📚 Year 9 CAIE Further Mathematics: Case Study Practical Exercises | Year 9 CAIE 进阶数学：案例分析实战演练
In further mathematics, real‑world problems are rarely presented as neat textbook exercises. This case study takes you through the entire process of tackling a multi‑step scenario that blends algebra, geometry, probability and data analysis – just as you would in a genuine CAIE‑style investigation. You will learn how to set up models, solve equations, work with inequalities, find maxima, calculate expected values and interpret results.
在进阶数学里，真实世界的问题很少会像教科书习题那样规整。这个案例研究会带你走完处理一个多步骤情景的完整过程，把代数、几何、概率和数据分析融合到一起——就像你在真正的 CAIE 式探究中会做的那样。你将学会如何建立模型、解方程、处理不等式、寻找最大值、计算期望值并解释结果。
1. Case Introduction and Problem Statement | 案例介绍与问题定义
A group of Year 9 students is planning a fundraising stall at the school fair. They will set up a safe playing area enclosed by rope and a dartboard game. The stall needs to maximise profit while respecting the budget and space limitations. The main question is: what dimensions for the enclosure and what game pricing will yield at least a £100 profit after covering all costs?
一群九年级学生正在为学校游园会策划一个筹款摊位。他们将用绳子围出一个安全活动区，并安排一个飞镖游戏。摊位需要在尊重预算和空间限制的同时使利润最大化。核心问题是：围栏的哪种尺寸以及游戏如何定价，才能在覆盖所有成本后，获得至少 100 英镑的利润？
2. Setting Up Mathematical Models | 建立数学模型
the two sides perpendicular to the wall have length x metres. Then the side parallel to the wall is (20 − 2x) metres. The area A (in m²) available for customers is given by A = x(20 − 2x).
围栏利用学校建筑的直墙作为一边，因此只有三边需要绳子。团队刚好购买了 20 米绳子。设垂直墙的两边长度均为 x 米，那么平行于墙的边为 (20 − 2x) 米。可用于接待顾客的面积 A（单位：平方米）由 A = x(20 − 2x) 给出。
The game has a fixed participation fee of £2 per player. The dartboard has three concentric prize zones. Historical data from similar events suggests the probability of hitting each zone can be modelled by the relative area of that zone on the board. The cost of prizes per player is a random variable; its expected value must be calculated.
游戏对每位参与者收取 2 英镑的固定费用。飞镖盘有三个同心的奖品区域。从类似活动的历史数据来看，命中每一个区域的概率可以用该区域在盘面上的相对面积来建模。每名玩家的奖品成本是一个随机变量；必须计算它的期望值。
The total profit P depends on the number of players n, the fixed rope cost of £15, the unit cost of rope per metre (£0.50, already bought) and the expected prize cost per player. The model must obey the constraint that the enclosure area is at least 30 m² for safety.
总利润 P 取决于玩家人数 n、固定的绳子成本 15 英镑、每米绳子单价（0.50 英镑，已购买）和每名玩家的预期奖品成本。模型必须遵守安全要求：围栏面积至少为 30 平方米。
3. Solving Equations | 解方程
20x − 30 = 0. Divide through by −2: x² − 10x + 15 = 0. Use the quadratic formula x = [10 ± √100 − 60] / 2 = [10 ± √40] / 2 = 5 ± √10. Thus x ≈ 5 ± 3.16, giving x ≈ 8.16 or x ≈ 1.84 metres.
要求出恰好得到 30 平方米的尺寸，解方程 x(20 − 2x) = 30。展开得 −2x² + 20x = 30，于是 −2x² + 20x − 30 = 0。两边除以 −2：x² − 10x + 15 = 0。使用求根公式 x = [10 ± √100 − 60] / 2 = [10 ± √40] / 2 = 5 ± √10。因此 x ≈ 5 ± 3.16，得到 x ≈ 8.16 或 x ≈ 1.84 米。
Both solutions must also satisfy the physical constraint that 20 − 2x > 0, so x < 10. They are both valid. The equation tells us that if x is between these two roots, the area is greater than 30 m².
这两个解还必须满足物理约束 20 − 2x > 0，即 x < 10。它们都是有效的。从方程可知，若 x 在这两根之间，面积便大于 30 平方米。
4. Working with Inequalities | 不等式分析
opening upward, so it is ≤ 0 between its roots. Therefore the acceptable range of x is 5 − √10 ≤ x ≤ 5 + √10, approximately [1.84, 8.16] metres.
安全规则要求 A ≥ 30，即 −2x² + 20x ≥ 30，化简为 x² − 10x + 15 ≤ 0。该二次函数图像是开口向上的抛物线，因此在两根之间其值 ≤ 0。所以 x 的可接受范围为 5 − √10 ≤ x ≤ 5 + √10，约 [1.84, 8.16] 米。
Additionally, x must be positive and less than 10. The intersection of these conditions is just the interval already found. The team can therefore choose any x in [1.84, 8.16] and still meet the minimum area requirement.
此外，x 必须为正且小于 10。这些条件的交集正是刚才求出的区间。因此团队可在 [1.84, 8.16] 中选择任意 x，仍能满足最小面积要求。
5. Maximising Revenue: Quadratic Functions | 最大化收入：二次函数
n for 5 hours. The number of players n = 5 × 2A = 10A. Substituting A(x) = −2x² + 20x gives n(x) = 10(−2x² + 20x) = −20x² + 200x.
假设每小时玩家人数与围栏面积成正比，比例常数为每平方米每小时 2 名玩家。摊位将运营 5 小时。玩家人数 n = 5 × 2A = 10A。代入 A(x) = −2x² + 20x 得 n(x) = 10(−2x² + 20x) = −20x² + 200x。
The total revenue R from entry fees is £ −b/(2a) = −400/(2 × (−40)) = 5 metres. At x = 5, A = 50 m², n = 500 players, and revenue = £1000. So the maximum possible entry-fee revenue is £1000.
来自入场费的总收入 R 为 2 英镑 × n = 2(−20x² + 200x) = −40x² + 400x。这是一个二次函数，其最大值在 x = −b/(2a) = −400/(2 × (−40)) = 5 米处取得。当 x = 5 时，A = 50 平方米，n = 500 名玩家，收入 = 1000 英镑。所以入场费收入的最大可能值为 1000 英镑。
s not yet account for costs; it simply shows the best x for revenue under the proportional‑ ion. The team can use x = 5 to maximise revenue, which also satisfies the area inequality.
注意：这还没有计入成本；它只是表明在玩家比例假设下，获得最大收入的最佳 x 值。团队可选择 x = 5 使收入最大化，且这个值也满足面积不等式。
6. Probability Analysis of the Game | 游戏概率分析
The dartboard is circular with a total radius of 15 cm. It is divided into three concentric zones: the bullseye (0–5 cm), the middle ring (5–10 cm), and the outer ring (10–15 cm). The prize values are £5, £2 and £0.50 respectively. Assume a dart hits the board uniformly at random over the entire area.
飞镖盘是半径为 15 厘米的圆形，分为三个同心区域：靶心（0–5 厘米）、中环（5–10 厘米）和外环（10–15 厘米）。奖品价值分别为 5 英镑、2 英镑和 0.50 英镑。假设飞镖均匀地随机落在整个盘面区域。
The probability of hitting a zone is the ratio of its area to the total area.
Total area = π × 15² = 225π.
Bullseye area = π × 5² = 25π → probability = 25π / 225π = 1/9.
Middle ring area = π(10² − 5²) = 75π → probability = 75π / 225π = 1/3.
Outer ring area = π(15² − 10²) = 125π → probability = 125π / 225π = 5/9.
Expected prize cost per player E(C) = £5×(1/9) + £2×(1/3) + £0.50×(5/9) = £(5/9 + 2/3 + 2.5/9). Convert to ninths: 5/9 + 6/9 + 2.5/9 = 13.5/9 = £1.50.
命中一个区域的概率等于该区域面积与总面积之比。
总面积 = π × 15² = 225π。
靶心面积 = π × 5² = 25π → 概率 = 25π / 225π = 1/9。
中环面积 = π(10² − 5²) = 75π → 概率 = 75π / 225π = 1/3。
外环面积 = π(15² − 10²) = 125π → 概率 = 125π / 225π = 5/9。
每名玩家的预期奖品成本 E(C) = 5×(1/9) + 2×(1/3) + 0.50×(5/9) = (5/9 + 2/3 + 2.5/9) 英镑。化为九分之几：5/9 + 6/9 + 2.5/9 = 13.5/9 = 1.50 英镑。
7. Geometry of the Booth Design | 摊位设计几何
re dimensions are 5 m by (20 − 10) = 10 m, giving area 50 m². The dartboard area occupies only 2.25 m², leaving plenty of room for queues and safety. From a geometry perspective, this layout is feasible.
围栏内还必须容纳飞镖盘装置。飞镖盘安装在一个支架上，需要占地 1.5 米 × 1.5 米。当 x = 5 米时，围栏尺寸为 5 米 × (20 − 10) = 10 米，面积 50 平方米。飞镖盘区域仅占 2.25 平方米，为排队和安全留出了足够空间。从几何角度看，这个布局是可行的。
2 m), while x = 8.16 gives a wide but shallow shape (8.16 m by 3.68 m). Both satisfy area but may affect customer flow. The maximum‑revenue choice x = 5 offers a balanced rectangular shape.
若团队选择不同的 x，平行于墙的边便会变化。极端情况 x = 1.84 产生一个很窄但很长的围栏（1.84 米 × 16.32 米），而 x = 8.16 则是一个宽而浅的形状（8.16 米 × 3.68 米）。两者都满足面积要求，但可能影响顾客流动。最大化收入的选择 x = 5 提供了一个均衡的长方形形状。
8. Data Analysis and Interpretation | 数据分析与解释
yers with a standard deviation of 40. Using the model, the expected total revenue is 480 × £2 = £960. The expected total prize cost is 480 × £1.50 = £720. Fixed costs include the rope (£15) and a stall decoration fee of £35, totalling £50. Expected profit = £960 − £720 − £50 = £190.
来自类似摊位的过往参与数据表明，玩家人数会有变化，但对于 50 平方米的摊位面积，平均有 480 名玩家，标准差为 40。根据模型，预期总收入为 480 × 2 英镑 = 960 英镑。预期总奖品成本为 480 × 1.50 英镑 = 720 英镑。固定成本包括绳子（15 英镑）和摊位装饰费 35 英镑，共 50 英镑。预期利润 = 960 − 720 − 50 = 190 英镑。
Let us find the break‑even number of players n*: £2 × n* = £1.50 × n* + £ iven the data, even one standard deviation below the mean (440 players) still gives profit £170, well above zero. The stall is almost certain to be profitable.
我们来找盈亏平衡的玩家人数 n*：2 英镑 × n* = 1.50 英镑 × n* + 50 英镑 → 0.5n* = 50 → n* = 100 名玩家。达到 100 名玩家时利润为零。任何超过 100 人的参与都会产生利润。从数据来看，哪怕比平均值低一个标准差（440 名玩家），仍然可带来 170 英镑的利润，远高于零。这个摊位几乎肯定能盈利。
The target profit is £ xceeded under normal attendance, so the stall design comfortably meets the fundraising goal.
目标利润是 100 英镑。解 0.5n − 50 = 100 → 0.5n = 150 → n = 300 名玩家。在正常参与水平下，这个目标很容易被超过，因此摊位设计能轻松达成筹款目标。
9. Summary and Reflective Thinking | 总结与反思
fundraising problem. We chose x = 5 m to maximise revenue, confirmed the area constraint was met, computed expected prize costs and used past data to verify profitability. The process mirrors the type of extended investigation expected in CAIE further mathematics.
本案例研究展示了如何结合二次函数、不等式、概率、几何和基础统计，来解决一个真实的筹款问题。我们选择 x = 5 米来最大化收入，确认满足面积约束，计算了预期奖品成本，并利用历史数据验证了盈利能力。这一过程反映了 CAIE 进阶数学所期待的拓展探究类型。
ic function by completing the square or using x = −b/(2a), calculating geometric probabilities from areas, and performing break‑ ld reflect on how changing the rope length or the prize structure would affect the outcome – a true mathematical modelling mindset.
练习的关键技能：从几何情景中建立二次模型，求解二次方程和不等式，通过配方法或利用 x = −b/(2a) 最大化二次函数，根据面积计算几何概率，以及利用线性方程进行盈亏平衡分析。学生应当反思：改变绳子长度或奖品结构将如何影响结果——这才是真正的数学建模思维。
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