📚 Pre-U WJEC Mathematics: Case Study Practical Combat | Pre-U WJEC 数学：案例分析实战演练
In the Pre-U WJEC Mathematics qualification, the ability to transfer abstract mathematical concepts to real-world scenarios is essential. This article presents a series of practical case studies spanning pure mathematics, mechanics, probability and statistics, each structured to mirror the style of exam-style modelling questions. By working through these worked examples, you will sharpen your analytical thinking and learn how to approach unfamiliar problems with confidence.
在 Pre-U WJEC 数学课程中，将抽象的数学概念迁移到现实场景的能力至关重要。本文通过一系列涵盖纯数学、力学、概率与统计的实战案例，模拟考试风格的建模题型。通过这些解析示例，你将磨炼分析思维，学会如何自信地应对陌生问题。
1. Optimisation in Manufacturing | 制造业中的优化问题
A factory produces a component with total cost function C(x) = 200 + 8x + 0.02x² (in pounds), where x is the number of units produced per day. The selling price per unit is given by p = 25 − 0.005x. Find the production level that maximises daily profit and determine the maximum profit.
某工厂生产一种零件，总成本函数为 C(x) = 200 + 8x + 0.02x²（英镑），其中 x 为每日产量。每件售价为 p = 25 − 0.005x。求使每日利润最大化的产量及最大利润。
Revenue R(x) = x × p = 25x − 0.005x². Profit P(x) = R(x) − C(x).
收入 R(x) = x × p = 25x − 0.005x²。利润 P(x) = R(x) − C(x)。
P(x) = (25x − 0.005x²) − (200 + 8x + 0.02x²) = 17x − 200 − 0.025x²
Differentiate to find the stationary point: dP/dx = 17 − 0.05x. Set equal to zero → 17 − 0.05x = 0 → x = 340.
求导找驻点：dP/dx = 17 − 0.05x。令其为零 → 17 − 0.05x = 0 → x = 340。
Second derivative d²P/dx² = −0.05 < 0, confirming a maximum. Maximum profit P(340) = 17×340 − 200 − 0.025×(340)² = 5780 − 200 − 2890 = £2690.
二阶导数 d²P/dx² = −0.05 < 0，确认为极大值。最大利润 P(340) = 2690 英镑。
2. Differential Equations in Ecology | 生态学中的微分方程
A population of rabbits in a nature reserve grows logistically. The rate of change of the population P (in hundreds) after t years is modelled by dP/dt = 0.8P(1 − P/50). Initially, there are 400 rabbits. Find the population after 5 years.
某自然保护区内兔子数量（以百只计）的增长符合逻辑斯蒂模型：dP/dt = 0.8P(1 − P/50)。初始有 400 只兔子。求 5 年后的种群数量。
Initial condition: P(0) = 4 (since 400 rabbits = 4 hundreds). Solving the logistic equation yields P(t) = K / [1 + (K/P₀ − 1)e^(−rt)], with K = 50, r = 0.8, P₀ = 4.
初始条件：P(0) = 4。解逻辑斯蒂方程得通解 P(t) = K / [1 + (K/P₀ − 1)e^(−rt)]，其中 K = 50，r = 0.8，P₀ = 4。
P(t) = 50 / [1 + (50/4 − 1)e^(−0.8t)] = 50 / [1 + 11.5e^(−0.8t)]
Substitute t = 5: P(5) = 50 / [1 + 11.5e^(−4)]. e^(−4) ≈ 0.0183, so 11.5×0.0183 ≈ 0.21045, denominator ≈ 1.21045 → P(5) ≈ 41.3 hundreds, i.e. approximately 4130 rabbits.
代入 t = 5：P(5) = 50 / [1 + 11.5e^(−4)]，e^(−4) ≈ 0.0183，分母约 1.21045，得 P(5) ≈ 41.3，即约 4130 只兔子。
3. Vectors and Navigation | 向量与导航问题
An aircraft departs from an airport and flies with an airspeed of 250 km/h on a bearing of 060°. A steady wind is blowing from the north-west at 60 km/h. Find the resultant ground speed and the true bearing of the aircraft.
一架飞机以空速 250 公里/小时，航向 060° 飞离机场。有稳定的西北风，风速 60 公里/小时。求地速及真实航行方位角。
Take north as the j-direction and east as i-direction. Air velocity vector: 250 at 60° from north, so components: v_a = 250 sin 60° i + 250 cos 60° j = 250×(√3/2) i + 250×(1/2) j = 125√3 i + 125 j.
取北为 j 方向，东为 i 方向。空速向量：角度从北起算 60°，故分量 v_a = 250 sin 60° i + 250 cos 60° j = 125√3 i + 125 j。
Wind from north-west (315° true) means wind blowing towards south-east (135°). Wind vector: 60 towards 135°: v_w = 60 cos 45° i − 60 sin 45° j = 60×(1/√2) i − 60×(1/√2) j = 30√2 i − 30√2 j.
西北风（真方位 315°）吹向东南（135°）。风向量 v_w = 60×cos45° i − 60×sin45° j = 30√2 i − 30√2 j。
Resultant ground velocity v_g = v_a + v_w = (125√3 + 30√2) i + (125 − 30√2) j. Approx: 125×1.732 + 30×1.414 ≈ 216.5 + 42.4 = 258.9 i; 125 − 42.4 = 82.6 j.
地速向量 v_g = (125√3 + 30√2) i + (125 − 30√2) j。近似计算：i 分量约 258.9，j 分量约 82.6。
Ground speed = √(258.9² + 82.6²) ≈ √(67029 + 6823) ≈ √73852 ≈ 271.8 km/h. True bearing θ measured from north clockwise: tan α = 258.9/82.6 → α ≈ 72.3°, so bearing ≈ 072°.
地速大小 ≈ 271.8 公里/小时。真实方位角从北顺时针起算：tan α = 258.9/82.6 → α ≈ 72.3°，故方位角约 072°。
4. Probability and Insurance Premiums | 概率与保险定价
An insurance company analyses the number of claims per policyholder per year as a Poisson random variable with mean λ = 0.2. If a claim occurs, the average payout is £5000. The company wants the expected profit per policy to be at least £200. What minimum annual premium must be charged?
保险公司将每位投保人的年索赔次数建模为均值 λ = 0.2 的泊松分布。若发生索赔，平均赔付额为 5000 英镑。公司要求每份保单的期望利润至少为 200 英镑。求应收的最低年保费。
Let X ~ Po(0.2) be number of claims. Expected number of claims E(X) = 0.2. Expected payout per policy = E(X) × 5000 = 0.2 × 5000 = £1000.
设 X ~ Po(0.2) 为索赔次数。期望索赔次数 E(X) = 0.2。每份保单期望赔款 = 1000 英镑。
Let premium be £C. Expected profit = C − 1000. Requirement: C − 1000 ≥ 200 → C ≥ £1200. The minimum annual premium should be £1200.
设保费为 C 英镑，期望利润 = C − 1000。要求 C − 1000 ≥ 200 → C ≥ 1200。最低年保费为 1200 英镑。
What if the payout distribution is not fixed? If 5% of claims exceed £50 000, the company may need reinsurance. This simple model illustrates how probabilistic reasoning directly informs financial decisions.
若赔付分布并非固定值呢？比如 5% 的索赔超过 50000 英镑，公司就需再保险。此简例说明概率推理如何直接影响金融决策。
5. Sequences and Mortgage Repayments | 数列与按揭还款
A borrower takes out a loan of £150 000 at an annual interest rate of 4.5%, compounded monthly. The loan is to be repaid in equal monthly instalments over 25 years. Calculate the monthly payment.
某借款人获得一笔 15 万英镑贷款，年利率 4.5%，按月复利，分 25 年等额月供偿还。求每月还款额。
Monthly interest rate i = 0.045/12 = 0.00375. Number of payments n = 25×12 = 300. The present value of an annuity formula: PV = PMT × [1 − (1+i)^(−n)] / i.
月利率 i = 0.00375。还款期数 n = 300。年金现值公式：PV = PMT × [1 − (1+i)^(−n)] / i。
Rearranging: PMT = PV × i / [1 − (1+i)^(−n)] = 150000 × 0.00375 / [1 − (1.00375)^(−300)].
变换得：PMT = 150000 × 0.00375 / [1 − (1.00375)^(−300)]。
Compute (1.00375)^(−300) = 1 / (1.00375^300). Using a calculator, 1.00375^300 ≈ 3.0702, so (1.00375)^(−300) ≈ 0.3257. Denominator = 1 − 0.3257 = 0.6743. PMT = 562.5 / 0.6743 ≈ £834.12.
计算 (1.00375)^300 ≈ 3.0702，其倒数约 0.3257。分母 = 0.6743。PMT = 562.5 / 0.6743 ≈ 834.12 英镑。
Thus the monthly repayment is £834.12. Geometric series understanding here is crucial; a Pre-U candidate might also be asked to derive the formula from scratch.
因此月还款额为 834.12 英镑。此处对等比数列的理解至关重要；Pre-U 考生可能被要求从零推导该公式。
6. Complex Numbers in AC Circuits | 交流电路中的复数
An alternating current circuit contains a resistor of 50 Ω and an inductor of 0.2 H connected in series. The supply voltage is 230 V at 50 Hz. Using complex impedance, find the magnitude of the current and the phase angle between voltage and current.
一交流电路包含串联的 50 Ω 电阻和 0.2 H 电感。电源电压为 230 V，50 Hz。利用复阻抗，求电流幅值及电压与电流的相位差。
Angular frequency ω = 2πf = 100π rad/s. Inductive reactance X_L = ωL = 100π × 0.2 = 20π Ω ≈ 62.83 Ω. Total complex impedance Z = R + jX_L = 50 + j62.83 Ω.
角频率 ω = 2π×50 = 100π rad/s。感抗 X_L = ωL = 20π ≈ 62.83 Ω。总复阻抗 Z = 50 + j62.83 Ω。
Magnitude |Z| = √(50² + 62.83²) = √(2500 + 3948) = √6448 ≈ 80.30 Ω. Phase angle φ = arctan(X_L/R) = arctan(62.83/50) ≈ arctan(1.2566) ≈ 51.5° (lagging: current lags voltage).
阻抗幅值 |Z| ≈ 80.30 Ω。相位角 φ = arctan(62.83/50) ≈ 51.5°（电流滞后电压）。
Current magnitude I = V / |Z| = 230 / 80.30 ≈ 2.864 A. The complex representation elegantly captures both magnitude and phase in a single quantity.
电流幅值 I ≈ 2.864 A。复数表示法优雅地将幅值与相位合并在一个量中。
7. Mechanics: Motion on an Inclined Plane | 力学：斜面上的运动
A block of mass 8 kg rests on a rough plane inclined at 25° to the horizontal. The coefficient of friction between the block and the plane is 0.3. A force P parallel to the plane pulls the block up the slope. Find the minimum value of P required to just move the block up the plane.
质量为 8 kg 的物块静置于与水平面成 25° 的粗糙斜面上，摩擦系数为 0.3。一平行于斜面的力 P 向上拉动物块。求使物块恰好开始上移的最小拉力。
Resolve forces perpendicular to the plane: normal reaction R = 8g cos 25°. Taking g = 9.8, cos 25° ≈ 0.9063, so R ≈ 8×9.8×0.9063 ≈ 71.05 N.
分解垂直于斜面的力：法向反力 R = 8g cos 25° ≈ 71.05 N。
Maximum static friction F_max = μR = 0.3 × 71.05 ≈ 21.32 N, acting down the plane. Component of weight down the plane = 8g sin 25° ≈ 78.4 × 0.4226 ≈ 33.13 N.
最大静摩擦力 F_max = 0.3×71.05 ≈ 21.32 N，沿斜面向下。重力沿斜面分量 = 8g sin 25° ≈ 33.13 N。
For impending motion up the plane, P must overcome both: P = 33.13 + 21.32 = 54.45 N. Hence the minimum pulling force is about 54.5 N.
为恰好上移，P 需克服两者之和：P = 33.13 + 21.32 = 54.45 N。因此最小拉力约 54.5 N。
8. Hypothesis Testing in Clinical Trials | 临床试验中的假设检验
A pharmaceutical company claims its new drug is 90% effective. In a trial of 200 patients, 170 showed improvement. Test at the 5% significance level whether there is evidence that the drug’s effectiveness is lower than claimed.
某制药公司声称其新药有效率为 90%。在 200 名患者的试验中，170 人显示改善。以 5% 显著性水平检验是否有证据表明实际有效率低于声称值。
Let p be the true proportion of improvement. H₀: p = 0.9; H₁: p < 0.9. Under H₀, the sample proportion P̂ ~ N(0.9, (0.9×0.1)/200) approximately, i.e. N(0.9, 0.00045).
设 p 为真实改善比例。H₀: p = 0.9；H₁: p < 0.9。在 H₀ 下，样本比例 P̂ 近似服从 N(0.9, 0.00045)。
Observed proportion = 170/200 = 0.85. Test statistic z = (0.85 − 0.9)/√0.00045 ≈ −0.05/0.0212 ≈ −2.358.
观测比例 = 0.85。检验统计量 z = (0.85 − 0.9)/√0.00045 ≈ −2.358。
One-tailed critical value at 5% is about −1.645. Since −2.358 < −1.645, we reject H₀. There is sufficient evidence that the drug's effectiveness is below 90%.
单侧 5% 临界值约为 −1.645。由于 −2.358 < −1.645，拒绝 H₀。有充分证据表明该药有效率低于 90%。
Considering a continuity correction or using the exact binomial test would refine the p-value, but the normal approximation illustrates the procedure clearly.
若考虑连续性校正或精确二项检验可更精细，但正态近似清晰地展示了检验流程。
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