📚 Year 11 Eduqas Further Mathematics: Case Study Practical Exercises | Year 11 Eduqas 进阶数学：案例分析实战演练
Real-world applications turn mathematical techniques into powerful tools. This article presents twelve case study exercises drawn from the Year 11 Eduqas Further Mathematics syllabus, each designed to sharpen your problem-solving skills in algebra, calculus, trigonometry, vectors, probability and modelling. Work through each example methodically—translating a situation into equations, selecting the appropriate method, and interpreting the results—to build both confidence and fluency.
真实世界的应用能让数学方法变成强有力的工具。本文提供十二个基于 Year 11 Eduqas 进阶数学大纲的案例分析练习，每个都旨在提升你在代数、微积分、三角学、向量、概率和建模方面的解题能力。逐一练习，将情境转化为方程，选择合适的方法，并解释结果，从而建立信心与熟练度。
1. Optimising the Area of a Rectangular Pen | 矩形围栏面积最优化
A farmer has 40 metres of fencing and plans to build a rectangular pen against a long barn wall. The wall forms one side of the pen, so only three sides need fencing. We need to find the dimensions that give the maximum possible area.
一位农民有 40 米围栏，计划靠着一面长长的谷仓墙建一个矩形围栏。墙壁作为围栏的一边，因此只需要围三边。我们要找出使面积最大化的尺寸。
nstraint gives 2x + y = 40, so y = 40 − 2x. The area is A = x y = x(40 − 2x) = −2x² + 40x.
设围栏的宽度为 x 米，平行于谷仓的长度为 y 米。围栏的约束条件为 2x + y = 40，因此 y = 40 − 2x。面积 A = x y = x(40 − 2x) = −2x² + 40x。
0 m × 20 m = 200 m². The second derivative d²A/dx² = −4 is negative, confirming a maximum.
为了求最大值，我们可以使用微分：dA/dx = −4x + 40。令其等于零得到 −4x + 40 = 0 → x = 10。于是 y = 40 − 20 = 20，最大面积为 10 m × 20 m = 200 m²。二阶导数 d²A/dx² = −4 为负，确认为最大值。
attains 200 when x = 10. This approach reinforces quadratic optimisation without calculus.
配方提供了代数上的替代方法：A = −2(x² − 20x) = −2[(x − 10)² − 100] = 200 − 2(x − 10)²。面积绝不会超过 200，并且在 x = 10 时达到 200。此法在不使用微积分的情况下巩固了二次函数最优化的概念。
2. Modelling a Projectile's Height | 抛射物高度建模
s modelled by h = 19.6t − 4.9t². Find the maximum height reached and the time at which it occurs.
一个球以 19.6 m/s 的初速度从地面竖直向上抛出。忽略空气阻力，t 秒后它的高度 h 米由模型 h = 19.6t − 4.9t² 给出。求到达的最大高度及对应的时间。
→ 19.6 − 9.8t = 0 → t = 2 seconds. Substitute t = 2 into h: h = 19.6×2 − 4.9×4 = 39.2 − 19.6 = 19.6 metres.
该方程关于 t 是二次的，且 t² 的系数为负，因此抛物线有最大值。求导：dh/dt = 19.6 − 9.8t。令 dh/dt = 0 → t = 2 秒。将 t = 2 代入 h：h = 19.6×2 − 4.9×4 = 39.2 − 19.6 = 19.6 米。
2. The model confirms the symmetry of projectile motion under constant gravitational acceleration g = 9.8 m/s², where maximum height = u²/(2g).
我们也可以通过配方确定顶点：h = −4.9(t² − 4t) = −4.9[(t − 2)² − 4] = 19.6 − 4.9(t − 2)²。显然最大值为 19.6，发生在 t = 2 时。该模型印证了在恒定重力加速度 g = 9.8 m/s² 下抛射运动的对称性，最大高度 = u²/(2g)。
3. Solving a System of Non-linear Equations in Business | 商业场景中的非线性方程组求解
units of a product is R = 10x, while the total cost is C = x² + 2x + 8. All amounts are in £ puts (where revenue equals cost) and the range of x for which the business makes a profit.
某小公司销售 x 件产品的总收入为 R = 10x，总成本为 C = x² + 2x + 8，单位均为英镑。求盈亏平衡产量（收入等于成本时）以及公司能够盈利的 x 范围。
Set R = C: 10x = x² + 2x + 8 → 0 = x² − 8x + 8. Using the quadratic formula, x = [8 ± √64 − 32]/2 = [8 ± √32]/2 = [8 ± 4√2]/2 = 4 ± 2√2. Thus the break-even points are at x = 4 − 2√2 ≈ 1.17 and x = 4 + 2√2 ≈ 6.83.
令 R = C：10x = x² + 2x + 8 → 0 = x² − 8x + 8。使用二次公式，x = [8 ± √64 − 32]/2 = [8 ± √32]/2 = [8 ± 4√2]/2 = 4 ± 2√2。因此盈亏平衡点分别在 x = 4 − 2√2 ≈ 1.17 和 x = 4 + 2√2 ≈ 6.83。
ative between the two roots. Profit P = R − C = −x² + 8x − 8, which is positive when 4 − 2√2 < x < 4 + 2√2. In practical terms, producing and selling between 2 and 6 units gives a positive profit, but the maximum number is limited to near 6.83.
因为抛物线 x² − 8x + 8 开口向上，它在两根之间为负值。利润 P = R − C = −x² + 8x − 8，当 4 − 2√2 < x < 4 + 2√2 时为正。实际上，生产和销售 2 到 6 个单位会带来正利润，但最大盈利数量被限制在约 6.83 以内。
4. Exponential Growth in a Bacteria Culture | 细菌培养的指数增长
ry hour. Assuming ideal conditions, how long will it take for the population to reach 5000? Model the population P at time t hours by P = 100 × 2t.
一个实验室培养物最初含有 100 个细菌，并且每小时数量翻倍。假设为理想条件，需要多长时间种群才能达到 5000？用 P = 100 × 2t 对 t 小时后种群建模。
Set 100 × 2t = 5000 → 2t about 5 hours 38 minutes). You can also write log₂ 50 = log₂(2×25) = 1 + log₂ 25; since 2⁴=16 and 2⁵=32, 25 lies between, so t ≈ 1 + 4.64 = 5.64.
令 100 × 2t = 5000 → 2t = 50。取对数（任意底）得 t = log₂ 50。使用自然对数：t = ln 50 / ln 2。用计算器，ln 50 ≈ 3.9120，ln 2 ≈ 0.6931，因此 t ≈ 5.64 小时（约 5 小时 38 分钟）。你也可以写成 log₂ 50 = log₂(2×25) = 1 + log₂ 25；由于 2⁴=16，2⁵=32，25 介于之间，所以 t ≈ 1 + 4.64 = 5.64。
Exponential growth problems often require solving ax rithmic form. In Further Mathematics, you should be comfortable switching between index and log equations and using the change-of-base formula.
指数增长问题通常需要解 ax = b。关键步骤是转换为对数形式。在进阶数学中，你应该能熟练地在指数方程和对数方程之间转换，并使用换底公式。
5. Trigonometric Height of a Ferris Wheel | 摩天轮高度的三角学
ime in seconds after boarding at the lowest point. Find the maximum and minimum heights, the time for one full revolution, and the height after 10 seconds.
摩天轮上一个客舱的高度 h（米）由 h(t) = 30 sin(π t / 15) + 35 给出，其中 t 是自最低点登舱以来经过的秒数。求最大和最小高度、完成一整圈所需的时间，以及 10 秒后的高度。
he period of the sine function: period = 2π / (π/15) = 30 seconds. After 10 seconds, h(10) = 30 sin(π×10/15) + 35 = 30 sin(2π/3) + 35 = 30×(√3/2) + 35 ≈ 25.98 + 35 = 60.98 m.
正弦函数在 −1 到 1 之间振荡，因此 h 的最大值为 30×1 + 35 = 65 m，最小值为 30×(−1) + 35 = 5 m。一整圈对应正弦函数的周期：周期 = 2π / (π/15) = 30 秒。10 秒后，h(10) = 30 sin(π×10/15) + 35 = 30 sin(2π/3) + 35 = 30×(√3/2) + 35 ≈ 25.98 + 35 = 60.98 m。
ertical shift and period is essential for trigonometric modelling. Always check the domain—here t ≥ 0— ontext, such as determining when the capsule reaches a specific height for the first time.
理解振幅、垂直位移和周期对于三角学建模至关重要。务必检查定义域——这里 t ≥ 0——并准备好在上下文中解释数值，例如确定客舱何时首次达到特定高度。
6. Using Velocity-Time Graphs for Skydiving | 跳伞运动的速度-时间图
locity remains constant at 78.4 m/s for another 12 seconds before the parachute opens. Sketch the velocity-time graph and find the total distance fallen in these 20 seconds.
跳伞者从静止跳下。前 8 秒他们以 9.8 m/s² 的加速度匀加速，直至达到终极速度。从 t = 8 s 起，速度保持在 78.4 m/s 不变，又经过 12 秒才打开降落伞。画出速度-时间图，并求这 20 秒内总下落距离。
lacement. For the acceleration phase, the area is a triangle: ½ × base × height = ½ × 8 × 78.4 = 313.6 m. For the constant-velocity phase, the area is a rectangle: 12 × 78.4 = 940.8 m. Total distance = 313.6 + 940.8 = 1254.4 m.
该图由一条从 (0,0) 到 (8, 78.4) 的直线段，然后从 (8, 78.4) 到 (20, 78.4) 的水平线段组成。v-t 图下的面积表示位移。加速阶段，面积是一个三角形：½ × 底 × 高 = ½ × 8 × 78.4 = 313.6 m。匀速阶段，面积是一个矩形：12 × 78.4 = 940.8 m。总距离 = 313.6 + 940.8 = 1254.4 m。
slope of the first segment to confirm the acceleration (gradient = 9.8). Area problems in kinematics often split motion into simple geometric shapes.
检查单位：加速度 9.8 m/s² 持续 8 s 得到末速度 9.8×8 = 78.4 m/s，一致。画图时，明确标注坐标轴，并用第一段的斜率来确认加速度（梯度 = 9.8）。运动学中的面积问题常将运动分割为简单的几何形状。
7. Inequalities in Resource Allocation | 资源分配中的不等式
A workshop produces two types of trophy, A and B. Each trophy A requires 2 hours of machining and 1 hour of finishing. Each trophy
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