二阶导数检验法确认
Many problems in Exercise 6F.3 can be efficiently verified using the second derivative f''(x). If f'(c) = 0 and f''(c) < 0, the function is concave down at that point, confirming a local maximum. If f'' derivative is easy to compute and eliminates the need for a sign chart. Be cautious: if f''(c) = 0, the test is inconclusive, and you must return to the first derivative test.
练习 6F.3 中的许多问题可以用二阶导数 f''(x) 高效地验证。如果 f'(c) = 0 且 f''(c) < 0，函数在该点处是凹口向下的，从而确认局部最大值。如果 f''(c) > 0，则是局部最小值。当二阶导数容易计算时，这种检验法尤其简洁，能省去符号表。但要注意：如果 f''(c) = 0，检验法失效，你就必须回到一阶导数检验法。
In IB exams, using the second derivative test shows a deeper understanding of calculus concepts. When you complete Exercise 6F.3, aim to apply both methods at least once to build confidence. The second derivative also plays a role in discussions about "rate of change of the rate of change," which is central to interpreting acceleration in kinematics problems that sometimes appear in the same chapter.
在 IB 考试中，使用二阶导数检验法能体现你对微积分概念的更深理解。在完成练习 6F.3 时，力求至少各用一次这两种方法以建立信心。二阶导数在讨论“变化率的变化率”时也扮演重要角色，这是解释同一章中可能出现的运动学问题里加速度概念的核心。
4. The Five-Step Optimization Framework | 五步优化框架
Every question in Exercise 6F.3 can be tackled systematically using a five-step approach. First, read the problem and identify the quantity to be maximized or minimized (the objective function). Second, express this quantity as a function of one variable by using given constraints to eliminate other variables. Third, compute the derivative and find critical points. Fourth, determine the nature of these points using the first or second derivative test. Fifth, interpret your mathematical result in the original context and write a concluding statement with units.
练习 6F.3 中的每一道题都可以用五步法系统解决。第一步，读题并识别需要最大化或最小化的量（目标函数）。第二步，利用给定的约束条件消去其他变量，将该量表示为单变量的函数。第三步，计算导数并寻找临界点。第四步，使用一阶或二阶导数检验法确定这些点的性质。第五步，在原始情境中阐释你的数学结果，并写出带有单位的结论性陈述。
ms. Practicing this structure will make even the most complex-looking problems manageable.
这一框架能防止常见错误，比如忘记检查端点，或忽略了确认临界点确实对应最大值而非最小值。在练习 6F.3 中，约束方程常常是几何方面的（例如，周长 2l + 2w = 100）或基于固定总和的。练习这种结构将使看上去最复杂的问题也变得易于应对。
5. Example 1: Maximizing Area with a Fixed Perimeter | 示例 1：固定周长下最大化面积
he length parallel to the wall be y and the two equal perpendicular sides be x. The constraint is 2x + y = 200, so y = 200 - 2x. The objective function is area A = x · y = x(200 - 2x) = 200x - 2x².
设想练习 6F.3 中的一个经典问题：一位农场主有 200 米围栏，想靠着一面笔直的墙围成一个矩形场地（因此只有三边需要围栏）。用平行于墙的边长表示面积 A，然后找出使面积最大化的尺寸。设平行于墙的边长为 y，两条相等的垂直边长为 x。约束条件是 2x + y = 200，所以 y = 200 - 2x。目标函数是面积 A = x · y = x(200 - 2x) = 200x - 2x²。
Now differentiate: A' = 200 - 4x. Set A' = 0 gives x = 50. The second derivative A'' The maximum area is A = 50 × 100 = 5000 m². Always include units and a clear final answer: "The field should be 50 m by 100 m to achieve a maximum area of 5000 m²."
现在求导：A' = 200 - 4x。令 A' = 0 解得 x = 50。二阶导数 A'' = -4，为负值，确认在 x = 50 处取得最大值。当 x = 50 m 时，y = 200 - 100 = 100 m。最大面积 A = 50 × 100 = 5000 m²。记得始终包含单位并给出清晰的最终答案：“场地的尺寸应为 50 m 乘 100 m，以获得 5000 m² 的最大面积。”
6. Example 2: Minimizing Material Cost | 示例 2：最小化材料成本
l material cost. Let the cost per cm² of the side material be k. Then the cost function C = 2πr h · k + 2πr² · (2k). Use the volume constraint πr²h = 1000 to substitute for h.
练习 6F.3 经常引入经济情境。一个典型问题：要制造一个容积为 1 升（1000 cm³）的圆柱形罐子。顶部和底部材料的单位面积成本是弯曲侧面材料的两倍。求使总材料成本最小化的半径 r 和高 h。设侧面材料每 cm² 的成本为 k。那么成本函数 C = 2πr h · k + 2πr² · (2k)。利用体积约束条件 πr²h = 1000 消去 h。
) = 2πr · k · (1000/(πr²)) + 4kπr² = (2000k)/r + 4kπr². Differentiate with respect to r: C'(r) = -2000k/r² + 8kπr. Set this to zero: 8kπr = 2000k/r² → 8πr³ = 2000 → r = ∛ find h and the cost. This problem demonstrates how to handle multiple constants elegantly.
代入 h = 1000 / (πr²) 得到 C(r) = 2πr · k · (1000/(πr²)) + 4kπr² = (2000k)/r + 4kπr²。对 r 求导：C'(r) = -2000k/r² + 8kπr。令其为零：8kπr = 2000k/r² → 8πr³ = 2000 → r = ∛(250/π)。使用二阶导数确认其为最小值。然后求出 h 和成本。这一问题展示了如何优雅地处理多个常数。
7. Example 3: Volume Optimization with Constraints | 示例 3：带约束的体积优化
e the cuts must be positive and less than half the side, the domain is 0 < x < 15. Differentiate to find the x that maximizes V.
练习 6F.3 中另一种常见题型涉及优化通过从平板上切去小正方形制成的无盖盒子的体积。假设有一张边长为 30 cm 的正方形卡纸，从每个角切去边长为 x 的小正方形，然后折起边壁制成无盖盒子。体积 V = x(30 - 2x)²。由于切去的部分必须为正且小于边长的一半，定义域为 0 < x < 15。求导以找出使 V 最大化的 x。
Expand V = x(900 - 120x + 4x²) = 4x³ - 120x² + 900x. Then V' = 12x² - 240x + 900. Set V' - 5)(x - 15) = 0. The critical points are x = 5 and x = 15 (rejected as boundary). Using V'' = 24x - 240; at x = 5, V'' = -120 < 0, so maximum volume when x = 5 cm. The maximum volume is 5 × 20 × 20 = 2000 cm³.
展开 V = x(900 - 120x + 4x²) = 4x³ - 120x² + 900x。于是 V' = 12x² - 240x + 900。令 V' = 0 并除以 12：x² - 20x + 75 = 0 → (x - 5)(x - 15) = 0。临界点为 x = 5 和 x = 15（作为边界舍去）。使用 V'' = 24x - 240；在 x = 5 处，V'' = -120 < 0，因此在 x = 5 cm 时取得最大体积。最大体积为 5 × 20 × 20 = 2000 cm³。
8. Navigating Common Pitfalls in Exercise 6F.3 | 规避练习 6F.3 中的常见陷阱
One frequent error is differentiating too early before simplifying the objective function. If you apply the product rule to an unsimplified expression like x(30 - 2x)², you risk algebraic errors. Always expand or simplify where possible. Another pitfall is ignoring the physical domain: a critical point may lie outside the feasible interval, so always check the endpoints and reject extraneous solutions. Students also sometimes give only the x-value and forget to compute the maximum area or volume.
一个常见错误是在简化目标函数之前过早求导。如果你对像 x(30 - 2x)² 这样的未简化表达式应用乘积法则，就有可能出现代数错误。因此要尽可能先展开或化简。另一个陷阱是忽视实际定义域：临界点可能位于可行区间之外，所以要始终检查端点并舍去无关解。学生们有时只给出 x 值，而忘记计算最大面积或体积。
Additionally, when using the second derivative test, ensure you state clearly "f''(c) < 0, therefore maximum" IB marking, communication is assessed. Finally, always interpret the solution in context: "the maximum profit is $1200 when 15 units are produced" is a complete answer; "x = 15" alone is not.
此外，在使用二阶导数检验法时，一定要清晰地陈述“f''(c) < 0，因此为最大值”，而不是仅仅写一个负数。在 IB 评分中，表述也是评核的一部分。最后，始终要在情境下解释解：“当生产 15 件产品时，最大利润为 1200 美元”是一个完整答案；仅仅“x = 15”则不是。
9. Connecting to Graphical Interpretation | 联系图形解释
rms your algebraic result visually. In IB exams, you may be asked to sketch a derivative graph or discuss the significance of intercepts. Recognizing that the maximum corresponds to the vertex of a quadratic is a quick mental check for polynomial optimization problems.
将你正在优化的函数图形化可以加深你对练习 6F.3 的概念性把握。例如，绘制 A(x) = 200x - 2x² 的图像会展现出一条开口向下的清晰抛物线，顶点在 x = 50 处。这从视觉上印证了你的代数结果。在 IB 考试中，你可能被要求绘制导数图像或讨论截距的意义。认识到最大值对应二次函数的顶点，是对多项式优化问题快速进行心理检验的方法。
r², a sketch helps identify asymptotic behavior and confirm that the critical point is indeed a minimum. Using graphing technology (GDC) allowed in Paper 2 can help you verify answers, but the analytical work in Exercise 6F.3 remains essential for Paper 1, where no calculator is permitted.
对于非多项式函数，例如成本函数 C(r) = 2000/r + 4πr²，绘制草图有助于识别渐近行为并确认临界点确实是最小值。在试卷二中允许使用图形计算器（GDC）来验证答案，但练习 6F.3 中的分析工作对于不允许使用计算器的试卷一仍然至关重要。
10. Problem-Solving Strategies for Success | 成功的解题策略
When tackling the later parts of Exercise 6F.3, begin by highlighting the quantity to optimize and the given fixed value. Write a clear "Let" statement to define variables. For example, "Let r be the radius and h be the height." Then write the constraint equation and the objective function separately. Perform dimensional checks: if the objective is volume, its derivative should be in terms of area, which can help catch algebraic slips.
在处理练习 6F.3 的后半部分时，首先高亮出要优化的量和给定的固定值。写下一个清晰的 “设” 语句来定义变量。例如，“设 r 为半径，h 为高。” 然后分别写出约束方程和目标函数。进行量纲检查：如果目标是体积，它的导数应当与面积有关，这有助于发现代数上的疏忽。
Practice by attempting the problems without looking at the solution first, then compare your reasoning step-by-step with the markscheme. Exercise 6F.3 is deliberately structured to build up difficulty gradually. The early questions often provide the function to differentiate, while the later ones require you to construct it from scratch. Embrace the challenge, and remember that each completed problem engrains the optimization mindset deeper.
尝试先不看解答自己做题，然后逐步将你的推理与评分方案比较。练习 6F.3 被有意识地设计为逐渐增加难度。早期题目通常会提供需要求导的函数，而后面则要求你从头构建。拥抱挑战，并记住每完成一道题都会将优化思维更深地刻入脑中。
11. Exam-Style Tips for Exercise 6F.3 | 应对练习 6F.3 的考试风格技巧
tion questions similar to Exercise 6F.3 often appear as longer structured problems worth 7– nature of the optimum and stating the conclusion. Ensure you do not skip the verification step, as it often carries a mark. Use the notation f'(x) = 0 and present your working logically.
在 IB 考试中，类似于练习 6F.3 的优化题通常以较长的结构化问题出现，价值 7–12 分。分数被分配在建立函数、正确求导、求解临界点，以及最后验证最值的性质和陈述结论上。确保不要跳过验证步骤，因为它常常占一分。使用符号 f'(x) = 0，并逻辑清晰地展示你的运算过程。
If the problem asks for the maximum profit, do not just find the number of items; compute the actual profit and write it in currency format. When a question states "justify your answer," you must use a derivative test or a sign diagram explicitly. A table summarizing the first derivative test can be a concise way to present your justification.
如果问题要求找出最大利润，不要仅仅求出产品数量；要计算实际利润并以货币格式写出。当问题表述为“证明你的答案”时，你必须明确使用导数检验法或符号示意图。一个总结了一阶导数检验法的表格可以是呈现你证明的一种简洁方式。
12. Consolidating Skills for Further Topics | 巩固技能以应对后续主题
Mastering Exercise 6F.3 equips you with analytical tools that extend far beyond a single exercise. The logic of optimization seamlessly transitions into kinematics, where you minimize travel time or maximize displacement, and into the calculus of exponential growth and decay. The discipline of expressing a problem in mathematical language and systematically solving it is a transferable skill that will support your work in the internal assessment and beyond.
掌握练习 6F.3 将为你配备远超单一练习的分析工具。优化的逻辑无缝衔接到运动学中，在运动学里你最小化旅行时间或最大化位移，还会用到指数增长和衰减的微积分。用数学语言表述问题并系统解决它的自律性是一项可迁移的技能，将支持你在内部评估及以后的学习与工作。
Return to Exercise 6F.3 periodically even after moving on to integration or probability. The refresher will keep your differentiation skills sharp and your problem-solving instinct finely tuned. Re-attempt the most challenging problem from the set every few weeks to measure your growth and to solidify the neural pathways that make optimization second nature.
在继续学习积分或概率之后，也要定期回看练习 6F.3。这种复习将使你的求导技能保持锐利，并使你的解题直觉得到精细调整。每隔几周重新尝试练习集中最具挑战性的那道题，以衡量自己的成长，并巩固使优化成为第二天性的神经通路。
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