📚 Case Study Practical Exercise | 案例分析实战演练
In CCEA A-Level Mathematics, applying theoretical knowledge to real-world scenarios is essential. This case study walkthrough demonstrates how pure and applied mathematics come together to solve a business optimisation problem involving revenue, cost, and statistical testing.
在CCEA A-Level数学中，将理论知识应用于实际情境至关重要。本篇案例分析演练展示了纯数学和应用数学如何结合起来，解决一个涉及收益、成本和统计检验的商业优化问题。
1. Introduction to Case Studies in CCEA Mathematics | CCEA数学中的案例分析简介
CCEA's A-Level Mathematics specification encourages students to model realistic situations using techniques from Pure Mathematics, Mechanics, and Statistics. A case study often begins with a loosely defined problem, which you must translate into precise mathematical language, solve, and then critically evaluate.
CCEA的A-Level数学大纲鼓励学生运用纯数学、力学和统计学的技巧对现实情境进行建模。案例分析通常从一个模糊定义的问题开始，你必须将其转化为精确的数学语言，求解，然后进行批判性评估。
In Year 13, relevant topics include optimisation using calculus (C3/C4) and statistical inference such as t‑tests for a population mean (S2). Integrating these areas is key to a successful case study response.
在Year 13阶段，相关主题包括利用微积分进行优化（C3/C4）以及统计推断，例如总体均值的t检验（S2）。整合这些领域是成功解决案例分析的关键。
2. Defining the Problem: Maximising Profit for a New Product | 定义问题：新产品利润最大化
a new fitness tracker. The marketing department estimates that the unit selling price p (£ demand x (units) are related by p = 120 − 0.2x, while the total weekly production cost C (£ er a promotional campaign has significantly increased average daily sales beyond 50 units.
一家公司计划推出一款新的健身追踪器。市场部门估计单位售价p（英镑）与周需求量x（件）满足关系p = 120 − 0.2x，而总周生产成本C（英镑）由C(x) = 0.05x² + 20x + 5000建模。目标是确定使周利润最大化的产量，并检验一次促销活动是否将日均销量显著提高到50件以上。
This is a rich problem combining revenue and cost modelling with a one‑sample t‑test. It mirrors the multi‑step reasoning expected in CCEA coursework or extended exam questions.
这是一个丰富的问题，将收益和成本建模与单样本t检验相结合。它反映了CCEA课程作业或扩展考试题目中所期望的多步骤推理。
3. Data Collection and Scenario Assumptions | 数据收集与情景假设
600. Cost parameters are derived from historical production data and are treated as fixed. For the statistical test, we assume that daily sales figures follow a normal distribution, and we have a random sample of 8 days after the campaign.
在建模之前，我们必须清楚地陈述假设。需求函数假设为线性，并在0 ≤ x ≤ 600范围内有效。成本参数来自历史生产数据，并视为固定值。对于统计检验，我们假设日销售量服从正态分布，并且我们在促销活动后获得了8天的随机样本。
The sample data (units sold per day): 52, 48, 55, 60, 49, 53, 58, 56. The mean x̄ = 53.875 and the sample standard deviation s = 4.051 (to 3 d.p.). We will use a 5% significance level for the test.
样本数据（每日销量）：52, 48, 55, 60, 49, 53, 58, 56。均值x̄ = 53.875，样本标准差s = 4.051（精确到3位小数）。我们将使用5%的显著性水平进行检验。
4. Building a Mathematical Model: Revenue and Cost Functions | 建立数学模型：收益与成本函数
given as a quadratic. Therefore, weekly profit Π(x) is the difference: Π(x) = R(x) − C(x).
周收益R(x)为价格乘以数量：R(x) = p · x = (120 − 0.2x)x = 120x − 0.2x²。周成本C(x)以二次函数给出。因此，周利润Π(x)为两者之差：Π(x) = R(x) − C(x)。
20x − 5000 = 100x − 0.25x² − 5000. This is a quadratic profit model open to optimisation.
代入给定函数得到： 0.2x²) − (0.05x² + 20x + 5000) = 120x − 0.2x² − 0.05x² − 20x − 5000 = 100x − 0.25x² − 5000。这是一个可以优化的二次利润模型。
5. Applying Differentiation to Find Maximum Profit | 应用微分求最大利润
ard differentiation rules, Π′(x) = 100 − 0.5x. The stationary point occurs when Π′(x) = 0.
为了最大化Π(x)，我们求出一阶导数并令其为零。使用标准微分法则，Π′(x) = 100 − 0.5x。驻点出现在Π′(x) = 0时。
The equation and solution are displayed below.
方程和求解如下所示。
100 − 0.5x = 0 → x = 200
ng maximum weekly profit is Π(200) = 100(200) − 0.25(200)² − 5000 = 20000 − 10000 − 5000 = £5,000.
因此，根据模型，每周生产200件可使利润最大化。相应的最大周利润为Π(200) = 100(200) − 0.25(200)² − 5000 = 20000 − 10000 − 5000 = 5,000英镑。
6. Analysing Sensitivity with Second Derivatives | 使用二阶导数进行敏感性分析
derivative. Π″(x) = −0.5, which is negative for all x. This confirms a maximum at x = 200.
我们通过计算二阶导数来确认驻点的性质。Π″(x) = −0.5，该值对所有x均为负。这证实了在x = 200处取得最大值。
The constancy of the second derivative also tells us that the profit function is not highly sensitive to small changes in output around the optimum — a desirable robustness in business, although the linear demand assumption limits the real-world accuracy.
二阶导数的恒定性也告诉我们，利润函数在最优值附近对产量的微小变动不太敏感——这在商业上是可取的稳健性，尽管线性需求假设限制了现实的准确性。
7. Statistical Testing: Is the Marketing Campaign Effective? | 统计检验：营销活动是否有效？
aims the campaign raised average daily sales above 50 units. To test this, we set up a one‑tailed t‑ on mean μ. Let H₀: μ = 50 and H₁: μ > 50. The sample size n = 8, degrees of freedom ν = 7.
公司声称促销活动将日均销量提高到50件以上。为了检验这一说法，我们对总体均值μ建立单尾t检验。设H₀: μ = 50，H₁: μ > 50。样本量n = 8，自由度ν = 7。
The test statistic is computed using the formula t = (x̄ − μ₀) / (s / √n), where x̄ is the sample mean, μ₀ is the hypothesised mean, s is the sample standard deviation, and n is the sample size.
检验统计量使用公式 t = (x̄ − μ₀) / (s / √n) 计算，其中x̄是样本均值，μ₀是假设均值，s是样本标准差，n是样本量。
t = (53.875 − 50) / (4.051 / √8) = 3.875 / 1.432 = 2.706 (to 3 d.p.)
Using the t‑distribution table at the 5% significance level (one‑tail), the critical value for ν = 7 is tc = 1.895. Since our calculated t = 2.706 > 1.895, we reject H₀.
使用t分布表，在5%显著性水平（单尾）下，自由度ν = 7的临界值为tc = 1.895。由于我们计算的t = 2.706 > 1.895，我们拒绝H₀。
8. Hypothesis Testing Using the t‑test | 使用t检验进行假设检验
There is sufficient evidence at the 5% level to conclude that the mean daily sales after the campaign exceed 50 units. This provides statistical backing for the marketing department’s claim.
在5%的显著性水平下，有足够的证据表明促销活动后的平均日销售额超过50件。这为市场部门的说法提供了统计支持。
), a normal probability plot or prior knowledge of sales distributions is needed. If normality is in doubt, a non‑parametric test would be more appropriate.
我们还应该检查正态性假设。对于小样本（n = 8），需要正态概率图或关于销售分布的预先知识。如果正态性存疑，非参数检验会更加合适。
9. Interpreting Results and Making Recommendations | 解释结果并提出建议
The deterministic optimisation suggests producing 200 units per week for a profit of £5,000. However, this profit figure does not account for uncertainty in demand or cost parameters. The t‑test result suggests the campaign is effective, which might shift the demand curve upward.
确定性优化建议每周生产200件，利润为5,000英镑。然而，这一利润数字并未考虑需求或成本参数的不确定性。t检验结果表明促销活动是有效的，这可能会使需求曲线上移。
A more comprehensive model could incorporate the effect of the campaign on the price‑demand relationship. For example, if the campaign increases the intercept of the demand function to 130, the optimal output changes.
一个更全面的模型可以纳入促销活动对价格-需求关系的影响。例如，如果促销活动将需求函数的截距提高到130，最优产量就会改变。
New Π(x) = 110x − 0.25x² − 5000 → x = 220, Π(220) = £7,100
This linkage between statistical evidence and model revision is a hallmark of high‑level case study thinking expected by CCEA.
统计证据与模型修订之间的这种联系是CCEA所期望的高级案例分析思维的标志。
10. Critical Evaluation of the Model | 模型的批判性评估
Every mathematical model has limitations. The linear demand and quadratic cost functions are simplifications; in practice, demand may be non‑linear and cost structures may exhibit economies of scale. The assumption that all units produced are sold ignores market saturation.
每个数学模型都有局限性。线性需求和二次成本函数是简化的；在实践中，需求可能是非线性的，成本结构可能表现出规模经济效应。假设所有生产的单位都能售出，忽略了市场饱和。
The t‑test assumes independence and randomness of the sample. If the 8 days were consecutive, autocorrelation might be present. Additionally, the significance level of 5% implies a 5% risk of Type I error — claiming an effect when none exists.
t检验假设样本的独立性和随机性。如果这8天是连续的，可能存在自相关。此外，5%的显著性水平意味着5%的第一类错误风险——在效果不存在时却声称存在效果。
Addressing these limitations would involve more advanced calculus (e.g., constrained optimisation) and time‑series analysis, which are beyond C3/S2 but illustrate the iterative nature of mathematical modelling.
解决这些局限性需要更高级的微积分（例如，约束优化）和时间序列分析，这超出了C3/S2的范围，但说明了数学建模的迭代性质。
11. Conclusion: Integrating Pure and Applied Mathematics | 结论：整合纯数学与应用数学
This case study demonstrates how CCEA Year 13 topics — differentiation for optimisation and hypothesis testing with the t‑distribution — can be woven together to solve a realistic business problem. The dual approach of calculus and statistics provides both a prescriptive optimum and inferential evidence to support strategic decisions.
本案例研究展示了如何将CCEA Year 13的主题——用于优化的微分和基于t分布的假设检验——结合在一起，解决一个现实的商业问题。微积分和统计学的双重方法既提供了规定性的最优值，又提供了支持战略决策的推断证据。
When approaching your own case study, always define the problem clearly, state assumptions, carry out calculations with precision, and finish with a reflective evaluation. This structured method will serve you well in both coursework and examinations.
当你处理自己的案例分析时，务必清晰地定义问题，陈述假设，精确地进行计算，并以反思性评估结束。这种结构化的方法将在课程作业和考试中为你带来良好成绩。
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