📚 Year 10 SQA Maths: Case Study Analysis in Practice | Year 10 SQA 数学：案例分析实战演练
Case studies in SQA Mathematics are designed to mirror real-life problems, blending algebra, geometry, statistics and numerical reasoning into one extended task. This step-by-step workout walks you through a realistic school garden project, showing you how to build mathematical models, maximise area, work with percentages and handle sensitivity analysis – all essential skills for your assessments.
SQA 数学中的案例分析旨在模拟现实问题，将代数、几何、统计和数字推理融于一个综合性任务中。这份分步演练将带你完成一个真实的校园花园项目，展示如何建立数学模型、最大化面积、处理百分比以及进行敏感性分析——这些都是你评估中必不可少的能力。
1. The Scenario: School Garden Project | 案例情景：学校花园项目
Eco Club plans to build a rectangular garden against a long school wall. The garden will be fenced on the three remaining sides using exactly 30 metres of fencing. A diagonal gravel path will be cut across the garden, and the remaining soil area will be planted with three types of flowers according to a student survey. Your job is to design the garden dimensions, allocate planting zones and cost all materials while staying within budget.
生态俱乐部计划沿着学校的一面长墙建造一个长方形花园。花园用30米长的篱笆围起其余三边。一条沿对角线的碎石小径将穿过花园，剩余的土壤区域将根据学生调查种植三种花卉。你的任务是设计花园尺寸、分配种植区域并计算所有材料费用，同时不超出预算。
2. Setting Up Variables and Constraints | 设定变量与约束条件
be positive, we require 0 < x < 15. The garden area A (in m²) is given by A = x (30 − 2x).
设垂直于墙的边长为 x 米。则平行于墙的边长为 (30 − 2x) 米。由于长度必须为正，因此 0 < x < 15。花园面积 A (单位 m²) 由 A = x (30 − 2x) 给出。
3. Maximising the Garden Area | 面积最大化
−2(x² − 15x). Complete the square inside the bracket: x² − 15x = (x − 7.5)² − 56.25. Therefore,
A = −2(x − 7.5)² + 112.5
e, the parallel side is 30 − 2×7.5 = 15 m. The optimal garden is 15 m long and 7.5 m wide.
为求最大面积，我们采用配方法或使用二次函数的顶点公式。展开得 A = −2x² + 30x。提取 −2：A = −2(x² − 15x)。括号内配方：x² − 15x = (x − 7.5)² − 56.25。因此，最大面积 112.5 m² 出现在 x = 7.5 时。此时平行边长度为 30 − 2×7.5 = 15 m。最优花园长 15 米，宽 7.5 米。
4. Diagonal Path: Length Calculation | 对角线小径：长度计算
The garden will feature a straight gravel path along one diagonal. Using Pythagoras’ theorem, the length L of the diagonal is:
L = √7.5² + 15² = √56.25 + 225 = √281.25 ≈ 16.77 m
This result will be rounded to two decimal places in subsequent calculations.
花园将铺设一条沿对角线的笔直碎石小径。根据勾股定理，对角线长度 L 为：L = √7.5² + 15² = √56.25 + 225 = √281.25 ≈ 16.77 m。在后续计算中该结果保留两位小数。
5. Path Area and Net Planting Space | 小径面积与净种植空间
The path will have a constant width of 0.5 metres. Its area Apath 0.5 = 8.385 m² (using 8.39 m² when rounded to two decimal places). The net planting area Aplant is the total garden area minus the path area:
Aplant = 112.5 − 8.39 = 104.11 m²
This is the area that will be filled with soil and flowers.
小径将保持 0.5 米的固定宽度。因此其面积 A小径 ≈ 16.77 × 0.5 = 8.385 m²（四舍五入取 8.39 m²）。净种植面积 A种植 为花园总面积减去小径面积：A种植 = 112.5 − 8.39 = 104.11 m²。这部分区域将填土并种花。
6. Survey Data on Flower Preferences | 花卉偏好的调查数据
A survey of 200 students gave the following preferences for three flower types. We can convert these into actual numbers and angles for a pie chart.
| Flower | Percentage | Number of students | Pie chart angle |
|---|---|---|---|
| Roses | 40% | 80 | 144° |
| Lavender | 35% | 70 | 126° |
| Sunflowers | 25% | 50 | 90° |
These proportions will guide how the planting area is split among the three varieties.
对 200 名学生的调查得出对三种花卉的偏好如下。我们可将其转化为实际人数和饼图角度。这些比例将指导种植面积如何分配给三种花卉。
7. Allocating Planting Areas by Proportion | 按比例分配种植面积
Using the net planting area of 104.11 m², we apply the survey percentages:
- Roses: 40% of 104.11 = 0.40 × 104.11 ≈ 41.64 m²
- Lavender: 35% of 104.11 = 0.35 × 104.11 ≈ 36.44 m²
- Sunflowers: 25% of 104.11 = 0.25 × 104.11 ≈ 26.03 m²
Check total: 41.64 + 36.44 + 26.03 = 104.11 m².
利用净种植面积 104.11 m²，我们按调查百分比分配：玫瑰 40% → 约 41.64 m²，薰衣草 35% → 约 36.44 m²，向日葵 25% → 约 26.03 m²。合计为 104.11 m²。
8. Soil Volume Calculation | 土壤体积计算
eds will be filled with nutrient-rich topsoil to a depth of 0.4 m. Total soil volume V = Aplant er type: roses need 41.64 × 0.4 = 16.656 m³, lavender 14.576 m³, and sunflowers 10.412 m³.
种植区将铺上 0.4 米深的营养表土。总体积 V = 净种植面积 × 深度 = 104.11 × 0.4 = 41.644 m³。如需按花卉划分：玫瑰需土 16.656 m³，薰衣草 14.576 m³，向日葵 10.412 m³。
9. Cost Analysis and Budgeting | 成本分析与预算
Let us estimate the total project cost. The topsoil supplier charges £12 per cubic metre. Soil cost = 41.644 × 12 ≈ £499.73. The gravel path is priced at £ re, which includes the required width and depth of gravel base. Path cost = 16.77 × 3.50 ≈ £58.70. Assuming the fencing is already funded, the preliminary total is £499.73 + £58.70 = £ ional costs for plants or edging can be added later but are excluded here for simplicity.)
我们来估算项目总成本。表土供应商计价为每立方米 12 英镑。土壤费用 = 41.644 × 12 ≈ £499.73。碎石小径按每延米 £3.50 计价，已包含所需的宽度和基础深度。小径费用 = 16.77 × 3.50 ≈ £58.70。假设篱笆已购置，初步总费用为 £499.73 + £58.70 = £558.43。（植物或边饰等额外成本可后续添加，此处为简化不计。）
10. Sensitivity Analysis: What If the Fencing Budget Increases? | 敏感性分析：若篱笆预算增加？
Suppose the Eco Club receives extra funding, allowing 35 metres of fencing instead of 30. Then the parallel side becomes 35 − 2x, and area Anew = x(35 − 2x) = −2x² + 35x. Completing the square:
Anew = −2(x − 8.75)² + 153.125
now 153.125 m² when x = 8.75 m, with a parallel side of 17.5 m. The new diagonal length = √8.75² + 17.5² ases to roughly 9.78 m², leaving a net planting area of about 143.34 m². Soil volume becomes 57.34 m³, costing about £688, and the path cost rises to about £ lly, but the project team can see that the extra fencing dramatically increases the garden's productive space.
假设生态俱乐部获得额外拨款，可购买 35 米篱笆而非 30 米。此时平行边变为 35 − 2x，面积 A新 = x(35 − 2x) = −2x² + 35x。配方得：A新 = −2(x − 8.75)² + 153.125。最大面积变为 153.125 m²，此时 x = 8.75 m，平行边 17.5 m。新对角线长度 ≈ 19.57 m，小径面积增加到约 9.78 m²，净种植面积约 143.34 m²。土方量变为 57.34 m³，成本约 £688，小径成本升至约 £68.50。总费用明显增加，但项目团队可以看到额外篱笆显著扩大了花园的生产空间。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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