📚 Pre-U AQA Mathematics: Case Study Practical Exercise | Pre-U AQA 数学：案例分析实战演练
Case studies are a vital component of the AQA Pre-U Mathematics assessment, requiring you to apply pure and applied mathematical techniques to a realistic scenario. This practical exercise will guide you through a complete optimisation problem, modelling profit for a new product launch, and will demonstrate how to structure your reasoning, perform calculations, and present justified conclusions—key skills for success in the examination.
案例研究是 AQA Pre-U 数学考核的重要组成部分，要求你将纯数学与应用数学技巧应用于实际情境。本次实战演练将带你完整经历一个新产品上市的利润最优化建模过程，展示如何组织推理、进行计算并呈现有据可依的结论——这些都是考试成功的关键技能。
1. Understanding the Case Study Scenario | 理解案例场景
TechNova Ltd is about to launch a new smartwatch. The marketing team has estimated that at a price of £0, the demand would be 5000 units, and for every £1 increase in price, the demand decreases by 20 units. The company faces fixed costs of £20 000 and a variable cost of £30 per unit produced. As the lead analyst, your task is to determine the selling price that maximises total profit, calculate the maximum profit, and advise the management on the optimal production quantity.
TechNova 有限公司即将推出一款新型智能手表。市场团队估计，当售价为 £0 时需求量为 5000 件，价格每提高 £1，需求减少 20 件。公司面临 £20 000 的固定成本和每件 £30 的可变成本。作为首席分析师，你的任务是确定能使总利润最大的售价，计算最大利润，并就最佳生产量向管理层提出建议。
Key assumptions include a linear demand curve, no competitor reaction, and cost structures that remain constant over the relevant range of production. These simplifications allow us to build a manageable model; in a real-world report, you would discuss the impact of relaxing them.
关键假设包括线性需求曲线、无竞争对手反应以及成本在相关产量范围内保持恒定。这些简化让我们可以构建一个易于处理的模型；在现实报告中，你需要讨论放宽这些假设可能带来的影响。
2. Defining the Variables and Parameters | 定义变量与参数
Let x (in £) be the selling price per unit. The quantity demanded, q, is given by the linear relationship q = 5000 – 20x, where x ≥ 0 and q ≥ 0. The total revenue will depend on both x and q, while costs will be expressed in terms of q, and subsequently in terms of x.
令 x（单位：英镑）为每件的售价。需求量 q 由线性关系 q = 5000 – 20x 给出，其中 x ≥ 0 且 q ≥ 0。总收入将取决于 x 和 q，而成本则用 q 表示，再进一步转换为 x 的函数。
q = 5000 – 20x
q = 5000 – 20x
3. Building the Revenue Function R(x) | 建立收入函数 R(x)
Total revenue is the product of the selling price and the quantity sold. Substituting the demand function yields a quadratic revenue function in terms of price alone.
总收入是售价与销售量的乘积。代入需求函数后，得到仅关于价格的二次收入函数。
R(x) = x · q = x(5000 – 20x) = 5000x – 20x²
R(x) = x · q = x(5000 – 20x) = 5000x – 20x²
Notice that the revenue function is concave (the coefficient of x² is negative), meaning there will be a maximum point within the feasible domain. Marginal revenue is the derivative R'(x) = 5000 – 40x, which becomes zero when x = 125; however, revenue maximisation is not our goal—profit must also account for costs.
请注意，收入函数是凹的（x² 的系数为负），这意味着在可行域内会有一个极大值点。边际收入为导数 R'(x) = 5000 – 40x，当 x = 125 时边际收入为零；然而，收入最大化并非我们的目标——利润还需考虑成本。
4. Determining the Cost Function C(x) | 确定成本函数 C(x)
The total cost consists of a fixed cost of £20 000 and a variable cost of £30 per unit. Since production quantity equals demand q, we can write C(q) = 20 000 + 30q. By expressing q in terms of x using the demand function, we obtain the cost as a function of price.
总成本由 £20 000 的固定成本和每件 £30 的可变成本组成。由于生产量等于需求量 q，我们可以写成 C(q) = 20 000 + 30q。利用需求函数将 q 用 x 表示，便得到成本关于价格的函数。
C(x) = 20 000 + 30(5000 – 20x) = 20 000 + 150 000 – 600x = 170 000 – 600x
C(x) = 20 000 + 30(5000 – 20x) = 20 000 + 150 000 – 600x = 170 000 – 600x
This linear cost function in x has a negative slope—as the price increases and fewer units are demanded, the variable cost component falls, which partially offsets the decline in revenue.
这个关于 x 的线性成本函数具有负斜率——随着价格上升，需求减少，可变成本部分下降，这在一定程度上抵消了收入的减少。
5. Formulating the Profit Function P(x) | 构建利润函数 P(x)
Profit is defined as total revenue minus total cost. By combining the expressions from the previous sections, we arrive at a clean quadratic function that encapsulates the entire scenario.
利润定义为总收入减去总成本。将前面部分的表达式组合起来，我们得到一个简洁的二次函数，它概括了整个情景。
P(x) = R(x) – C(x) = (5000x – 20x²) – (170 000 – 600x) = –20x² + 5600x – 170 000
P(x) = R(x) – C(x) = (5000x – 20x²) – (170 000 – 600x) = –20x² + 5600x – 170 000
Because the coefficient of x² is negative (–20), the parabola opens downward, confirming that a unique maximum profit exists. The domain of interest is where both price and demand are non-negative, i.e. 0 ≤ x ≤ 250.
由于 x² 的系数为负（–20），抛物线开口向下，确认存在唯一的最大利润点。我们关心的定义域是价格和需求量均非负的区域，即 0 ≤ x ≤ 250。
6. Applying Differentiation to Find the Optimal Price | 应用微分求最优价格
To locate the price that maximises profit, we differentiate P(x) and set the first derivative equal to zero. This gives the stationary point.
为了找到使利润最大化的价格，我们对 P(x) 求导，并令一阶导数等于零，从而得到驻点。
P'(x) = –40x + 5600
P'(x) = –40x + 5600
Setting P'(x) = 0 yields –40x + 5600 = 0, so x = 140. At this price, the demand is q = 5000 – 20(140) = 2200 units. We must now verify that this stationary point indeed corresponds to a maximum.
令 P'(x) = 0，得到 –40x + 5600 = 0，因此 x = 140。在此价格下，需求量为 q = 5000 – 20(140) = 2200 件。现在我们必须验证这个驻点确实对应最大值。
7. Second Derivative Test for Maximum | 二阶导数检验最大值
The second derivative of the profit function tells us about the concavity at the stationary point. A negative second derivative confirms a local maximum.
利润函数的二阶导数告诉我们驻点处的凹凸性。负的二阶导数确认是局部极大值。
P”(x) = –40
P”(x) = –40
Since P”(x) < 0 for all x, the profit function is concave everywhere, and the stationary point at x = 140 is a global maximum on the domain. No further boundary checks are needed, but it is always good practice to evaluate P(x) at x = 0 and x = 250 to confirm: P(0) = –170 000, P(250) = –20(62 500) + 5600×250 – 170 000 = –1 250 000 + 1 400 000 – 170 000 = –20 000, both lower than the profit at x = 140.
由于对所有 x 都有 P”(x) < 0，利润函数处处凹，x = 140 处的驻点即为定义域内的全局最大值。无需额外的边界检验，但通常建议计算一下 P(0) 和 P(250) 加以确认：P(0) = –170 000，P(250) = –20(62 500) + 5600×250 – 170 000 = –1 250 000 + 1 400 000 – 170 000 = –20 000，均低于 x = 140 时的利润。
8. Calculating the Maximum Profit and Demand | 计算最大利润与需求量
Substituting the optimal price x = 140 back into the profit function gives the highest achievable profit under the model. The associated demand informs the production volume that should be planned.
将最优价格 x = 140 代回利润函数，得出在此模型下可实现的最高利润。相应的需求量则为应规划的生产量提供依据。
P(140) = –20(140)² + 5600(140) – 170 000 = –392 000 + 784 000 – 170 000 = 222 000
P(140) = –20(140)² + 5600(140) – 170 000 = –392 000 + 784 000 – 170 000 = 222 000
Thus the maximum profit is £222 000, achieved by selling 2200 units at £140 each. The average profit per unit is approximately £100.91. The company would break even at two prices found by solving P(x) = 0, but the only relevant one in the feasible range is near £51.6; above that, profit is positive until the demand vanishes.
因此，最大利润为 £222 000，通过以 £140 的单价销售 2200 件产品实现。平均每件利润约为 £100.91。公司盈亏平衡可通过求解 P(x) = 0 得到两个价格，但在可行范围内相关的盈亏平衡点约为 £51.6；高于此价格，利润为正，直到需求消失。
9. Sensitivity Analysis: Exploring Changes in Fixed Costs | 敏感性分析：探讨固定成本变化
Real-world conditions are rarely static. A sensitivity analysis tests how the optimal decision changes if a key parameter, such as the fixed cost, is altered. Suppose a rent increase pushes fixed costs up by 10% to £22 000.
现实情况很少一成不变。敏感性分析用于测试当关键参数（例如固定成本）发生变化时，最优决策会如何改变。假设租金上涨使固定成本提高 10%，达到 £22 000。
The new cost function becomes C_new(x) = 22 000 + 30(5000 – 20x) = 172 000 – 600x, and the new profit is P_new(x) = –20x² + 5600x – 172 000. Differentiating gives the same first-order condition: P’_new(x) = –40x + 5600, so the optimal price remains £140. The maximum profit becomes P_new(140) = 222 000 – 2 000 = 220 000.
新成本函数变为 C_new(x) = 22 000 + 30(5000 – 20x) = 172 000 – 600x，新利润为 P_new(x) = –20x² + 5600x – 172 000。求导后得到相同的一阶条件：P’_new(x) = –40x + 5600，因此最优价格仍为 £140。最大利润变为 P_new(140) = 222 000 – 2 000 = 220 000。
The table below summarises the impact of different fixed cost assumptions while keeping all other parameters unchanged.
下表总结了在保持其他参数不变的前提下，不同固定成本假设对结果的影响。
| Fixed Cost (£) | Optimal Price (£) | Demand (units) | Max Profit (£) |
|---|---|---|---|
| 18 000 | 140 | 2200 | 224 000 |
| 20 000 (base) | 140 | 2200 | 222 000 |
| 22 000 | 140 | 2200 | 220 000 |
This demonstrates that the optimal price does not depend on the fixed cost in this linear cost model; profit simply shifts by the amount of the fixed cost change. A similar analysis could be performed for the variable cost or the demand coefficient.
这表明在此线性成本模型中，最优价格与固定成本无关；利润仅随固定成本的变化而等额增减。类似分析也可针对可变成本或需求系数进行。
10. Presenting and Justifying Your Conclusions | 展示并论证结论
A well-structured case study response must go beyond calculations. You should communicate your findings in plain language, justify the use of a quadratic model, acknowledge its limitations, and suggest next steps. For example, you would recommend setting a price of £140
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