📚 Case Study Drill: Maximizing Box Volume with Calculus | 案例分析实战演练：用微积分求最大体积
In OCR Level 2 Further Maths, applying differentiation to real-world optimisation problems is a key skill. This case study takes you step by step through a classic packaging task: cutting squares from the corners of a rectangular card to form an open box, and finding the cut size that maximises the volume. Work through each stage to strengthen your modelling, algebra and calculus techniques.
在 OCR Level 2 进阶数学中，将微分应用于现实优化问题是一项关键技能。本案例分析将带你逐步完成一个经典的包装任务：从一张矩形卡片的角上切去正方形，制成一个无盖盒子，并找出能使体积最大化的切口尺寸。通过每个阶段的练习，强化你的建模、代数和微积分技巧。
1. Understanding the Case Context | 了解案例背景
A manufacturer has a rectangular sheet of card measuring 100 cm by 80 cm. Identical squares of side length x cm are cut from each corner, and the edges are folded up to create an open-top box. The goal is to determine the value of x that gives the greatest possible volume.
制造商有一张 100 cm × 80 cm 的矩形卡片。从每个角上剪去边长为 x cm 的相同正方形，然后将边缘折起，形成一个无盖盒子。目标是求出能使体积尽可能大的 x 值。
This type of problem appears in packaging design, where material costs are fixed and storage capacity must be optimised. The mathematical model must respect physical constraints: x must be positive, and the cut cannot exceed half the shorter side length, so x < 40 cm.
这类问题出现在包装设计中，其材料成本固定，而存储容量需要优化。数学模型必须尊重物理约束：x 必须为正，且切口不能超过较短边长的一半，因此 x < 40 cm。
2. Building the Volume Function | 构建体积函数
ox will be x cm. Therefore, the volume V (in cm³) is given by V(x) = x(100 − 2x)(80 − 2x).
剪去边长为 x 的正方形后，底部尺寸变为 (100 − 2x) cm 和 (80 − 2x) cm。盒子的高度为 x cm。因此，体积 V（单位 cm³）表示为 V(x) = x(100 − 2x)(80 − 2x)。
Always define your variables clearly. Here, x is the cut length, V is the volume. Writing a function in fully factorised form is helpful for seeing the zeros, but for differentiation it is easier to expand.
始终要明确定义变量。这里 x 是切口长度，V 是体积。将函数写成完全因式分解的形式有助于观察零点，但为进行微分运算，展开会更方便。
3. Expanding and Simplifying | 展开并化简
8000 − 360x + 4x². Then multiply by x: V(x) = x(8000 − 360x + 4x²) = 8000x − 360x² + 4x³.
首先将两个线性括号相乘：(100 − 2x)(80 − 2x) = 8000 − 200x − 160x + 4x² = 8000 − 360x + 4x²。再乘以 x：V(x) = x(8000 − 360x + 4x²) = 8000x − 360x² + 4x³。
= 4x³ − 360x² + 8000x. This form is ready for applying the power rule of differentiation.
标准做法是按 x 的降幂书写多项式：V(x) = 4x³ − 360x² + 8000x。这样便可用幂法则进行微分。
4. Applying Differentiation | 应用微分
vative represents the instantaneous rate of change of volume with respect to the cut size.
对于多项式项 axⁿ，其导数为 naxⁿ⁻¹。应用此法则：dV/dx = 12x² − 720x + 8000。该导数表示体积随切口尺寸变化的瞬时变化率。
ionary points, which could be maximum, minimum or points of inflection. At a maximum volume, the derivative changes from positive to negative.
令 dV/dx = 0 可求出驻点，这些点可能是极大值、极小值或拐点。在体积最大处，导数由正变负。
5. Solving the Quadratic Equation | 求解二次方程
mon factor 4 to simplify: 3x² − 180x + 2000 = 0. Now apply the quadratic formula x = [−b ± √b² − 4ac] / (2a) with a = 3, b = −180, c = 2000.
令导数为零：12x² − 720x + 8000 = 0。除以公因子 4 进行化简：3x² − 180x + 2000 = 0。现应用二次公式 x = [−b ± √b² − 4ac] / (2a)，其中 a = 3, b = −180, c = 2000。
(−180)² − 4 × 3 × 2000 = 32400 − 24000 = 8400. Hence x = [180 ± √8400] / 6. Since √8400 = √100 × 84 = 10√84 o solutions are x₁ ≈ (180 + 91.6515)/6 ≈ 45.275 cm and x₂ ≈ (180 − 91.6515)/6 ≈ 14.725 cm.
判别式 Δ = (−180)² − 4 × 3 × 2000 = 32400 − 24000 = 8400。因此 x = [180 ± √8400] / 6。因为 √8400 = √100 × 84 = 10√84 ≈ 91.6515，两个解为 x₁ ≈ (180 + 91.6515)/6 ≈ 45.275 cm 和 x₂ ≈ (180 − 91.6515)/6 ≈ 14.725 cm。
6. Evaluating Feasibility | 评估可行性
≈ 45.275 cm is invalid because it makes the width negative. Only x₂ ≈ 14.725 cm lies within the feasible domain 0 < x < 40.
回顾物理约束：较短边为 80 cm，因此切口 x 必须小于 40 cm 才能保证底部宽度为正（80 − 2x > 0）。解 x₁ ≈ 45.275 cm 无效，因为它会使宽度成为负值。只有 x₂ ≈ 14.725 cm 位于可行域 0 < x < 40 内。
Always check your candidate values against the real-world context. A mathematically correct root can be nonsensical in the original problem, so domain considerations are critical in case study analysis.
务必根据实际情境检查你的候选值。数学上正确的根在原问题中可能毫无意义，因此在案例分析中，定义域的考量至关重要。
7. Testing for Maximum | 检验极大值
bstitute x = 14.725: d²V/dx² ≈ 24(14.725) − 720 ≈ 353.4 − 720 = −366.6, which is negative.
为确认 x ≈ 14.725 cm 给出最大体积，使用二阶导数检验。再次对 dV/dx 求导：d²V/dx² = 24x − 720。代入 x = 14.725：d²V/dx² ≈ 24(14.725) − 720 ≈ 353.4 − 720 = −366.6，结果为负值。
Since the second derivative is negative at the stationary point, the function is concave down here, indicating a local maximum. This is the optimal cut size for the given card.
由于驻点处的二阶导数为负，函数在该点上凹向下，表明为局部极大值。这就是给定卡片的最佳切口尺寸。
8. Calculating the Maximum Capacity | 计算最大容量
25) = 14.725 × (70.55) × (50.55) ≈ 14.725 × 3566.3 ≈ 52,504 cm³ (to nearest whole number).
将 x = 14.725 cm 代回体积函数。为精确起见，使用因式形式： − 2×14.725) × (80 − 2×14.725) = 14.725 × (70.55) × (50.55) ≈ 14.725 × 3566.3 ≈ 52,504 cm³（精确到整数）。
Using the expanded polynomial gives the same result. The box can hold about 52.5 litres. Expressing the answer in a meaningful unit helps the manufacturer understand the storage benefit.
使用展开的多项式也能得到相同结果。盒子可容纳约 52.5 升。用有意义的单位表达答案有助于制造商理解存储效益。
9. Modelling Sensitivity and Limits | 建模灵敏度和限制
Small changes in x around the optimal value produce little change in volume because the derivative is zero at the peak. This is useful for manufacturing tolerances: if the cut is slightly off, the volume remains close to maximum. But deviating too far reduces capacity sharply.
在最优值附近，x 的微小变化几乎不引起体积变化，因为导数在峰值处为零。这对制造公差很有用：若切口有轻微偏差，体积仍接近最大值。但偏离过多则会导致容量急剧下降。
However, the model assumes perfect rigidity, no material loss from seams, and a perfectly rectangular net. In practice, folding and joining require extra material and alter the effective volume, so a prototype trial is still needed.
然而，该模型假设完全刚性、接缝处无材料损耗且平面图为完美矩形。在实践中，折叠和接合需要额外材料并会改变有效体积，因此仍需进行原型试验。
10. Extension: Altering the Sheet Dimensions | 扩展：改变卡片尺寸
olutions x = L/6 (max) and x = L/2 (min, discarded). For L = 80 cm, x = 13.33 cm yields the maximum box.
如果制造商使用边长为 L 的正方形卡片，其体积为 V = x(L − 2x)²。微分可得 dV/dx = (L − 2x)(L − 6x)，从而得到最优解 x = L/6（最大值）和 x = L/2（极小值，舍去）。对于 L = 80 cm，x = 13.33 cm 可得到最大盒子。
This generalisation shows that the optimal cut is one-sixth of the side length for a square sheet. You can extend the case study further by considering a lid, a trapezoidal fold, or adding cost constraints per unit area.
这一推广表明，对于正方形卡片，最优切口为边长的六分之一。你可以进一步扩展案例，考虑加上盖子、梯形折叠，或添加单位面积的成本约束。
11. Key Takeaways from the Drill | 演练要点总结
This case study demonstrates the full modelling cycle: translate a practical problem into algebraic form, apply differentiation, solve equations, interpret results within physical limits, and validate with a second derivative test. Such structured reasoning is exactly what OCR Further Maths examiners look for.
本案例分析展示了完整的建模循环：将实际问题转化为代数形式，应用微分，求解方程，在物理限制下解读结果，并通过二阶导数检验确认。OCR 进阶数学考官所看重的正是这种结构化的推理过程。
Mastering case drills like this builds confidence in tackling unfamiliar contexts. Always define variables, state your domain, show clear algebraic steps, and provide a concluding statement that answers the original question.
掌握此类案例演练能增强应对陌生情境的信心。始终定义变量，声明定义域，展示清晰的代数步骤，并给出回答原始问题的总结陈述。
Published by TutorHao | Further Maths Revision Series | aleveler.com
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