Year 9 Edexcel Further Maths: Case Study Practice | Year 9 Edexcel 进阶数学：案例分析实战演练

📚 Year 9 Edexcel Further Maths: Case Study Practice | Year 9 Edexcel 进阶数学：案例分析实战演练

In Year 9 Edexcel Further Maths, students are expected to move beyond routine exercises and tackle multi-step, real-world problems that combine several areas of the specification. This article presents a complete worked case study of a manufacturing company, guiding you through the construction of cost, revenue and profit functions, the use of quadratic modelling, inequality solving and graphical analysis. By following each stage, you will see how algebraic and graphical techniques are woven together in an exam-style investigation.

在 Year 9 Edexcel 进阶数学中，学生需要跳出常规练习，解决融合了多个考点的多步骤现实问题。本文通过一个制造企业的完整案例分析，带领你一步步建立成本、收入和利润函数，运用二次模型、不等式求解和图形分析。跟随每个阶段，你将看到代数与图形技巧如何在考试风格的探究中交织在一起。

1. Setting Up the Cost Function | 建立成本函数

A business producing electronic accessories has a fixed daily cost of £500 (rent, salaries) and a variable cost of £20 per unit for materials and energy. The total cost C (in £) of producing x units per day is the sum of the fixed and variable parts: C(x) = 20x + 500. This linear function tells us that even if no items are made, the company still pays £500, and every extra unit adds £20 to the total cost.

一家生产电子配件的企业每天有固定成本 500 英镑（租金、工资）和每单位 20 英镑的变动成本（材料与能源）。每天生产 x 件产品的总成本 C（单位：英镑）是固定与变动部分之和：C(x) = 20x + 500。这个线性函数说明，即使一件都不做，公司仍需支付 500 英镑，而每多生产一件，总成本增加 20 英镑。

In Further Maths, you must clearly define your variables and state any assumptions, such as 'the variable cost per unit stays constant' for the range of production we consider.

在进阶数学中，你必须明确地定义变量并陈述假设，例如在我们考虑的产量范围内“每单位变动成本保持不变”。

2. Revenue and Profit Functions | 收入与利润函数

The selling price of each accessory is set at £40. Therefore, the revenue R (in £) from selling x units is simply R(x) = 40x. Profit P is the difference between revenue and cost: P(x) = R(x) - C(x) = 40x - (20x + 500) = 20x - 500. Profit is also a linear function, with a gradient of £20 per unit.

每件配件的售价定为 40 英镑。因此，销售 x 件的收入 R（英镑）就是 R(x) = 40x。利润 P 是收入与成本之差：P(x) = R(x) - C(x) = 40x - (20x + 500) = 20x - 500。利润也是一个线性函数，斜率为每件 20 英镑。

This initial model assumes that the company can sell as many units as it produces at £40 each. Later we will refine this assumption to make the case more realistic.

这个初始模型假设公司可以按每件 40 英镑的价格卖出所有生产的产品。稍后我们会修正这个假设，使案例更贴近现实。

3. Finding the Break-Even Point | 寻找盈亏平衡点

l exactly 25 units to cover all costs. For x < 25, a loss is made; for x > 25, a profit is earned. This linear model suggests that profit increases without limit as x grows, provided capacity allows.

盈亏平衡点是利润为零的位置。解 P(x) = 0： 20x - 500 = 0 → x = 25。公司需要恰好售出 25 件产品才能覆盖所有成本。当 x < 25 时亏损，当 x > 25 时盈利。这个线性模型表明，只要产能允许，利润随着 x 增长而无限制增加。

Always show your solving steps clearly in an exam, and include units in the final answer: 'The break-even point is 25 units.'

在考试中务必清晰地展示求解步骤，并在最终答案中给出单位：“盈亏平衡点为 25 件。”

4. Introducing Price–Demand Relationship | 引入价格–需求关系

In reality, demand changes with price. Market research shows that for every £1 increase in the selling price, the daily demand falls by 5 units. At the original price of £40, the demand is 100 units. If we let p be the selling price in £, the demand quantity q can be expressed as q = 100 - 5(p - 40), which simplifies to q = 300 - 5p. This linear demand curve reflects the law of demand: higher price leads to lower quantity demanded.

现实中，需求会随价格变化。市场调研显示，售价每提高 1 英镑，日需求量就减少 5 件。在原始售价 40 英镑时，需求为 100 件。若用 p 表示售价（英镑），需求量 q 可表示为 q = 100 - 5(p - 40)，化简得 q = 300 - 5p。这条线性需求曲线反映了需求定律：价格越高，需求量越低。

5. Maximising Profit with Quadratic Functions | 利用二次函数最大化利润

Now revenue depends on both price and quantity: R(p) = p × q = p(300 - 5p) = 300p - 5p². The total cost now uses the demand quantity q: C(p) = 20q + 500 = 20(300 - 5p) + 500 = 6500 - 100p. Profit as a function of price is P(p) = R(p) - C(p) = (300p - 5p²) - (6500 - 100p) = -5p² + 400p - 6500. This is a quadratic function with a negative coefficient of p², so it has a maximum value.

现在收入同时依赖于价格和销量：R(p) = p × q = p(300 - 5p) = 300p - 5p²。总成本使用需求量 q：C(p) = 20q + 500 = 20(300 - 5p) + 500 = 6500 - 100p。利润作为价格的函数为 P(p) = R(p) - C(p) = (300p - 5p²) - (6500 - 100p) = -5p² + 400p - 6500。这是一个二次函数，且 p² 的系数为负，因此存在最大值。

To find the optimal price, complete the square: P(p) = -5(p² - 80p) - 6500 = -5[(p - 40)² - 1600] - 6500 = -5(p - 40)² + 8000 - 6500 = -5(p - 40)² + 1500. The vertex is at p = 40, and the maximum profit is £ y, use the vertex formula p = -b/(2a) with a = -5, b = 400, giving p = -400/(2 × -5) = 40.

为了找到最优价格，配方： P(p) = -5(p² - 80p) - 6500 = -5[(p - 40)² - 1600] - 6500 = -5(p - 40)² + 8000 - 6500 = -5(p - 40)² + 1500。 顶点位于 p = 40，最大利润为 1500 英镑。也可以使用顶点公式 p = -b/(2a)，其中 a = -5，b = 400，得 p = -400/(2 × -5) = 40。

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Interestingly, the price that maximises profit is the original £ he demand model. The corresponding demand is q = 300 - 5×40 = 100 units, and the profit is £1500.

有趣的是，最大化利润的价格正是最初的 40 英镑；不过，这一分析证实了在该需求模型下，这个价格确实带来了最高利润。对应的需求量为 q = 300 - 5×40 = 100 件，利润为 1500 英镑。

6. Solving Profit Inequalities | 求解利润不等式

A manager may wish to know the price range that guarantees a profit of at least £1000. Set up the inequality: -5p² + 400p - 6500 ≥ 1000 p² + 400p - 7500 ≥ 0. Divide both sides by -5, remembering to reverse the inequality sign: p² - 80p + 1500 ≤ 0. Factorise the quadratic: p² - 80p + 1500 = (p - 30)(p - 50). So the inequality becomes (p - 30)(p - 50) ≤ 0.

管理者可能想知道能够保证利润至少 1000 英镑的价格区间。建立不等式：-5p² + 400p - 6500 ≥ 1000。将 1000 移到左边：-5p² + 400p - 7500 ≥ 0。两边同除以 -5，注意不等式方向要改变：p² - 80p + 1500 ≤ 0。对二次式进行因式分解：p² - 80p + 1500 = (p - 30)(p - 50)。于是不等式变为 (p - 30)(p - 50) ≤ 0。

Using a sign diagram or critical values, the product is negative or zero between the roots, so 30 ≤ p ≤ 50. Therefore, any selling price from £30 to £50 inclusive yields a profit of at least £1000, with the maximum £1500 at p = 40.

利用符号表或关键值，乘积在两根之间为负或零，因此 30 ≤ p ≤ 50。所以，含端点在内，从 30 英镑到 50 英镑的任何售价都能带来至少 1000 英镑的利润，在 p = 40 时达到最大值 1500 英镑。

7. Graphical Analysis of the Profit Function | 利润函数的图形分析

t is zero when -5p² + 400p - 6500 = 0, i.e. p² - 80p + 1300 = 0. Solving gives p = 40 ± 10√3 xis at approximately p ≈ 22.68 and p ≈ 57.32. Below these prices the company makes a loss.

绘制带标注的 P(p) = -5p² + 400p - 6500 图像有助于直观理解。抛物线开口向下，顶点位于 (40, 1500)。当 -5p² + 400p - 6500 = 0 时利润为零，即 p² - 80p + 1300 = 0。解得 p = 40 ± 10√3 ≈ 40 ± 17.32，因此图像在 p ≈ 22.68 和 p ≈ 57.32 处与 p 轴相交。低于这些价格时公司亏损。

p (£)	P(p) (£)
25	-875
30	1000
40	1500
50	1000
55	-125

In the exam, you might be asked to sketch such a graph and label intercepts, vertex and the region where profit exceeds £1000.

表格验证了计算出的值；关于 p = 40 的对称性清晰可见。在考试中，你可能需要画出这样的草图，并标出截距、顶点以及利润超过 1000 英镑的区域。

8. Sensitivity Analysis: Fixed Cost Increase | 敏感性分析：固定成本增加

Suppose the fixed cost rises by £200 to £700 due to higher rent. In the linear model with fixed price £ x - 700. The new break-even point is x = 35 units. The gradient of the profit line remains £0 per unit, but the whole line shifts down, requiring more sales to start making a profit.

假设由于租金上涨，固定成本增加 200 英镑至 700 英镑。在售价固定为 40 英镑的线性模型中，成本函数变为 C(x) = 20x + 700，利润 P(x) = 20x - 700。新的盈亏平衡点为 x = 35 件。利润线的斜率仍为每件 20 英镑，但整条线向下平移，需要销售更多产品才能开始盈利。

0(300 - 5p) + 700 = 6700 - 100p. Profit is P(p) = -5p² + 400p - 6700. Complete the square: P(p) = -5(p - 40)² + 8000 - 6700 = -5(p - 40)² + 1300. The maximum profit drops to £ ing at p = 40. The break-even prices come from -5p² + 400p - 6700 = 0, leading to p = 40 ± √1040/5? More systematically, p² - 80p + 1340 = 0, giving p = 40 ± √1600 - 1340 √260 ≈ 40 ± 16.12, so about p ≈ 23.88 and p ≈ 56.12. The profitable price window narrows.

现在用新的固定成本重新审视二次模型。总成本变为 C(p) = 20(300 - 5p) + 700 = 6700 - 100p。利润为 P(p) = -5p² + 400p - 6700。配方得：P(p) = -5(p - 40)² + 8000 - 6700 = -5(p - 40)² + 1300。最大利润下降至 1300 英镑，依然在 p = 40 处取得。盈亏平衡价格由 -5p² + 400p - 6700 = 0 解得 p² - 80p + 1340 = 0，p = 40 ± √260 ≈ 40 ± 16.12，即约为 23.88 英镑和 56.12 英镑。盈利的价格区间变窄了。

This analysis shows that an increase in fixed costs does not alter the optimum price if the demand function is unchanged, but it

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