Case Studies in Further Maths: Practical Problem-Solving for OCR Year 10 | OCR Year 10 进阶数学案例分析实战演练

📚 Case Studies in Further Maths: Practical Problem-Solving for OCR Year 10 | OCR Year 10 进阶数学案例分析实战演练

This article presents a series of real-world case studies designed to build problem-solving skills for the OCR Year 10 Further Maths course. Each case demonstrates how to apply core topics such as quadratic functions, trigonometry, vectors, matrices, and calculus to practical situations. Working through these examples will deepen your understanding and prepare you for examination-style questions that require logical reasoning and multi-step solutions.

本文通过一系列贴近现实问题的案例研究，帮助 OCR Year 10 进阶数学课程的学生提升解题能力。每个案例都展示了如何将二次函数、三角学、向量、矩阵和微积分等核心知识应用到实际场景中。通过研习这些例子，你将加深对概念的理解，并为需要逻辑推理与多步骤求解的考试题型做好充分准备。

1. Quadratic Optimisation: Maximising Enclosure Area | 二次函数优化：最大化围栏面积

ncing. Let the side parallel to the wall be length x metres and each of the two equal sides be length y metres. The total fencing used is x + 2y = 80. The area A = x × y. Express A in terms of one variable, find its maximum value and the corresponding dimensions.

一位农民有 80 米长的围栏，想靠着一面笔直的墙围出一个矩形区域，以墙作为其中一边。设平行于墙的边长为 x 米，两个等长的侧边各为 y 米。所用围栏总长为 x + 2y = 80。面积 A = x × y。将 A 表示为单一变量的函数，并求其最大值及相应的尺寸。

From the constraint, y = (80 – x)/2. Substituting: A(x) = x(80 – x)/2 = 40x – th coefficient of x² negative, so it has a maximum at the vertex. The vertex occurs at x = –b/(2a) with a = –0.5, b = 40, giving x = –40/(2 × –0.5) = 40. Then y = (80 – 40)/2 = 20. Maximum area = 40 × 20 = 800 m².

由约束条件得 y = (80 – x)/2。代入面积公式：A(x) = x(80 – x)/2 = 40x – 0.5x²。此为二次函数，x² 系数为负，故在顶点处取得最大值。顶点横坐标 x = –b/(2a)，其中 a = –0.5, b = 40，得 x = –40/(2 × –0.5) = 40。于是 y = (80 – 40)/2 = 20。最大面积为 40 × 20 = 800 平方米。

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2. Simultaneous Equations in Business Contexts | 商业情境中的联立方程

A company produces two types of gift boxes: standard and deluxe. Each standard box requires 2 hours of assembly and 1 hour of packaging, while a deluxe box requires 3 hours of assembly and 2 hours of packaging. In one week the assembly department works 100 hours and the packaging department works 60 hours. Determine the number of each type of box produced, assuming full utilisation of available hours.

一家公司生产两种礼品盒：标准版和豪华版。每个标准盒需要 2 小时的装配和 1 小时的包装，每个豪华盒需要 3 小时装配和 2 小时包装。某一周内，装配部门工作 100 小时，包装部门工作 60 小时。求在完全利用可用工时的条件下，每种盒子的生产数量。

iply the second equation by 2 to get 2s + 4d = 120. Subtract the first equation: (2s + 4d) – (2s + 3d) = 120 – ck: s + 2(20) = 60 → s = 20. Thus 20 standard and 20 deluxe boxes can be produced exactly.

设标准盒数量为 s，豪华盒数量为 d。装配：2s + 3d = 100。包装：s + 2d = 60。用消元法求解：将第二个方程乘以 2 得 2s + 4d = 120。减去第一个方程：(2s + 4d) – (2s + 3d) = 120 – 100，解得 d = 20。代回：s + 2(20) = 60，得 s = 20。因此可正好生产 20 个标准盒和 20 个豪华盒。

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3. Coordinate Geometry: Locus of Points | 坐标几何：点的轨迹

A point P moves such that its distance from the point A(2, 3) is always twice its distance from the point B(8, 3). Find the equation of the locus of P and describe its geometric shape.

动点 P 到点 A(2, 3) 的距离恒等于它到点 B(8, 3) 距离的两倍。求点 P 的轨迹方程，并描述其几何形状。

Let P be (x, y). Distance PA = √[(x – 2)² + (y – 3)²], PB = √[(x – 8)² + (y – 3)²]. Condition: PA = 2 × PB. Square both sides: (x – 2)² + (y – 3)² = 4[(x – 8)² + (y – 3)²]. Expand: x² – 4x + 4 + (y – 3)² = 4(x² – 16x + 64) + 4(y – 3)². Collect terms: 0 = 3x² – 60x + 252 + 3(y – 3)². Divide by 3: x² – 20x + 84 + (y – 3)² = 0. Complete the square for x: (x – 10)² – 100 + 84 + (y – 3)² = 0 → (x – 10)² + (y – 3)² = 16. This is a circle with centre (10, 3) and radius 4.

设 P 坐标为 (x, y)。PA = √[(x – 2)² + (y – 3)²]，PB = √[(x – 8)² + (y – 3)²]。条件为 PA = 2 × PB。两边平方：(x – 2)² + (y – 3)² = 4[(x – 8)² + (y – 3)²]。展开：x² – 4x + 4 + (y – 3)² = 4(x² – 16x + 64) + 4(y – 3)²。移项得 0 = 3x² – 60x + 252 + 3(y – 3)²。除以 3：x² – 20x + 84 + (y – 3)² = 0。对 x 配方：(x – 10)² – 100 + 84 + (y – 3)² = 0，即 (x – 10)² + (y – 3)² = 16。这是圆心在 (10, 3)、半径为 4 的圆。

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4. Trigonometric Modelling: Tidal Height | 三角建模：潮汐高度

m and minimum water heights, the time of the first high tide after midnight, and the times when the height is exactly 6.5 m between t = 0 and t = 12.

某港口潮汐高度可用模型 h(t) = 5 + 3 cos(30t)° 表示，其中 h 单位为米，t 为午夜过后的小时数。求最高与最低水位、午夜后首次高潮的时间，以及在 t = 0 到 t = 12 之间水位恰好为 6.5 米的时刻。

Maximum h occurs when cos(30t)° = 1, giving h_max = 5 + 3 = 8 m. Minimum when cos(30t)° = –1, h_min = 5 – 3 = 2 m. High tide occurs when cos(30t)° = 1, i.e. 30t = 0, 360, 720… so first after midnight is at t = 0 (already high) or t = 12? (30t)° = 1.5 → cos(30t)° = 0.5. 30t = 60°, 300° in first 360°. Thus t = 2 or t = 10. So water height is 6.5 m at 2 am and 10 am.

当 cos(30t)° = 1 时水位最高，h_max = 5 + 3 = 8 米。最低水位在 cos(30t)° = –1 时，h_min = 5 – 3 = 2 米。高潮出现在 cos(30t)° = 1，即 30t = 0, 360, 720… 午夜后第一次高潮在 t = 0，下一次在 t = 12 小时（因 30×12 = 360°）。求 h = 6.5：5 + 3 cos(30t)° = 6.5 → 3 cos(30t)° = 1.5 → cos(30t)° = 0.5。在首个 360° 内 30t = 60° 或 300°，故 t = 2 或 t = 10。因此水位在凌晨 2 点和上午 10 点恰好为 6.5 米。

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5. Vector Proofs: Midpoints and Parallelograms | 向量证明：中点与平行四边形

In quadrilateral ABCD, the midpoints of the sides are P, Q, R and S. Prove using vectors that PQRS is a parallelogram.

在四边形 ABCD 中，各边中点依次为 P、Q、R、S。试用向量证明 PQRS 是平行四边形。

be a, b, c, d. Then P = ½(a + b), Q = ½(b + c), R = ½(c + d), S = ½(d + a). Vector PQ = Q – P = ½(b + c) – ½(a + b) = ½(c – a). Vector SR = R – S = ½(c + d) – ½(d + a) = ½(c – a). So PQ = SR. Similarly, PS = S – P = ½(d + a) – ½(a + b) = ½(d – b) and QR = R – Q = ½(c + d) – ½(b + c) = ½(d – so PS = QR. Hence opposite sides are parallel and equal, proving PQRS is a parallelogram.

设各顶点的位置向量为 a, b, c, d。则 P = ½(a + b)，Q = ½(b + c)，R = ½(c + d)，S = ½(d + a)。向量 PQ = Q – P = ½(b + c) – ½(a + b) = ½(c – a)。向量 SR = R – S = ½(c + d) – ½(d + a) = ½(c – a)。因此 PQ = SR。类似地，PS = S – P = ½(d + a) – ½(a + b) = ½(d – b)，QR = R – Q = ½(c + d) – ½(b + c) = ½(d – b)，得 PS = QR。故对边平行且相等，PQRS 为平行四边形。

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6. Matrix Transformations: Rotating a Triangle | 矩阵变换：旋转三角形

A triangle has vertices at A(2, 1), B(5, 1) and C(4, 4). The triangle is rotated 90° anticlockwise about the origin. Find the coordinates of the image vertices using a matrix transformation, and calculate the area of the transformed triangle.

一个三角形的顶点为 A(2, 1)、B(5, 1) 和 C(4, 4)。将三角形绕原点逆时针旋转 90°。用矩阵变换求出像点坐标，并计算变换后三角形的面积。

The rotation matrix for 90° anticlockwise is

[ 0 –1 ]
[ 1 0 ]

. Multiply each column vector of coordinates: A' = [0 –1; 1 0] [2; 1] = [–1; 2]; B' = [–1; 5]; C' = [–4; 4]. Area of original triangle: ½ | xA(yB – yC) + xB(yC – yA) + xC(yA – yB) | = ½ | 2(1 – 4) + 5(4 – 1) + 4(1 – 1) | = ½ | 2(–3) + 5(3) + 0 | = ½ | –6 + 15 | = 4.5. Rotation preserves area, so transformed triangle area is also 4.5 square units.

逆时针旋转 90° 的矩阵为

[ 0 –1 ]
[ 1 0 ]

。 乘以每个坐标列向量：A' = [0 –1; 1 0] [2; 1] = [–1; 2]；B' = [–1; 5]；C' = [–4; 4]。原三角形面积：½ | xA(yB – yC) + xB(yC – yA) + xC(yA – yB) | = ½ | 2(1 – 4) + 5(4 – 1) + 4(1 – 1) | = ½ | –6 + 15 | = 4.5。旋转不改变面积，故变换后三角形面积仍为 4.5 平方单位。

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7. Differentiation: Minimising Surface Area | 微分：最小化表面积

e V = πr²h, surface area S = 2πr² + 2πrh. Express S in terms of r only and differentiate.)

一个带盖的圆柱形罐子需容纳 500 cm³ 液体。求使总表面积最小的半径 r 和高 h。（使用体积公式 V = πr²h，表面积 S = 2πr² + 2πrh。将 S 表示为仅含 r 的函数并求导。）

titute into S: S(r) = 2πr² + 2πr × (500/(πr²)) = 2πr² + 1000/r. Differentiate: dS/dr = 4πr – 1000/r². Set derivative to zero: 4πr = 1000/r² → r³ = 250/π → r = ∛ × (4.30)²) ≈ 8.60 cm. The minimum area occurs when height equals 2r, a well-known result.

由体积得 h = 500/(πr²)。代入 S：S(r) = 2πr² + 2πr × (500/(πr²)) = 2πr² + 1000/r。求导：dS/dr = 4πr – 1000/r²。令导数等于零：4πr = 1000/r² → r³ = 250/π → r ≈ 4.30 cm。二阶导数 d²S/dr² = 4π + 2000/r³，对于 r>0 恒正，故为极小值。此时 h ≈ 8.60 cm。最小表面积出现在高度等于直径时，这是一个已知结论。

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8. Integration: Area Enclosed by a Curve and Line | 积分：曲线与直线围成的面积

Find the area of the region bounded by the curve y = 6x – x² and the line y = x.

求曲线 y = 6x – x² 与直线 y = x 所围成区域的面积。

Find intersection points: set 6x – x² = x → x² – 5x = 0 → x(x – 5) = 0, so x = 0 and x = 5. The area is ∫ from 0 to 5 of (top – bottom) dx = ∫₀⁵ [ (6x – x²) – x ] dx = ∫₀⁵ (5x – x²) dx = [ (5/2)x² – x³/3 ]₀⁵ = (5/2)(25) – (125/3) = 125/2 – 125/3 = 125(1/2 – 1/3) = 125(1/6) = 125/6 ≈ 20.83 square units.

求交点：令 6x – x² = x → x² – 5x = 0 → x(x – 5) = 0，得 x = 0 和 x = 5。面积 = ∫₀⁵ (上曲线 – 下曲线) dx = ∫₀⁵ [ (6x – x²) – x ] dx = ∫₀⁵ (5x – x²) dx = [ (5/2)x² – x³/3 ]₀⁵ = (5/2)(25) – (125/3) = 125/2 – 125/3 = 125(1/2 – 1/3) = 125(1/6) = 125/6 ≈ 20.83 平方单位。

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9. Sequences and Series: Compound Interest | 数列与级数：复利计算

An investment of £2000 earns 4% interest per annum, compounded monthly. Find the value after 3 years. Then determine the monthly deposit needed to reach £10,000 in 5 years with the same interest rate.

一笔 2000 英镑的投资，年利率 4%，按月复利。求 3 年后的价值。然后，在相同利率下，要 5 年内达到 10,000 英镑，每月需存入多少钱？

Monthly interest rate = 4%/12 = 0.333...% = 0.003333... Value after 3 years (36 months): A = 2000(1 + 0.04/12)³⁶ ≈ 2000 × 1.12727 = £ hly deposit problem, we use the future value of an annuity formula: FV = PMT × [( (1 + i)ⁿ – 1 ) / i], where i = 0.04/12, n = 60 months. Need FV = £10,000, so 10000 = PMT × [(1.003333...⁶⁰ – 1)/0.003333...]. Compute (1.003333...)⁶⁰ ≈ 1.22100, so the factor ≈ (1.22100 – 1)/0.003333... = 66.30. Thus PMT ≈ 10000 / 66.30 ≈ £150.85 per month.

月利率 = 4%/12 ≈ 0.3333%。3 年（36 个月）后价值：A = 2000(1 + 0.04/12)³⁶ ≈ 2000 × 1.12727 = 2254.54 英镑。对于每月存款问题，使用年金终值公式：FV = PMT × [( (1 + i)ⁿ – 1 ) / i]，其中 i = 0.04/12，n = 60 个月。欲使 FV = 10,000，故 10000 = PMT × [(1.003333...⁶⁰ – 1)/0.003333...]。计算 (1.003333...)⁶⁰ ≈ 1.22100，因子 ≈ (1.22100 – 1)/0.003333... = 66.30。因此 PMT ≈ 10000/66.30 ≈ 150.85 英镑/月。

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10. Inequalities and Linear Programming: Production Planning | 不等式与线性规划：生产计划

A factory produces chairs (x) and tables (y). Each chair requires 1 hour of labour and 3 m² of wood; each table requires 2 hours of labour and 1 m² of wood. Available labour: 40 hours; wood: 42 m². The profit per chair is £30 and per table is £40. Determine the number of chairs and tables that maximise profit, and state the maximum profit.

某工厂生产椅子 (x) 和桌子 (y)。每把椅子需要 1 小时人工和 3 平方米木材；每张桌子需要 2 小时人工和 1 平方米木材。可用人工 40 小时；木材 42 平方米。椅子利润 30 英镑/把，桌子利润 40 英镑/张。求使利润最大化的椅子与桌子数量，并给出最大利润。

20); Wood (14,0) and (0,42). Feasible region is bounded by these lines and axes. Find vertices: (0,0), (14,0), (0,20), and intersection of labour and wood lines. Solve x + 2y = 40 and 3x + y = 42: from second, y = 42 – 3x. Substitute into first: x + 2(42 – 3x) = 40 → x + 84 – 6x = 40 → –5x = –44 → x = 8.8, then y = 42 – 3×8.8 = 42 – ; (14,0)=420; (0,20)=800; (8.8,15.6)=30×8.8 + 40×15.6 = 264 + 624 = 888. Maximum profit is £ gers may be required, but the optimal continuous solution is useful for rounding analysis.

约束条件：x ≥ 0, y ≥ 0。人工：x + 2y ≤ 40。木材：3x + y ≤ 42。利润 P = 30x + 40y。绘制可行域：人工线截距为 (40,0) 和 (0,20)；木材线截距为 (14,0) 和 (0,42)。可行域由这两条线与坐标轴围成。顶点包括 (0,0)、(14,0)、(0,20)，以及两条直线的交点。解方程组 x + 2y = 40 和 3x + y = 42：由第二式得 y = 42 – 3x，代入第一式：x + 2(42 – 3x) = 40 → x + 84 – 6x = 40 → –5x = –44 → x = 8.8，则 y = 42 – 3×8.8 = 15.6。计算各顶点利润：(0,0)=0；(14,0)=420；(0,20)=800；(8.8,15.6)=30×8.8 + 40×15.6 = 264 + 624 = 888。最大利润为 888 英镑，生产 8.8 把椅子和 15.6 张桌子。如需整数解可四舍五入分析，但连续解指明了最优方向。

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