Case Study: Maximising Profit Using Calculus | 案例分析：利用微积分实现利润最大化

📚 Case Study: Maximising Profit Using Calculus | 案例分析：利用微积分实现利润最大化

In this case study we explore a classic business optimisation problem: a company wishes to determine the production level that maximises its profit. You will learn how to translate a real-world scenario into mathematical equations, use differentiation to find the maximum point, and interpret your results. This exercise mirrors the type of modelling questions found in the CAIE AS Mathematics (9709) syllabus, where functions, equations and calculus come together.

在本案例研究中，我们探讨一个经典的商业优化问题：某公司希望确定能够实现利润最大化的生产水平。你将学习如何将现实情景转化为数学方程，利用微分法求最大值，并解读你的结果。此练习与 CAIE AS 数学（9709）大纲中的建模类题目类似，是函数、方程与微积分的综合应用。

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1. Introduction to the Case Study | 案例介绍

A local manufacturing firm, ProTech Ltd., produces designer widgets. The market price per widget depends on the quantity produced, following the demand equation p = 200 − 0.5x (in £), where x is the number of widgets. The total cost of production is given by C = 2500 + 50x, consisting of fixed costs of £2500 and a variable cost of £50 per widget. Management needs to find the output level that yields the highest profit.

本地制造企业 ProTech 有限公司生产设计师小配件。每个小配件的市场价格取决于产量，遵循需求方程 p = 200 − 0.5x（英镑），其中 x 为小配件数量。生产总成本由 C = 2500 + 50x 给出，包括固定成本 2500 英镑和每个配件 50 英镑的可变成本。管理层需要找到产生最大利润的产量水平。

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2. Translating the Problem into Mathematics | 将问题转化为数学语言

To optimise profit, we must express profit as a function of x. Identify the key components: Revenue (R) is income from sales, Cost (C) is total expenditure, Profit (P) = R − C. The demand equation relates price p and quantity x, so revenue will be price times quantity.

为了优化利润，我们必须将利润表示为 x 的函数。确认关键组成部分：收益（R）为销售收入，成本（C）为总支出，利润（P）= R − C。需求方程给出了价格 p 与数量 x 的关系，因此收益为价格乘以数量。

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3. Writing the Revenue Function | 写出收益函数

Revenue R(x) = p × x. Substitute the demand equation: R(x) = (200 − 0.5x)x = 200x − 0.5x². This yields a quadratic revenue function opening downward, typical when a firm faces a downward-sloping demand curve.

收益 R(x) = p × x。代入需求方程：R(x) = (200 − 0.5x)x = 200x − 0.5x²。这得到一个开口向下的二次收益函数，在公司面临向下倾斜的需求曲线时很典型。

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4. Deriving the Cost Function | 推导成本函数

The cost function is already provided in linear form: C(x) = 2500 + 50x. Note that 2500 is the fixed cost (e.g. rent, insurance) and 50x is the variable cost (raw materials, labour). This cost structure is typical for short-run analysis.

成本函数已以线性形式给出：C(x) = 2500 + 50x。注意 2500 是固定成本（例如租金、保险），50x 为可变成本（原材料、劳动）。这种成本结构在短期分析中很典型。

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5. Formulating the Profit Function | 建立利润函数

Profit P(x) = R(x) − C(x) = (200x − 0.5x²) − (2500 + 50x) = 150x − 0.5x² − 2500. This quadratic function models the profit of the company. Careful expansion and collection of like terms are essential to avoid sign errors.

利润 P(x) = R(x) − C(x) = (200x − 0.5x²) − (2500 + 50x) = 150x − 0.5x² − 2500。这一二次函数模拟了公司的利润。仔细展开和合并同类项对于避免符号错误至关重要。

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6. Finding the First Derivative | 求一阶导数

To locate the maximum, differentiate P(x) with respect to x:

P'(x) = 150 − x

Using the power rule: derivative of 150x is 150, derivative of −0.5x² is −x, and the constant −2500 gives 0.

要确定最大值，对 P(x) 关于 x 求导：运用幂法则：150x 的导数为 150，−0.5x² 的导数为 −x，常数 −2500 导数为 0。

P'(x) = 150 − x

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7. Solving for Stationary Points | 求解驻点

Set the first derivative equal to zero: 150 − x = 0 ⇒ x = 150. This is the only stationary point, where the slope of the profit function is horizontal.

令一阶导数等于零：150 − x = 0 ⇒ x = 150。这是唯一的驻点，此时利润函数的斜率为水平。

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8. Using the Second Derivative to Confirm Maximum | 利用二阶导数确认最大值

Differentiate P'(x) to get the second derivative: P”(x) = −1. Since P”(150) = −1 < 0, the stationary point is a maximum by the second derivative test. A negative second derivative confirms the profit function is concave down at that point.

对 P'(x) 求导得到二阶导数：P”(x) = −1。由于 P”(150) = −1 < 0，根据二阶导数判别法，该驻点为极大值。负的二阶导数确认了利润函数在该点是下凹的。

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9. Calculating the Maximum Profit | 计算最大利润

Substitute x = 150 into P(x):

P(150) = 150(150) − 0.5(150)² − 2500 = 22500 − 11250 − 2500 = 8750

Thus the maximum profit is £8750. The company should produce 150 widgets to achieve this optimal result.

将 x = 150 代入 P(x)：因此最大利润为 8750 英镑。公司应生产 150 个小配件以实现这一最优结果。

P(150) = 8750

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10. Interpreting the Results and Business Insights | 解读结果与商业洞察

The optimal strategy is to produce 150 widgets, generating a profit of £8750. At this output, the marginal revenue (MR = R'(x) = 200 − x) equals the marginal cost (MC = C'(x) = 50). Indeed, MR(150) = 50, confirming the profit-maximising condition MR = MC, a fundamental concept in economics. Beyond 150 units, the extra cost of producing one more widget exceeds the extra revenue, reducing profit.

最优策略是生产 150 个小配件，获得利润 8750 英镑。在这一产量下，边际收益（MR = R'(x) = 200 − x）等于边际成本（MC = C'(x) = 50）。事实上，MR(150) = 50，验证了利润最大化的条件 MR = MC，这是经济学中的基本概念。超出 150 件时，多生产一个配件的额外成本超过额外收益，导致利润下降。

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11. Break-Even Analysis (Optional Extension) | 盈亏平衡分析（选讲扩展）

The break-even points occur when profit is zero: P(x) = 0 → 150x − 0.5x² − 2500 = 0. Multiply both sides by −2 to simplify: x² − 300x + 5000 = 0. Use the quadratic formula:

x = [300 ± √(300² − 4×5000)] / 2 = [300 ± √(90000 − 20000)] / 2 = [300 ± √70000] / 2 = 150 ± 50√7

Approximating: 150 − 50√7 ≈ 17.7 and 150 + 50√7 ≈ 282.3. So the company breaks even at about 18 widgets and 282 widgets. Producing between these values yields a profit; outside that range leads to a loss. This interval (18, 282) is the profitable region.

盈亏平衡点出现在利润为零时：P(x) = 0 → 150x − 0.5x² − 2500 = 0。两边乘以 −2 化简：x² − 300x + 5000 = 0。使用求根公式：x = [300 ± √(300² − 4×5000)] / 2 = [300 ± √(90000 − 20000)] / 2 = [300 ± √70000] / 2 = 150 ± 50√7。近似计算：150 − 50√7 ≈ 17.7，150 + 50√7 ≈ 282.3。因此公司在生产约 18 个和 282 个小配件时达到盈亏平衡。在此数量之间生产可获利；超出该范围则亏损。区间 (18, 282) 是盈利区域。

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12. Common Pitfalls and Examiner Tips | 常见错误与考官提示

Avoid these frequent mistakes when solving optimisation and break‑even problems:

• Forgetting to subtract the full cost function, leading to sign errors in the profit expression.
• Misapplying the power rule: the derivative of −0.5x² is −x, not −0.5x.
• Failing to confirm the nature of the stationary point – always use the second derivative test or a sign diagram.
• Giving an answer without units or context; clearly state ‘maximum profit is £8750 when 150 units are produced’.
• In break‑even analysis, careful handling of negative coefficients is required when applying the quadratic formula.

在解决优化和盈亏平衡问题时避免以下常见错误：

• 忘记完整减去成本函数，导致利润表达式出现符号错误。
• 错误应用幂法则：−0.5x² 的导数是 −x，而不是 −0.5x。
• 未确认驻点的性质——务必使用二阶导数检验或符号表。
• 答题时遗漏单位或背景说明；清楚表述“当生产 150 件时，最大利润为 8750 英镑”。
• 在盈亏平衡分析中，应用求根公式时需小心处理负系数。

Mastering these techniques will sharpen your mathematical modelling skills and prepare you for exam questions that blend algebra, differentiation and real‑world interpretation.

掌握这些技巧将提高你的数学建模能力，为你应对融合了代数、微分和实际解释的考试题目做好准备。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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