Year 10 AQA Further Maths: Case Study Practice | Year 10 AQA 进阶数学：案例分析实战演练

📚 Year 10 AQA Further Maths: Case Study Practice | Year 10 AQA 进阶数学：案例分析实战演练

In AQA Level 2 Further Mathematics, theoretical knowledge becomes truly powerful when applied to real-world scenarios. This article presents ten authentic case studies that blend algebra, calculus, matrices, trigonometry and geometry, guiding you through each solution step-by-step. By working through these practical exercises, you will sharpen your problem-solving skills and build the confidence to tackle any exam question.

在 AQA Level 2 进阶数学中，理论知识只有应用于实际情境才能真正发挥威力。本文提供十个真实的案例分析，融合代数、微积分、矩阵、三角和几何，逐步引导你完成每个解答。通过演练这些实战练习，你将提升解题能力，并建立应对任何考试题目的信心。

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1. Optimising a Box Volume | 盒子体积优化

A packaging manufacturer intends to create an open box from a rectangular cardboard sheet measuring 20 cm by 16 cm. Four identical squares of side length x cm are cut from the corners, and the flaps are folded up. The company wants to determine the value of x that yields the maximum possible volume.

一家包装制造商计划用一张 20 cm × 16 cm 的长方形卡纸制作一个无盖盒子。从四个角各剪去边长为 x cm 的正方形，再将四周折起。公司希望找出使盒子体积最大的 x 值。

he length becomes (20−2x), the width (16−2x), and the height x. Hence V = x(20−2x)(16−2x).

步骤1：用 x 表示体积 V。剪切折叠后，长变为 (20−2x)，宽变为 (16−2x)，高为 x。因此 V = x(20−2x)(16−2x)。

320 − 40x − 32x + 4x² = 320 − 72x + 4x². Then V = x(320 − 72x + 4x²) = 4x³ − 72x² + 320x.

步骤2：展开成多项式。先计算 (20−2x)(16−2x) = 320 − 40x − 32x + 4x² = 320 − 72x + 4x²。于是 V = x(320 − 72x + 4x²) = 4x³ − 72x² + 320x。

Step 3: Find the derivative dV/dx to locate stationary points.

dV/dx = 12x² − 144x + 320

步骤3：求导数 dV/dx，以确定驻点。

dV/dx = 12x² − 144x + 320

lve the quadratic. Divide by 4: 3x² − 36x + 80 = 0. Using the quadratic formula, x = [36 ± √1296 − 960] / 6 = [36 ± √336] / 6 = [36 ± 4√21] / 6 = 6 ± (2√21)/3.

步骤4：令 dV/dx = 0 并解二次方程。除以 4：3x² − 36x + 80 = 0。利用求根公式，x = [36 ± √1296 − 960] / 6 = [36 ± √336] / 6 = [36 ± 4√21] / 6 = 6 ± (2√21)/3。

Step 5: The feasible domain is 0 < x < 8. Compute approximate values: 6 + (2√21)/3 ≈ 6 + 3.06 ≈ 9.06 (reject, too large), 6 − (2√21 second derivative d²V/dx² = 24x − 144. At x ≈ 2.94, d²V/dx² < 0, confirming a maximum.

步骤5：实际定义域为 0 < x < 8。计算近似值：6 + (2√21)/3 ≈ 6 + 3.06 ≈ 9.06（超出范围，舍去），6 − (2√21)/3 ≈ 6 − 3.06 ≈ 2.94。验证二阶导数 d²V/dx² = 24x − 144，当 x ≈ 2.94 时 d²V/dx² < 0，确认为极大值。

size is x ≈ 2.94 cm, giving a maximum volume of about V ≈ 2.94 × 14.12 × 10.12 ≈ 420 cm³.

步骤6：因此，最佳裁剪尺寸为 x ≈ 2.94 cm，最大体积约为 V ≈ 2.94 × 14.12 × 10.12 ≈ 420 cm³。

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2. Solving Simultaneous Equations with Matrices | 使用矩阵解方程组

A shop sells two types of gift bags. On Monday, 2 premium bags and 3 standard bags cost £7 in total. On Tuesday, 5 premium and 2 standard bags cost £8. Model the situation as a matrix equation and use the inverse matrix to find the individual prices.

一家商店出售两种礼品袋。周一，2 个精品袋和 3 个标准袋总价 £7。周二，5 个精品袋和 2 个标准袋总价 £8。将情景建模为矩阵方程，并利用逆矩阵求单价。

bag and q the price of a standard bag. Write the linear system: 2p + 3q = 7, 5p + 2q = 8.

步骤1：设精品袋单价为 p，标准袋单价为 q。列出线性方程组：2p + 3q = 7，5p + 2q = 8。

Step 2: Expressed in matrix form AX = B,

[2 3] [p] = [7]
[5 2] [q] [8]

步骤2：用矩阵形式 AX = B 表示，

[2 3] [p] = [7]
[5 2] [q] [8]

) = (2)(2) − (3)(5) = 4 − 15 = −11. Since the determinant is non-zero, the inverse exists.

步骤3：求矩阵 A 的行列式：det(A) = (2)(2) − (3)(5) = 4 − 15 = −11。由于行列式不为零，逆矩阵存在。

4: The inverse of a 2×2 matrix [a b; c d] is 1/(ad−bc) [d −b; −c a]. Thus A⁻¹ = 1/(-11) [2 −3; −5 2].

步骤4：对于 2×2 矩阵 [a b; c d]，其逆为 1/(ad−bc) [d −b; −c a]。因此 A⁻¹ = 1/(-11) [2 −3; −5 2]。

·8; (−5)·7 + 2·8] = (1/(-11)) [14 − 24; −35 + 16] = (1/(-11)) [−10; −19] = [10/11; 19/11].

步骤5：两边左乘 A⁻¹： )·8; (−5)·7 + 2·8] = (1/(-11)) [14 − 24; −35 + 16] = (1/(-11)) [−10; −19] = [10/11; 19/11]。

Step 6: So the premium bag costs £10/11 ≈ £0.91 and the standard bag £19/11 ≈ £1.73. Check: 2×0.91 + 3×1.73 ≈ 7.

步骤6：因此精品袋单价 £10/11 ≈ £0.91，标准袋单价 £19/11 ≈ £1.73。验证：2×0.91 + 3×1.73 ≈ 7。

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3. Solving Trigonometric Equations | 解三角方程

les in the range 0° ≤ θ ≤ 360°. Determine all possible angles θ that satisfy the equation.

一位工程师监测一个周期信号，其模型为方程 2sin²θ − cosθ − 1 = 0，角度范围为 0° ≤ θ ≤ 360°。求所有满足该方程的角 θ。

ity sin²θ = 1 − cos²θ to express everything in terms of cosθ. 2(1 − cos²θ) − cosθ − 1 = 0.

步骤1：利用恒等式 sin²θ = 1 − cos²θ，将方程全部用 cosθ 表示。2(1 − cos²θ) − cosθ − 1 = 0。

Step 2: Simplify: 2 − 2cos²θ − cosθ − 1 = 0 → −2cos²θ − cosθ + 1 = 0. Multiply by −1:

2cos²θ + cosθ − 1 = 0

步骤2：化简：2 − 2cos²θ − cosθ − 1 = 0 → −2cos²θ − cosθ + 1 = 0。乘以 −1 得：

2cos²θ + cosθ − 1 = 0

Step 3: Factorise the quadratic in cosθ: (2cosθ − 1)(cosθ + 1) = 0.

步骤3：对 cosθ 的二次式进行因式分解：(2cosθ − 1)(cosθ + 1) = 0。

Step 4: Set each factor to zero: 2cosθ − 1 = 0 → cosθ = ½, or cosθ + 1 = 0 → cosθ = −1.

步骤4：令每个因式为零：2cosθ − 1 = 0 → cosθ = ½，或 cosθ + 1 = 0 → cosθ = −1。

Step 5: For cosθ = ½ in the specified interval, θ = 60° and 300°. For cosθ = −1, θ = 180°.

步骤5：在指定区间内，cosθ = ½ 的解为 θ = 60° 和 300°。cosθ = −1 的解为 θ = 180°。

Step 6: The complete solution set is {60°, 180°, 300°}.

步骤6：完整解集为 {60°, 180°, 300°}。

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4. Intersection of a Circle and a Line | 圆与直线的交点

lows the line y = 2x + 1. Find the coordinates where the path meets the fountain boundary.

一个规划部门采用坐标网格，其中圆形喷泉可用 (x−3)² + (y+2)² = 25 表示，一条直行小径沿直线 y = 2x + 1 延伸。求小径与喷泉边界的交点坐标。

= 2x + 1 into the circle equation: (x−3)² + (2x + 1 + 2)² = 25 → (x−3)² + (2x + 3)² = 25.

步骤1：将 y = 2x + 1 代入圆方程：(x−3)² + (2x + 1 + 2)² = 25 → (x−3)² + (2x + 3)² = 25。

Step 2: Expand both brackets: x² − 6x + 9 + 4x² + 12x + 9 = 25 → 5x² + 6x + 18 = 25.

步骤2：展开平方项：x² − 6x + 9 + 4x² + 12x + 9 = 25 → 5x² + 6x + 18 = 25。

Step 3: Rearrange to standard form: 5x² + 6x − 7 = 0.

步骤3：整理为标准式：5x² + 6x − 7 = 0。

Step 4: Apply the quadratic formula: x = [−6 ± √36 + 140] / 10 = [−6 ± √176] / 10 = [−6 ± 4√11] / 10 = [−3 ± 2√11] / 5.

Published by TutorHao | Year 10 进阶数学 Revision Series | aleveler.com

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