📚 Case Study Practice for Year 10 Edexcel Further Maths | Year 10 Edexcel 进阶数学：案例分析实战演练
In Year 10 Edexcel Further Maths, simply memorising formulas is not enough. You need to apply your knowledge to real-world scenarios and multi‑step problems. This article presents a series of case studies that mirror the style of exam questions. Each case integrates topics such as calculus, algebra, trigonometry, sequences, and vectors, challenging you to think critically and link different areas of mathematics. Work through each example carefully, studying the bilingual explanations to strengthen both your problem‑solving skills and your mathematical vocabulary.
在 Year 10 Edexcel 进阶数学中，仅仅记住公式是不够的。你需要将知识应用到现实情境和多步骤问题中。本文提供了一系列模拟考试题型风格的案例分析。每个案例都融合了微积分、代数、三角学、数列和向量等主题，挑战你的批判性思维并让你将不同数学领域联系起来。仔细研究每个例题，学习双语解释，以增强你的解题能力和数学词汇量。
1. Introduction to Case Study Practice | 案例分析实战概述
A case study in Further Maths usually describes a practical situation, provides some numerical data, and asks you to formulate a mathematical model. You might need to set up an equation, differentiate to find a maximum, solve a trigonometric identity, or interpret the meaning of a gradient. The key skill is translating words into mathematical expressions and then solving them systematically. Always define your variables clearly and check whether your answer makes sense in the original context. Units and rounding are essential parts of the final answer.
进阶数学中的案例分析通常会描述一个实际情境，提供一些数值数据，并要求你建立数学模型。你可能需要建立方程、求导以找到最大值、证明三角恒等式或解释斜率的实际意义。核心技能是将文字转化为数学表达式，然后系统地求解。始终清晰地定义变量，并检查答案在原始情境中是否有意义。单位和四舍五入是最终答案的重要组成部分。
We will now explore ten detailed case studies. Each one is introduced in English first, followed by the Chinese translation, and the working steps follow the same pattern. Try to solve each problem yourself before reading the solution.
现在我们将详细探讨十个案例。每个案例先用英文介绍，随后附中文翻译，解题步骤也遵循相同模式。在阅读解答之前，尽量自己先尝试解题。
2. Case 1: Maximising the Area of a Rectangular Enclosure | 案例一：最大化矩形围场面积
A farmer has 100 metres of fencing and wants to enclose a rectangular area against a straight wall. The wall serves as one side of the rectangle, so fencing is only needed for the other three sides. Find the dimensions that maximise the enclosed area.
一位农民有 100 米长的围栏，想要靠着一面笔直的墙围成一个矩形区域。墙作为矩形的一边，因此只需要为另外三边设置围栏。求使所围面积最大化的尺寸。
0, so x = 100 − 2y. The area A is given by A = x y = (100 − 2y) y = 100y − 2y². To maximise A, differentiate with respect to y: dA/dy = 100 − 4y. Set the derivative equal to zero: 100 − 4y = 0 → y = 25. Then x = 100 − 2(25) = 50. The maximum area is A = 50 × 25 = 1250 m². A quick check using the second derivative gives d²A/dy² = −4 < 0, confirming a maximum.
设平行于墙的边长为 x 米，两条垂直于墙的边长均为 y 米。所用围栏总量为 x + 2y = 100，因此 x = 100 − 2y。面积 A 为 A = x y = (100 − 2y) y = 100y − 2y²。为了最大化 A，对 y 求导：dA/dy = 100 − 4y。令导数为零：100 − 4y = 0 → y = 25。于是 x = 100 − 2(25) = 50。最大面积为 A = 50 × 25 = 1250 平方米。用二阶导数检验得 d²A/dy² = −4 < 0，确认是最大值。
This classic optimisation problem shows how calculus can be applied to practical resource allocation. Always remember to check that your solution lies within the feasible domain (y > 0 and x > 0).
这个经典的优化问题展示了微积分如何应用于实际的资源分配。始终记住检查解是否位于可行域内（y > 0 且 x > 0）。
3. Case 2: Projectile Motion Trajectory | 案例二：抛体运动轨迹
the trajectory in the form y = f(x) and determine the horizontal range. Take g = 9.8 m/s².
一个球从地面以 20 米/秒的初速率与水平方向成 30°角抛出。忽略空气阻力，推导出轨迹方程，形式为 y = f(x)，并确定水平射程。取 g = 9.8 米/秒²。
Horizontal component of velocity: uₓ = 20 cos30° = 20 × (√3 y = 10 t − ½ × 9.8 t² = 10 t − 4.9 t². From x = 17.32 t, we get t = x / 17.32. Substitute into y: y = 10 (x / 17.32) − 4.9 (x / 17.32)² = (10/17.32) x − (4.9/300) x² approximately. The range is found by setting y = 0 (return to ground): 10 t − 4.9 t² = 0 → t(10 − 4.9 t) = 0 → t = 0 or t = 10/4.9 ≈ 2.04 s. Then range = 17.32 × 2.04 ≈ 35.35 m. Alternatively, use the formula R = (u² sin2θ)/g = (20² × sin60°)/9.8 = (400 × 0.8660)/9.8 ≈ 35.35 m.
速度的水平分量：uₓ = 20 cos30° = 20 × (√3/2) ≈ 17.32 米/秒。垂直分量：uᵧ = 20 sin30° = 20 × 0.5 = 10 米/秒。利用 s = ut + ½at²，水平位移：x = 17.32 t。垂直位移：y = 10 t − ½ × 9.8 t² = 10 t − 4.9 t²。由 x = 17.32 t，得 t = x / 17.32。代入 y：y = 10 (x / 17.32) − 4.9 (x / 17.32)² = (10/17.32) x − (4.9/300) x² 近似。射程通过设 y = 0（返回地面）求得：10 t − 4.9 t² = 0 → t(10 − 4.9 t) = 0 → t = 0 或 t = 10/4.9 ≈ 2.04 秒。然后射程 = 17.32 × 2.04 ≈ 35.35 米。另一种方法，使用公式 R = (u² sin2θ)/g = (20² × sin60°)/9.8 = (400 × 0.8660)/9.8 ≈ 35.35 米。
4. Case 3: Compound Interest and Exponential Growth | 案例三：复利与指数增长
An investment of £5000 earns interest at a nominal annual rate of 4.8% compounded monthly. Calculate the value of the investment after 6 years. Determine how long it will take for the investment to double.
一笔 5000 英镑的投资以年名义利率 4.8% 按月复利计息。计算 6 年后的投资价值。并确定该投资翻倍所需的时间。
The monthly interest rate is 4.8% / 12 = 0.4% 0.004, n = 72: A = 5000 × (1.004)⁷². Calculate (1.004)⁷² ≈ 1.3310, so A ≈ 5000 × 1.3310 = £ 0.693147 / 0.003992 = 173.6 months ≈ 14.47 years. So it takes about 14.5 years to double.
月利率为 4.8% / 12 = 0.4% = 0.004。6 年中的复利期数：n = 6 × 12 = 72。使用公式 A = P (1 + i)ⁿ，其中 P = 5000，i = 0.004，n = 72：A = 5000 × (1.004)⁷²。计算 (1.004)⁷² ≈ 1.3310，因此 A ≈ 5000 × 1.3310 = 6655 英镑。为求翻倍时间，设 A = 10,000：5000 × (1.004)ⁿ = 10000 → (1.004)ⁿ = 2。取自然对数：n ln(1.004) = ln 2 → n = ln 2 / ln(1.004) ≈ 0.693147 / 0.003992 = 173.6 个月 ≈ 14.47 年。因此约需 14.5 年翻倍。
Exponential growth models appear frequently in finance, population studies, and radioactive decay. The natural logarithm is a key tool for solving for the exponent.
指数增长模型经常出现在金融、人口研究和放射性衰变中。自然对数是求解指数问题的关键工具。
5. Case 4: Symmetry of Polynomial Roots | 案例四：多项式根的对称性
³ − 15x² + 24x + k = 0, two of the roots are equal. Find the value of k and all the roots.
对于三次方程 2x³ − 15x² + 24x + k = 0，有两个根相等。求 k 的值及所有的根。
Let the three roots be α, α, and β (repeated root α). By Vieta' 24/2 = 12. This gives α² + 2αβ = 12. Substitute β: α² + 2α(15/2 − 2α) = 12 → α² + 15α − 4α² = 12 → −3α² + 15α − 12 = 0 → divide by −3: α² − 5α + 4 = 0 → (α − 1)(α − 4) = 0. So α = 1 or α = 4. If α = 1, β = 7.5 − 2 = 5.5. If α = 4, β = 7.5 − 8 = −0.5. For α = 1, the roots are 1, 1, 5.5. The product of roots = −d/a = −k/2 = 1×1×5.5 = 5.5 → −k/2 = 5.5 → k = −11. For α = 4, product = 4×4×(−0.5) = −8 → −k/2 = −8 → k = 16. So two possible values: k = −11 or k = 16, giving root sets {1, 1, 5.5} and {4, 4, −0.5}.
设三个根为 α、α 和 β（重根 α）。对于三次方程 ax³ + bx² + cx + d = 0，使用韦达定理：根的和 = −b/a = 15/2。于是 2α + β = 15/2 → β = 15/2 − 2α。两两根乘积之和 = c/a = 24/2 = 12。这给出 α² + 2αβ = 12。代入 β：α² + 2α(15/2 − 2α) = 12 → α² + 15α − 4α² = 12 → −3α² + 15α − 12 = 0 → 除以 −3：α² − 5α + 4 = 0 → (α − 1)(α − 4) = 0。因此 α = 1 或 α = 4。若 α = 1，则 β = 7.5 − 2 = 5.5。若 α = 4，则 β = 7.5 − 8 = −0.5。对于 α = 1，根为 1、1、5.5。根的乘积 = −d/a = −k/2 = 1×1×5.5 = 5.5 → −k/2 = 5.5 → k = −11。对于 α = 4，乘积 = 4×4×(−0.5) = −8 → −k/2 = −8 → k = 16。因此两个可能的 k 值：k = −11 或 k = 16，对应的根集合为 {1, 1, 5.5} 和 {4, 4, −0.5}。
6. Case 5: Optimising Production Cost | 案例五：优化生产成本
+ 5000. Find the production level that minimises the average cost per unit. Verify your result using marginal cost.
一家工厂每天生产 x 单位产品。总成本（英镑）建模为 C(x) = 0.001x³ − 0.3x² + 40x + 5000。求使单位平均成本最小的生产水平。并用边际成本验证结果。
Average cost A(x) = C(x)/x = 0.001x² − 0.3x + 40 + 5000/x. Differentiate: A'(x) = 0.002x − 0.3 − 5000/x². Set A' s. Try x = 200: 8,000,000 − 150×40,000 = 8,000,000 − 6,000,000 = 2,000,000 − 2,500,000 = −500,000 (not zero). x = 250: 15,625,000 − 150×62,500 = 15,625,000 − 9,375,000 = 6,250,000 − 2,500,000 = 3,750,000. x = 220: 10,648,000 − 150×48,400 = 10,648,000 − 7,260,000 = 3,388,000 − 2,500,000 = 888,000. x = 210: 9,261,000 − 150×44,100 = 9,261,000 − 6,615,000 = 2,646,000 − 2,500,000 = 146,000. x = 205: 8,615,125 − 150×42,025 = 8,615,125 − 6,303,750 = 2,311,375 − 2,500,000 = −188,625. So root between 205 and 210. Using further trial or a numerical method we can approximate x ≈ 208. The marginal cost is C' .3x − 5000/x = 0, which is equivalent to the same equation. So x ≈ 208 units. Check A(208) and C'(208) to confirm.
平均成本 A(x) = C(x)/x = 0.001x² − 0.3x + 40 + 5000/x。求导：A'(x) = 0.002x − 0.3 − 5000/x²。令 A'(x) = 0：0.002x − 0.3 − 5000/x² = 0。两边乘 x²：0.002x³ − 0.3x² − 5000 = 0 → 乘以 500：x³ − 150x² − 2,500,000 = 0。我们可以测试整数值。尝试 x = 200：8,000,000 − 150×40,000 = 8,000,000 − 6,000,000 = 2,000,000 − 2,500,000 = −500,000（不等于零）。x = 250：15,625,000 − 150×62,500 = 15,625,000 − 9,375,000 = 6,250,000 − 2,500,000 = 3,750,000。x = 220：10,648,000 − 150×48,400 = 10,648,000 − 7,260,000 = 3,388,000 − 2,500,000 = 888,000。x = 210：9,261,000 − 150×44,100 = 9,261,000 − 6,615,000 = 2,646,000 − 2,500,000 = 146,000。x = 205：8,615,125 − 150×42,025 = 8,615,125 − 6,303,750 = 2,311,375 − 2,500,000 = −188,625。因此根在 205 到 210 之间。通过进一步试算或数值方法可近似得 x ≈ 208。边际成本为 C'(x) = 0.003x² − 0.6x + 40。在平均成本最小处，边际成本等于平均成本：0.003x² − 0.6x + 40 = 0.001x² − 0.3x + 40 + 5000/x → 0.002x² − 0.3x − 5000/x = 0，这与原方程等价。因此 x ≈ 208 单位。验证 A(208) 和 C'(208) 以确认。
7. Case 6: Trigonometric Identity Proof and Modelling | 案例六：三角恒等式证明与建模
) cos(πt/6) + 3 cos(2πt/12), where t is hours after midnight. Simplify h(t) using a double‑angle identity and find the times of high and low tides within the first 12 hours.
某港口潮汐高度建模为 h(t) = 5 + 2 sin(πt/6) cos(πt/6) + 3 cos(2πt/12)，其中 t 为午夜后的小时数。用倍角恒等式化简 h(t)，并求出前 12 小时内的高潮和低潮时刻。
πt/12 = πt/6. So h(t) = 5 + sin(πt/3) + 3 cos(πt/6). We want to combine into a single sinusoidal form if possible, or differentiate. To find extrema, differentiate: h'(t) = (π/3) cos(πt/3) − 3 × (π/6) sin(πt/6) = (π/3) cos(πt/3) − (π/2) sin(πt/6). Set h' ss h(t) = 5 + sin(πt/3) + 3 cos(πt/6). Let’s observe that πt/3 = 2 × πt/6. So we can rewrite sin(πt/3) = sin(2·πt/6) = 2 sin(πt/6) cos(πt/6). Then h(t) = 5 + 2 sin(πt/6) cos(πt/6) + 3 cos(πt/6) = 5 + cos(πt/6) [2 sin(πt/6) + 3]. The derivative approach might still be messy. We can instead note that high and low tides occur when the derivative is zero. Using a graphical or numerical approach, we can test key times: t = 0, 3, 6, 9, 12. At t = 0: h = 5 + sin0 + 3 cos0 = 5 + 0 + 3 = 8. t = 3: h = 5 + sin(π) + 3 cos(π/2) = 5 + 0 + 0 = 5. t = 6: h = 5 + sin(2π) + 3 cos(π) = 5 + 0 − 3 = 2. t = 9: h = 5 + sin(3π) + 3 cos(3π/2) = 5 + 0 + 0 = 5. t = 12: h = 5 + sin(4π) + 3 cos(2π) = 5 + 0 + 3 = 8. So high tides at t = 0 and t = 12, low tide at t = 6. Another high might occur in between? Derivative analysis confirms these are extrema. The model shows a high tide of 8 m at midnight and noon, low tide of 2 m at 6 am.
首先，注意 2 sin A cos A = sin 2A。这里 A = πt/6，所以 2 sin(πt/6) cos(πt/6) = sin(2 × πt/6) = sin(πt/3)。同时，第二个余弦里的角度：2πt/12 = πt/6。因此 h(t) = 5 + sin(πt/3) + 3 cos(πt/6)。我们可以将表达式重写为 h(t) = 5 + cos(πt/6)[2 sin(πt/6) + 3]。求极值时求导：h'(t) = (π/3) cos(πt/3) − (π/2) sin(πt/6)。设 h'(t) = 0，解析求解较复杂。我们可测试关键时间点：t = 0：h = 5 + sin0 + 3 cos0 = 8。t = 3：h = 5 + sin(π) + 3 cos(π/2) = 5。t = 6：h = 5 + sin(2π) + 3 cos(π) = 2。t = 9：h = 5 + sin(3π) + 3 cos(3π/2) = 5。t = 12：h = 5 + sin(4π) + 3 cos(2π) = 8。因此高潮在 t = 0 和 t = 12，低潮在 t = 6。模型显示午夜和正午潮高 8 米，清晨 6 点潮低 2 米。
8. Case 7: Vectors in Navigation | 案例七：向量在导航中的应用
A boat sails from a point O. It travels at a speed of 8 km/h in a direction given by the vector 3i + 4j relative to the water. There is a current with velocity vector 2i − j km/h. Find the resultant velocity of the boat relative to the seabed and the distance from O after 2 hours.
一艘船从 O 点出发。它相对于水以 8 千米/时的速度沿向量 3i + 4j 的方向航行。水流速度向量为 2i − j 千米/时。求船相对于海床的合速度以及 2 小时后离 O 点的距离。
The direction vector of the boat's own velocity has magnitude |3i + 4j| = √3² + 4² = 5. So the unit direction vector is (3i + 4j)/5. The boat's velocity vector through the water is its speed times the unit vector: vboat 4.8i + 6.4j km/h. The current velocity is v_current = 2i − j. The resultant velocity v = vboat v_current = (4.8 + 2)i + (6.4 − 1)j = 6.8i + 5.4j km/h. Magnitude of resultant velocity = √6.8² + 5.4² = √46.24 + 29.16 = √754 ≈ 8.683 km/h. After 2 hours, displacement = 2 × v = 13.6i + 10.8j. Distance from O = 2 × |v| ≈ 2 × 8.683 = 17.366 km. Alternatively, calculate distance = |2v| = √13.6² + 10.8² = √184.96 + 116.64 = √301.6 ≈ 17.37 km.
船自身速度的方向向量大小为 |3i + 4j| = √3² + 4² = 5。因此单位方向向量为 (3i + 4j)/5。船在水中的速度向量为其速率乘以单位向量：vboat = 8 × (3i + 4j)/5 = (24/5)i + (32/5)j = 4.8i + 6.4j 千米/时。水流速度为 v_current = 2i − j。合速度 v = vboat + v_current = (4.8 + 2)i + (6.4 − 1)j = 6.8i + 5.4j 千米/时。合速度的大小 = √6.8² + 5.4² = √46.24 + 29.16 = √75.4 ≈ 8.683 千米/时。2 小时后，位移 = 2 × v = 13.6i + 10.8j。距 O 点的距离 = 2 × |v| ≈ 2 × 8.683 = 17.366 千米。或者计算距离 = |2v| = √13.6² + 10.8² = √184.96 + 116.64 = √301.6 ≈ 17.37 千米。
9. Case 8: Matrix Transformations and Image Processing | 案例八：矩阵变换与图像处理
by M followed by a reflection in the line y = x. Find the coordinates of the final image.
一个几何图形通过矩阵 M = [[2, 0], [0, 1]] 进行变换。描述该变换。然后将点 P(3, 4) 先经 M 变换，再关于直线 y = x 反射。求最终像点的坐标。
The matrix M multiplies the x‑coordinate by 2 and leaves the y‑coordinate unchanged. This is a stretch parallel to the x‑axis with scale factor 2. For point P(3, 4), applying M: P' flection in the line y = x is represented by the matrix R = [[0, 1], [1, 0]]. Apply R to P' roduct R M = [[0,1],[1,0]] × [[2,0],[0,1]] = [[0,1],[2,0]], which sends (x, y) to (y, 2x).
矩阵 M 将 x 坐标乘以 2，而 y 坐标保持不变。这是平行于 x 轴、缩放因子为 2 的拉伸变换。对于点 P(3, 4)，应用 M：P' = M × P = [[2,0],[0,1]] × [3, 4] = (2×3 + 0×4, 0×3 + 1×4) = (6, 4)。关于直线 y = x 的反射由矩阵 R = [[0, 1], [1, 0]] 表示。将 R 应用于 P'：R × [6, 4] = (0×6 + 1×4, 1×6 + 0×4) = (4, 6)。因此最终像点为 (4, 6)。这两个变换的组合对应于矩阵乘积 R M = [[0,1],[1,0]] × [[2,0],[0,1]] = [[0,1],[2,0]]，该变换将 (x, y) 映射到 (y, 2x)。
10. Case 9: Sequences and Limits | 案例九：数列与极限
etermine the limit of the sequence as n → ∞, if it converges. Prove your limit conjecture.
一个数列由 u₁ = 3 和 uₙ₊₁ = 0.5uₙ + 4（n ≥ 1）
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