📚 Case Study Practice: Edexcel Year 11 Further Maths | Edexcel 进阶数学案例分析实战演练
In Edexcel Year 11 Further Mathematics, applying theoretical knowledge to real-world case studies is essential for deepening understanding and exam success. This article presents a series of practical case studies covering key topics such as optimisation with quadratics, differentiation, trigonometry, vectors, exponential growth, and matrices. Each case study is solved step-by-step, showing how mathematical techniques model and solve authentic problems.
在Edexcel 11年级进阶数学中，将理论知识应用于实际案例分析对于深化理解和取得考试成功至关重要。本文呈现一系列实战案例，涵盖二次函数优化、微分、三角学、向量、指数增长和矩阵等关键主题。每个案例逐步求解，展示如何运用数学方法建模并解决真实问题。
1. Optimising Area with a Fixed Perimeter | 固定周长下的面积优化
A farmer has 100 metres of fencing and wants to enclose a rectangular area against an existing wall, so only three sides need fencing. Find the dimensions that maximise the area.
一位农民有100米长的围栏，想靠着一面墙围出一个矩形区域，因此只需要围三边。求使面积最大化的尺寸。
lel to the wall is (100 − 2x) metres. The area A is given by A = x(100 − 2x) = 100x − 2x².
设垂直于墙的宽度为 x 米，则平行于墙的长度为 (100 − 2x) 米。面积 A = x(100 − 2x) = 100x − 2x²。
aximise A, differentiate: dA/dx = 100 − 4x. Set derivative to zero: 100 − 4x = 0 → x = 25.
为求最大化 A，求导：dA/dx = 100 − 4x。令导数为零：100 − 4x = 0 → x = 25。
a maximum. Thus width = 25 m, length = 100 − 50 = 50 m, maximum area = 25 × 50 = 1250 m².
二阶导数 d²A/dx² = −4 < 0，确认为最大值。因此宽 = 25 米，长 = 100 − 50 = 50 米，最大面积 = 25 × 50 = 1250 平方米。
Aₘₐₓ = 1250 m²
2. Projectile Motion: Maximum Height via Vectors | 抛体运动：向量法求最大高度
ere j is vertically upward. Using g = 9.8 m s⁻², find its maximum height above the ground.
一个小球从地面以初速度向量 (10i + 20j) 米/秒抛出，其中j为竖直向上方向。取 g = 9.8 米/秒²，求其离地面的最大高度。
height, vertical velocity vᵧ = 0. Using v = u + at: 0 = 20 − 9.8t → t = 20/9.8 ≈ 2.0408 s.
竖直运动：初速 uᵧ = 20 m s⁻¹。在最大高度处，竖直速度 vᵧ = 0。由 v = u + at：0 = 20 − 9.8t → t = 20/9.8 ≈ 2.0408 秒。
= 20t − 4.9t². Substitute t: s = 20(20/9.8) − 4.9(20/9.8)² = (400/9.8) − (4.9×400)/(9.8²).
高度 s = ∫ v dt = 20t − 4.9t²。代入 t：s = 20(20/9.8) − 4.9(20/9.8)² = (400/9.8) − (4.9×400)/(9.8²)。
4.9/9.8 = 0.5, so s = (400/9.8) × (1 − 0.5) = 200/9.8 ≈ 20.41 m. Maximum height ≈ 20.4 m.
注意到 4.9/9.8 = 0.5，因此 s = (400/9.8) × (1 − 0.5) = 200/9.8 ≈ 20.41 米。最大高度 ≈ 20.4 米。
hₘₐₓ = 200/9.8 ≈ 20.4 m
3. Maximising Profit Using Differentiation | 利用微分求最大利润
A company’s weekly profit £ 15x² − 24x + 50, for 0 ≤ x ≤ 6. Determine the production level that yields maximum profit.
一家公司的周利润 £P（千英镑）与生产的件数 x（百件）满足关系 P = −2x³ + 15x² − 24x + 50，其中 0 ≤ x ≤ 6。请确定使利润最大化的生产水平。
² + 30x − 24. Set derivative to zero: −6x² + 30x − 24 = 0 → divide by −6: x² − 5x + 4 = 0.
求导：dP/dx = −6x² + 30x − 24。令导数为零：−6x² + 30x − 24 = 0 → 两边除以 −6：x² − 5x + 4 = 0。
Factorise: (x − 1)(x − 4) = 0, so critical points are x = 1 and x = 4.
因式分解：(x − 1)(x − 4) = 0，因此临界点为 x = 1 和 x = 4。
= 4, d²P/dx² = −18 < 0 (maximum). Hence maximum profit occurs when 400 units are produced.
二阶导数：d²P/dx² = −12x + 30。当 x = 1 时，d²P/dx² = 18 > 0（极小值）；当 x = 4 时，d²P/dx² = −18 < 0（极大值）。因此生产400件时利润最大。
t x = 4: P(4) = −2(64) + 15(16) − 24(4) + 50 = −128 + 240 − 96 + 50 = 66. Maximum profit = £66 000.
计算 x = 4 时的 P：P(4) = −2(64) + 15(16) − 24(4) + 50 = −128 + 240 − 96 + 50 = 66。最大利润为 £66 000。
Pₘₐₓ = 66 (thousand £)
4. Logarithmic Scale: Comparing Sound Intensities | 对数尺度：声音强度比较
decibel level. If two identical machines operate together, what is the combined decibel reading?
声音强度级 L（分贝）由 L = 10 log₁₀(I / I₀) 给出，其中 I₀ = 10⁻¹² 瓦/米²。一台机器发出的声音强度为 I₁ = 10⁻⁵ 瓦/米²。求其分贝值。如果两台相同的机器同时工作，总的分贝读数是多少？
Single machine: L₁ = 10 log₁₀(10⁻⁵ / 10⁻¹²) = 10 log₁₀(10⁷) = 10 × 7 = 70 dB.
单个机器：L₁ = 10 log₁₀(10⁻⁵ / 10⁻¹²) = 10 log₁₀(10⁷) = 10 × 7 = 70 dB。
× 10⁻⁵ W m⁻². Then L₂ = 10 log₁₀(2 × 10⁻⁵ / 10⁻¹²) = 10 log₁₀(2 × 10⁷) = 10 (log₁₀2 + 7).
两台机器：总强度 I₂ = 2 × 10⁻⁵ 瓦/米²。则 L₂ = 10 log₁₀(2 × 10⁻⁵ / 10⁻¹²) = 10 log₁₀(2 × 10⁷) = 10 (log₁₀2 + 7)。
0 = 73.01 dB. The increase is not double but only about 3 dB due to the logarithmic scale.
利用 log₁₀2 ≈ 0.3010：L₂ ≈ 10 × 7.3010 = 73.01 dB。由于对数尺度，分贝值并未翻倍，仅增加了约3 dB。
L₂ ≈ 73.0 dB
5. Matrix Transformations in Image Processing | 图像处理中的矩阵变换
A graphic design tool applies transformations to a shape. Point A(2, 3) is first reflected in the y-axis, then rotated 90° anticlockwise about the origin. Find the final coordinates of the image point.
某图形设计工具对图形施行变换。点A(2, 3) 先关于 y 轴反射，再绕原点逆时针旋转90°。求最终像点的坐标。
iclockwise matrix: Q = [[0, -1], [1, 0]]. The combined transformation matrix is M = Q × R.
关于 y 轴反射矩阵：R = [[-1, 0], [0, 1]]；逆时针旋转90° 矩阵：Q = [[0, -1], [1, 0]]。组合变换矩阵 M = Q × R。
Compute M: [[0, -1], [1, 0]] × [[-1, 0], [0, 1]] = [[0, -1], [-1, 0]].
计算 M：[[0, -1], [1, 0]] × [[-1, 0], [0, 1]] = [[0, -1], [-1, 0]]。
× [2, 3]ᵗ = [0×2 + (-1)×3, (-1)×2 + 0×3] = [-3, -2]. The final image point is A′(-3, -2).
将 M 作用于点 A：[[0, -1], [-1, 0]] × [2, 3]ᵗ = [0×2 + (-1)×3, (-1)×2 + 0×3] = [-3, -2]。最终像点为 A′(-3, -2)。
A′( -3 , -2 )
6. Trigonometric Modelling of Daylight Hours | 日照时数的三角建模
ar (t = 1 on 1 Jan). Determine the longest day and how many hours of daylight it provides.
某城市的日照时数 H 可用模型 H = 12 + 3.5 sin(2π(t − 80)/365) 描述，其中 t 为年内天数（t = 1 表示1月1日）。求最长的白昼及其日照时数。
function reaches its maximum value of 1. Therefore maximum H = 12 + 3.5 × 1 = 15.5 hours.
正弦函数的最大值为1，因此最大 H = 12 + 3.5 × 1 = 15.5 小时。
− 80 = 91.25 → t = 171.25. This corresponds approximately to the 171st day, i.e., 20 June.
令 sin(2π(t − 80)/365) = 1 → 2π(t − 80)/365 = π/2 + 2kπ。取 k = 0 的第一个解：(t − 80)/365 = 1/4 → t − 80 = 91.25 → t = 171.25。约对应第171天，即6月20日。
Thus, the longest day has 15.5 hours of daylight, occurring around the summer solstice.
因此，最长白昼有15.5小时日照，出现在夏至前后。
Hₘₐₓ = 15.5 hours
7. Compound Interest and Geometric Series | 复利与几何级数
At the start of each year, £1000 is invested into an account paying 5% annual compound interest. Find the total value of the account immediately after the 10th deposit.
每年年初，将 £1000 投入一个年复利5%的账户。求第10次存款刚存入后账户的总价值。
nd so on. The last deposit earns no interest: 1000 × 1.05⁰. This forms a geometric series.
第一笔存款计息10年，本利和为 1000 × 1.05¹⁰；第二笔计息9年：1000 × 1.05
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