Case Study Practical Problem Solving: Real-world Applications of OCR Further Maths | 案例分析实战演练：OCR进阶数学的实际应用

📚 Case Study Practical Problem Solving: Real-world Applications of OCR Further Maths | 案例分析实战演练：OCR进阶数学的实际应用

This article walks through several practical case studies that bring OCR Level 2 Further Mathematics concepts to life. You will see how techniques like differentiation, matrices, sequences, and parametric equations are applied to real-world scenarios. Each case study breaks down the problem, identifies the relevant mathematics, and demonstrates a step-by-step solution. By working through these examples, you will strengthen your problem-solving skills and gain confidence in tackling unfamiliar questions.

本文将通过多个实际案例来展示 OCR Level 2 进阶数学概念的应用。你会看到微分、矩阵、数列和参数方程等技术如何解决现实问题。每个案例都包括问题分解、核心数学方法定位以及逐步推演。通过这些练习，你将提升解决问题的能力，并在面对陌生题型时更有信心。

1. Optimising a Product's Profit | 产品利润最大化

A company finds that the profit £ x - 800. The manager wants to know how many units should be produced to maximise profit and what that maximum profit will be.

一家公司发现，销售 x 百件小工具的利润 £P(x) 可以用 P(x) = -2x² + 120x - 800 表示。经理想知道生产多少件时利润最大，以及最大利润是多少。

Because this is a quadratic with a negative coefficient of x², the graph of P against x is a downward-opening parabola. The maximum occurs at the vertex. We can find the x-coordinate of the vertex by using -b/(2a) or by differentiating.

由于这是一个 x² 系数为负的二次函数，P 关于 x 的图像是开口向下的抛物线。最大值出现在顶点处。我们可以使用公式 -b/(2a) 或通过微分来找到顶点的 x 坐标。

Using differentiation: dP/dx = -4x + 120. Set dP/dx = 0 ⇒ -4x + 120 = 0 ⇒ x = 30. To confirm it is a maximum, note the second derivative d²P/dx² = -4 < 0.

使用微分：dP/dx = -4x + 120。令 dP/dx = 0 得 -4x + 120 = 0，解得 x = 30。要确认这是最大值，注意二阶导数 d²P/dx² = -4 < 0。

nits. Substituting x = 30 into P(x): P(30) = -2(30)² + 120×30 - 800 = -1800 + 3600 - 800 = £1000. So maximum profit is £1000.

数量为 30 百件，即 3000 件。将 x = 30 代入 P(x)：P(30) = -2(30)² + 120×30 - 800 = -1800 + 3600 - 800 = £1000。因此最大利润为 £1000。

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2. Designing a Safe Rollercoaster Loop | 设计安全的过山车回环

They need to find the height of the highest point above the ground (y-coordinate) and the width of the loop at the widest horizontal extent.

工程师使用参数曲线来模拟过山车回环的形状：x = 10 cos θ, y = 4 sin θ + 4，0 ≤ θ < 2π。他们需要找到最高点离地面的高度（y 坐标）以及回环水平方向的最大宽度。

This is an ellipse. The maximum y occurs when sin θ is maximum, which is 1. Then y_max = 4(1) + 4 = 8. The highest point is 8 units above the ground.

这是一个椭圆。当 sin θ 最大（即 1）时，y 达到最大值。y_max = 4(1) + 4 = 8。最高点离地面 8 个单位。

The widest horizontal extent is given by the maximum and minimum x values. Cos θ varies between -1 and 1, so x_min = -10, x_max = 10. The width is x_max - x_min = 20 units.

最大水平范围由 x 的最大值与最小值决定。cos θ 在 -1 到 1 之间变化，因此 x_min = -10，x_max = 10。宽度为 20 个单位。

This simple analysis lets designers ensure the loop fits within a certain vertical and horizontal space.

通过这个简单分析，设计师可以确保回环适合特定的垂直和水平空间。

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3. Cryptography with Matrices | 矩阵加密

Alice wants to send a secret message 'HELLO' rix M = [3 1; 2 1] (top row 3, 1; bottom row 2, 1). Multiply each plaintext vector by M and send the resulting numbers.

Alice 想用矩阵加密将秘密消息 'HELLO' 发送给 Bob。她把每个字母转换为数字（A=1, B=2，...，Z=26），并分成 2×1 的向量。她使用编码矩阵 M = [3 1; 2 1]（第一行 3, 1；第二行 2, 1）。将每个明文向量乘以 M，然后把结果数字发送出去。

number of letters, so add a dummy X=24 to make 6 letters: 8,5,12,12,15,24. Vectors: (8,5)ᵀ, (12,12)ᵀ, (15,24)ᵀ. Multiply M by each: For (8,5)ᵀ: [3 1; 2 1] × [8; 5] = [3×8+1×5; 2×8+1×5] = [29; 21]. For (12,12)ᵀ: [3×12+1×12; 2×12+1×12] = [48; 36]. For (15,24)ᵀ: [3×15+1×24; 2×15+1×24] = [69; 54]. So encrypted sequence: 29,21,48,36,69,54.

明文：H=8, E=5, L=12, L=12, O=15。但需要偶数个字母，故增加一个哑元 X=24 变为 6 个字母：8,5,12,12,15,24。向量为：(8,5)ᵀ, (12,12)ᵀ, (15,24)ᵀ。各乘 M：对于 (8,5)ᵀ：[3 1; 2 1] × [8; 5] = [29; 21]。对于 (12,12)ᵀ：[48; 36]。对于 (15,24)ᵀ：[69; 54]。加密后序列：29,21,48,36,69,54。

M⁻¹ = [1 -1; -2 3]. Multiplying the received vectors by M⁻¹ recovers the original numbers.

解密时，Bob 使用逆矩阵 M⁻¹。M 的行列式为 1，所以 M⁻¹ = [1 -1; -2 3]。将接收到的向量乘以 M⁻¹ 就能还原原始数字。

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4. Modelling Population Growth | 人口增长模型

A town's population is currently 50 000 and is growing at 2% per year. Write a model to predict the population after t years and find when the population will first exceed 60 000.

一个小镇目前人口为 50 000，每年增长 2%。写出一个模型来预测 t 年后的人口数，并求出人口首次超过 60 000 的时间。

th model is P(t) = P₀ × (1 + r)ᵗ = 50000 × (1.02)ᵗ. We need to solve 50000 × 1.02ᵗ > 60000 ⇒ 1.02ᵗ > 1.2.

指数增长模型为 P(t) = 50000 × (1.02)ᵗ。需要解不等式 50000 × 1.02ᵗ > 60000，即 1.02ᵗ > 1.2。

ral logs or common logs: t > log(1.2) / log(1.02). log(1.2) ≈ 0.07918, log(1.02) ≈ 0.00860 ⇒ 9.21 years. Therefore, the population first exceeds 60 000 during the 10th year (t = 10).

取对数：t log(1.02) > log(1.2)。log(1.2) ≈ 0.07918，log(1.02) ≈ 0.00860，得 t > 9.21 年。因此，人口会在第 10 年首次超过 60 000。

50000×1.02⁹ ≈ 50000×1.1951 = 59 755, P(10) ≈ 50000×1.2190 = 60 950, confirming the result.

验证：P(9) ≈ 59 755，P(10) ≈ 60 950，符合结果。

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5. Projectile Motion of a Football | 足球的运动轨迹

displacement y (metres) and horizontal displacement x (metres) after t seconds are given by: x = 20 cos 30° t, y = 20 sin 30° t - 4.9 t². Find the time when the ball hits the ground and the horizontal range.

一名足球运动员从地面以 20 m/s 的初速度、与水平面成 30° 角踢出足球。假设没有空气阻力，t 秒后的垂直位移 y（米）和水平位移 x（米）由下式给出：x = 20 cos 30° t，y = 20 sin 30° t - 4.9 t²。求足球落地时间及水平射程。

when y = 0: 20 sin 30° t - 4.9 t² = 0. Since sin 30° = 0.5, this becomes 10 t - 4.9 t² = 0 ⇒ t(10 - 4.9 t) = 0. Non-zero solution: t = 10 / 4.9 = 100/49 ≈ 2.041 seconds.

足球落地时 y = 0：20 sin 30° t - 4.9 t² = 0。sin 30° = 0.5，方程变为 10 t - 4.9 t² = 0，即 t(10 - 4.9 t) = 0。非零解：t = 10 / 4.9 = 100/49 ≈ 2.041 秒。

Horizontal range: x = 20 cos 30° × 2.041 = 20 × (√3/2) × 2.041 ≈ 20 × 0.8660 × 2.041 ≈ 35.36 metres.

水平射程：x = 20 cos 30° × 2.041 = 20 × (√3/2) × 2.041 ≈ 20 × 0.8660 × 2.041 ≈ 35.36 米。

half of g ≈ 9.8 m/s², hence the 4.9. This model is typical in OCR Further Maths problems.

注意我们使用了 g/2 ≈ 4.9（g = 9.8 m/s²）。这类模型在 OCR 进阶数学问题中很常见。

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6. Financial Forecasting with Geometric Sequences | 金融预测与等比数列

A young entrepreneur saves £2000 at the start of each year in an account earning 5% compound interest per annum. She wants to know how much her savings will be worth at the end of 10 years, just after the 10th deposit has been made.

一位年轻企业家在每年年初存入 £2000 到一个年利率为 5% 的复利账户中。她想知道在第 10 年末，第 10 笔存款刚存入后，她的储蓄总额是多少。

nt number of years: the first deposit earns interest for 10 years, the second for 9 years, …, the 10th for 0 years. The total value S = 2000×(1.05)¹⁰ + 2000×(1.05)⁹ + … = 2000 (the last deposit, 0 years growth) is actually 2000×1.05⁰ = 2000, but ordering: it's easier to write in reverse as S = 2000 + 2000×1.05 + … + 2000×1.05⁹? rest for 10 years. So the series is a geometric progression with first term a = 2000×1.05 (if we start counting from the first term?), let's set up properly: S = 2000×1.05¹⁰ + 2000×1.05⁹ + … + 2000×1.05¹. That' = 2100, common ratio r = 1.05, n = 10. Sum = a(rⁿ - 1)/(r - 1) = 2100×(1.05¹⁰ - 1)/(0.05).

每笔存款的增长年数不同：第一笔存款计息 10 年，第二笔 9 年，...，第 10 笔 0 年。总额 S = 2000×1.05¹⁰ + 2000×1.05⁹ + … + 2000×1.05¹。这是一个等比数列，首项 a = 2000×1.05 = 2100，公比 r = 1.05，项数 n = 10。求和公式为 S = a(rⁿ - 1)/(r - 1) = 2100×(1.05¹⁰ - 1)/0.05。

5¹⁰ ≈ 1.62889. Then S = 2100 × (1.62889 - 1)/0.05 = 2100 × 0.62889/0.05 = 2100 × 12.5778 = £26 413.38. So after 10 years, she will have about £26 413.

计算 1.05¹⁰ ≈ 1.62889。S = 2100 × (1.62889 - 1)/0.05 = 2100 × 0.62889/0.05 = 2100 × 12.5778 = £26 413.38。因此 10 年后她将拥有约 £26 413。

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7. Analysing an Electrical Network with Simultaneous Equations | 电路网络分析与联立方程组

In a simple DC circuit, Kirchhoff's laws give the following system for loop currents I₁, I₂, I₃ (amperes):
2I₁ - I₂ - I₃ = 10
-I₁ + 3I₂ - I₃ = 0
-I₁ - I₂ + 4I₃ = 5.
Find the currents using elimination or matrices.

在一个简单直流电路中，基尔霍夫定律给出回路电流 I₁、I₂、I₃（安培）的方程组：
2I₁ - I₂ - I₃ = 10
-I₁ + 3I₂ - I₃ = 0
-I₁ - I₂ + 4I₃ = 5。
使用消元法或矩阵求各电流。

Writing as augmented matrix: [ 2 -1 -1 | 10 ] [ -1 3 -1 | 0 ] [ -1 -1 4 | 5 ]. Perform row operations: R1+R3? We can solve stepwise. Add R2 to R3: new R3 = (-2, 2, 3 | 5)? Better systematic:

写成增广矩阵并执行行变换。我们可以通过逐步消元求解。先将第二个方程乘以 2 加到第一个？实际上我们可以直接解。从第二个方程得：3I₂ - I₃ = I₁。代入……更简单的方法：将三个方程相加：0·I₁ + I₂ + 2I₃ = 15，这不是直接的。我们使用矩阵或替代法。

₂ + 4I₃ = -4I₂ + 5I₃ = 5. Now system: 5I₂ - 3I₃ = 10; -4I₂ + 5I₃ = 5. Multiply first by 5: 25I₂ - 15I₃ = 50, second by 3: -12I₂ + 15I₃ = 15. Add: 13I₂ = 65 ⇒ I₂ = 5 A. Substitute: 5(5) - 3I₃ = 10 ⇒ 25 - 3I₃ = 10 ⇒ -3I₃ = -15 ⇒ I₃ = 5 A. Then I₁ = 3(5) - 5 = 10 A.

使用代入法：从第二个方程得 I₁ = 3I₂ - I₃。代入第一个方程：2(3I₂ - I₃) - I₂ - I₃ = 5I₂ - 3I₃ = 10。代入第三个方程：-(3I₂ - I₃) - I₂ + 4I₃ = -4I₂ + 5I₃ = 5。得到方程组：5I₂ - 3I₃ = 10；-4I₂ + 5I₃ = 5。第一个乘以 5，第二个乘以 3 然后相加解出 I₂ = 5 A，从而 I₃ = 5 A，I₁ = 10 A。

Thus, the currents are I₁ = 10 A, I₂ = 5 A, I₃ = 5 A.

所以电流为 I₁ = 10 A，I₂ = 5 A，I₃ = 5 A。

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8. Minimising Material for a Can | 罐头用料最小化

nd the radius r and height h that minimise the total surface area, given V = π r² h = 500.

一家制造商生产容积为 500 cm³ 的圆柱形罐头。为了降低成本，他们希望使用尽可能少的金属表面积。已知 V = π r² h = 500，求使总表面积最小的半径 r 和高 h。

V, express h = 500/(π r²). Substitute: A(r) = 2π r² + 2π r × 500/(π r²) = 2π r² + 1000/r.

表面积 A = 2π r² + 2π r h（两端和侧面）。由体积公式得 h = 500/(π r²)。代入得 A(r) = 2π r² + 2π r × 500/(π r²) = 2π r² + 1000/r。

Differentiate to find minimum: dA/dr = 4π r - 1000/r². Set to 0: 4π r = 1000/r² ⇒ 4π r³ = 1000 ⇒ r³ = 250/π ⇒ r = ∛(250/π). Using π ≈ 3.1416, r³ = 79.577 ⇒ r ≈ 4.30 cm. Then h = 500/(π × (4.30)²) ≈ 500/(3.1416×18.49) ≈ 500/58.09 ≈ 8.61 cm.

微分求最小值：dA/dr = 4π r - 1000/r²。令其为零：4π r = 1000/r² ⇒ r³ = 250/π ⇒ r = ∛(250/π)。用 π ≈ 3.1416，r³ ≈ 79.577，r ≈ 4.30 cm。然后 h ≈ 8.61 cm。

Check second derivative d²A/dr² = 4π + 2000/r³ > 0 for r > 0, so it's a minimum. The optimal can has a height roughly equal to the diameter, a classic result.

二阶导数 d²A/dr² = 4π + 2000/r³ > 0（r > 0），因此为最小值。最优罐头高度与直径大致相等，这是一个经典结果。

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9. Decoding a Caesar Cipher with Matrix Transformations | 用矩阵变换解密凯撒密码

d 26. But with a twist: a 2×2 encoding matrix is applied to pair vectors after conversion. Suppose plaintext letters are numbered 1-26, and matrix E = [1 2; 3 5] is used to encode pairs. Decode the received numbers 27, 58, 43, 92 representing two encoded vectors (27, 58) and (43, 92).

一条消息通过凯撒密码加密（每个字母按固定偏移量移位），同时再乘以一个 2×2 矩阵编码。假设明文字母编号 1-26，编码矩阵 E = [1 2; 3 5] 作用于成对向量。接收到的数字为 27, 58, 43, 92，代表两个编码向量 (27, 58) 和 (43, 92)。请解密。

First find the inverse of E. det(E) = 1×5 - 2×3 = 5 - 6 = -1. The inverse in modulo 26? ight require mod 26. However, OCR Further Maths encryption usually works in the real numbers and then numbers are rounded or just mapped. Let's assume it' antities map to letters. So E⁻¹ = (1/det) [5 -2; -3 1] = -1 × [5 -2; -3 1] = [-5 2; 3 -1].

首先求 E 的逆矩阵。det(E) = 1×5 - 2×3 = -1。E⁻¹ = (1/det) [5 -2; -3 1] = - [5 -2; -3 1] = [-5 2; 3 -1]。

econd vector (43,92): = [-5×43+2×92; 3×43-1×92] = [-215+184; 129-92] = [-31; 37]. Negative numbers can be shifted by adding multiples of 26 to fall into 1-26 range. -19 mod 26 = 7 (since -19+26=7), 23 stays 23, so first pair (7,23) = G, W. -31 mod 26 = 21 (since -31+52=21), 37 mod 26 = 11, so second pair (21,11) = U, K. The message might be 'GWUK' – possibly a name or code.

将每个编码向量乘以 E⁻¹。第一个向量 (27,58) 得 [-19; 23]。第二个向量 (43,92) 得 [-31; 37]。负数可以通过加上 26 的倍数映射到 1-26 范围。-19 mod 26 = 7 (G)，23 → 23 (W)，得到 (G,W)；-31 → 21 (U)，37 mod 26 = 11 (K)，得到 (U,K)。组合可能为 GWUK，可解读为某个单词或代号。

This illustrates the combined use of matrix algebra and modular arithmetic, both common in OCR Further Maths.

这展示了矩阵代数与模运算的结合使用，二者在 OCR 进阶数学中都很常见。

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10. Summary: Bringing It All Together | 总结：融会贯通

These case studies demonstrate that the topics in OCR Further Maths – quadratic optimisation, parametric curves, matrix algebra, exponential equations, projectile motion, geometric sequences, simultaneous equations, and calculus of solids – are not isolated textbook exercises. They form a toolkit for solving real-world engineering, business, and scientific problems. Practice with such multi-step problems will sharpen your ability to identify the right mathematical strategy, handle complex calculations, and communicate your reasoning clearly.

这些案例表明，OCR 进阶数学中的主题——二次优化、参数曲线、矩阵代数、指数方程、抛体运动、等比数列、联立方程组以及立体微积分——并非孤立的课本练习题。它们构成了解决实际工程、商业和科学问题的工具箱。通过练习这类多步骤问题，你将能更敏锐地识别正确的数学策略，处理复杂计算，并清晰地表达推理过程。

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