📚 Community Garden Design: A Case Study in Applied Mathematics | 社区花园设计：应用数学案例实战
Mathematics is not just about solving equations in a classroom; it is a powerful tool for planning and designing real-world projects. In this case study, we will apply a range of mathematical skills – from measurement and algebra to statistics and trigonometry – to design a community garden for our school. Each section presents a practical problem and guides you through the calculations, encouraging you to think like a mathematician and a project planner.
数学不仅仅是在课堂上解方程，它还是规划和设计现实项目的强大工具。在本案例研究中，我们将运用多种数学技能——从测量和代数到统计和三角学——来为我们学校设计一个社区花园。每一节都展示一个实际问题并引导你完成计算，鼓励你像数学家和项目规划者一样思考。
1. Site Measurement and Scale Drawing | 场地测量与比例尺绘图
First, we measure the available plot of land behind the school. Using a 30-metre tape, we find the length is 18.5 m and the width is 12.2 m. To create a clear plan, we convert these real dimensions to a scale drawing on A4 paper, choosing a scale of 1 : 100. This means 1 cm on the plan represents 100 cm (or 1 m) in reality.
首先，我们测量学校后面可用的地块。使用一把30米卷尺，测得长度为18.5 m，宽度为12.2 m。为了画出清晰的平面图，我们将实际尺寸转换为A4纸上的比例图，选择比例尺1 : 100。这意味着图上1 cm代表实际的100 cm（即1 m）。
The scaled length becomes 18.5 cm and the width 12.2 cm. We draw a rectangle with these dimensions, ensuring all corners are right angles. The drawing includes a reference arrow and a scale bar. This visual representation helps us plan the positioning of paths, flower beds and seating areas without needing to be on site all the time.
按比例缩放后的长度是18.5 cm，宽度是12.2 cm。我们画出这个尺寸的矩形，并确保所有角都是直角。图稿上包含指北箭头和比例尺。这种可视化展示帮助我们在不需要始终在现场的情况下规划路径、花坛和休息区的位置。
2. Calculating Areas of Garden Features | 计算花园各个区域的面积
The garden design splits the rectangular plot into three main zones: a central rectangular lawn, a semicircular flower bed against one wall, and a right-angled triangular herb corner. The lawn measures 8 m by 6 m, so its area is: Alawn = 8 × 6 = 48 m².
花园设计将矩形地块分成三个主要区域：中央的长方形草坪、靠墙的一个半圆形花坛，以及角落的一个直角三角形香草园。草坪尺寸为8 m × 6 m，因此它的面积为：Alawn = 8 × 6 = 48 m²。
The semicircular flower bed has a diameter of 3 m, giving a radius of 1.5 m. The area of a full circle is πr², so the semicircle area is: Asemi = ½πr² = ½ × π × (1.5)² ≈ 3.53 m² (using π ≈ 3.14).
半圆形花坛的直径为3 m，半径为1.5 m。整个圆的面积是πr²，因此半圆面积为：Asemi = ½πr² = ½ × π × (1.5)² ≈ 3.53 m²（使用π ≈ 3.14）。
The triangular herb garden is a right-angled triangle with legs of 2 m and 2.5 m. Its area is: Atri = ½ × base × height = ½ × 2 × 2.5 = 2.5 m². The total cultivated area becomes 48 + 3.53 + 2.5 = 54.03 m².
三角形香草园是一个两直角边分别为2 m和2.5 m的直角三角形。它的面积为：Atri = ½ × 底 × 高 = ½ × 2 × 2.5 = 2.5 m²。总种植面积为48 + 3.53 + 2.5 = 54.03 m²。
3. Perimeter and Fencing Costs | 周长与围栏成本
The school decides to fence the entire garden perimeter using wooden panels. The garden is rectangular, so the perimeter is: P = 2 × (18.5 + 12.2) = 2 × 30.7 = 61.4 m. We must subtract the 2-metre wide gate opening where no fencing is needed.
学校决定用木制围栏将整个花园围起来。花园是矩形的，因此周长为：P = 2 × (18.5 + 12.2) = 2 × 30.7 = 61.4 m。我们需要减去2米宽的大门开口，那里不需要围栏。
Effective fencing length = 61.4 - 2 = 59.4 m. Each fence panel covers 1.8 m and costs £ panels required is 59.4 ÷ 1.8 = 33 panels (rounding up). Total fencing cost = 33 × 24.50 = £808.50.
有效围栏长度 = 61.4 - 2 = 59.4 m。每块围栏板覆盖1.8 m，单价24.50英镑。所需围栏板数量为59.4 ÷ 1.8 = 33块（向上取整）。围栏总成本 = 33 × 24.50 = 808.50英镑。
4. Volume of Soil for Raised Beds | 抬高的花坛填土体积
e cylindrical with a diameter of 1.2 m and a depth of 0.6 m. The volume of a rectangular prism is Vrect = l × w × h = 2 × 1 × 0.5 = 1 m³.
设计了两个抬高的种植箱：一个是尺寸为2 m × 1 m × 0.5 m（长×宽×高）的长方体，另一个是直径为1.2 m、深度为0.6 m的圆柱体。长方体体积为 Vrect = l × w × h = 2 × 1 × 0.5 = 1 m³。
For the cylindrical planter, the radius is 0.6 m and the volume is: Vcyl = πr²h = π × (0.6)² × 0.6 = π × 0.36 × 0.6 ≈ 0.678 m³. Combined volume of soil needed is 1 + 0.678 = 1.678 m³.
对于圆柱形种植箱，半径为0.6 m，体积为：Vcyl = πr²h = π × (0.6)² × 0.6 = π × 0.36 × 0.6 ≈ 0.678 m³。所需土壤总体积为1 + 0.678 = 1.678 m³。
Topsoil is sold in 0.75 m³ bulk bags, each costing £55. Number of bags = 1.678 ÷ 0.75 ≈ 2.24, so we must purchase 3 bags. Cost = 3 × 55 = £165.
表土以0.75 m³的散装袋出售，每袋55英镑。所需袋数 = 1.678 ÷ 0.75 ≈ 2.24，因此必须购买3袋。成本 = 3 × 55 = 165英镑。
5. Ramp Design: Introducing Trigonometry | 坡道设计：引入三角学
To make the garden accessible, we need a ramp from the gate up to a raised deck 0.45 m high. Safety guidelines state the slope must not exceed a 1:12 ratio (1 cm rise per 12 cm horizontal run). The minimum horizontal length (run) is: run = 12 × 0.45 = 5.4 m.
为了让花园无障碍通行，我们需要从大门口到高0.45 m的升高平台上建一个坡道。安全指南规定坡度不得超过1:12的比例（每升高1 cm需要12 cm的水平长度）。最小水平长度（底边）为：run = 12 × 0.45 = 5.4 m。
The ramp itself is the hypotenuse of a right triangle. Using Pythagoras' theorem: ramp length = √0.45² + 5.4² = √0.2025 + 29.16 = √29.3625 ≈ 5.42 m .4 = 0.0833. Therefore, θ = tan⁻¹(0.0833) ≈ 4.76°, which is a gentle and safe inclination.
坡道本身是直角三角形的斜边。使用勾股定理：坡道长度 = √0.45² + 5.4² = √0.2025 + 29.16 = √29.3625 ≈ 5.42 m。仰角θ可通过 tan θ = 对边/邻边 = 0.45/5.4 = 0.0833 求得。因此，θ = tan⁻¹(0.0833) ≈ 4.76°，这是一个缓和且安全的倾斜角度。
6. Watering System and Circle Area | 灌溉系统与圆面积
A rotary sprinkler is placed at the centre of the lawn. It sprays water in a full circle with a radius of 5 m. The area it covers is: Acircle = πr² = π × 5² ≈ 78.5 m². Since the lawn is only 48 m², the sprinkler covers it completely but wastes some water outside the garden's boundary.
一个旋转洒水器安装在草坪中心。它以5 m的半径喷洒一个完整的圆形。覆盖的面积为：Acircle = πr² = π × 5² ≈ 78.5 m²。由于草坪只有48 m²，洒水器完全覆盖它，但会有一部分水喷到花园边界外造成浪费。
To improve efficiency, we consider two smaller sprinklers each covering a quarter-circle from waist height onto a flower bed. Each has a radius of 1.2 m, so the area of one quarter-circle is: A_quarter = ¼π × (1.2)² ≈ 1.13 m². Together they water 2.26 m² without overspray.
为了提高效率，我们考虑使用两个较小的洒水器，每个从齐腰高度向花坛喷洒四分之一圆。每个洒水器的半径为1.2 m，因此一个四分之一圆的面积为：A_quarter = ¼π × (1.2)² ≈ 1.13 m²。两个合起来可浇灌2.26 m²且无过量喷溅。
7. Survey and Data Handling | 调查与数据处理
We survey 120 students to choose their favourite additional garden feature: a pond, a bench, a bird table, or a vegetable patch. The results are recorded in a frequency table.
我们调查了120名学生，让他们选择最喜欢的额外花园特色：池塘、长椅、鸟桌或蔬菜种植区。结果记录在频率表中。
| Feature | Frequency |
|---|---|
| Pond | 36 |
| Bench | 42 |
| Bird table | 24 |
| Vegetable patch | 18 |
The mode is the 'Bench' 60° = 126°, Bird table = (24/120) × 360° = 72°, Vegetable patch = (18/120) × 360° = 54°. The mean number of votes per feature is 120 ÷ 4 = 30, though this is not particularly useful here.
众数是“长椅”类别，出现频次最高为42。要绘制饼图，我们计算每个扇形的角度：池塘 = (36/120) × 360° = 108°，长椅 = (42/120) × 360° = 126°，鸟桌 = (24/120) × 360° = 72°，蔬菜种植区 = (18/120) × 360° = 54°。每个特色的平均票数为120 ÷ 4 = 30，不过此处用处不大。
8. Probability of Preference | 偏好的概率
If a student is selected at random from the surveyed group, the probability they chose 'Bench' is: P(Bench) = 42/120 = 7/20 = 0.35. The probability of not choosing 'Pond' is 1 - (36/120) = 84/120 = 0.7. The probability of choosing either 'Bird table' or 'Vegetable patch' is (24+18)/120 = 42/120 = 0.35 as well.
如果从被调查群体中随机选择一名学生，他/她选择“长椅”的概率为：P(长椅) = 42/120 = 7/20 = 0.35。不选择“池塘”的概率是 1 - (36/120) = 84/120 = 0.7。选择“鸟桌”或“蔬菜种植区”的概率是 (24+18)/120 = 42/120 = 0.35。
We can also calculate the experimental probability that a student' 20 = 0.5. These probabilities help the design team understand likely visitor satisfaction.
我们还可以计算某位学生的首选不是生物类特色（即池塘或鸟桌）的经验概率。那就是长椅和蔬菜种植区的总和：42 + 18 = 60，所以 P = 60/120 = 0.5。这些概率帮助设计团队了解可能的访客满意度。
9. Optimising a Flower Bed Area | 优化花坛面积
The gardening club has 12 metres of wooden edging to create a rectangular flower bed against an existing wall (so only three sides need edging). Let the width perpendicular to the wall be x metres. Then the length parallel to the wall is (12 - 2x) metres. The area A is: A = x(12 - 2x) = 12x - 2x².
园艺社有12米长的木制边框，用来沿着已有墙壁围建一个长方形花坛（因此只有三条边需要边框）。设垂直于墙壁的宽度为 x 米。则平行于墙壁的长度为 (12 - 2x) 米。面积 A 为：A = x(12 - 2x) = 12x - 2x²。
This is a quadratic expression in the form -2x² + 12x. To find the maximum area, we complete the square: first factor out -2: A = -2(x² - 6x) = -2[(x - 3)² - 9] = -2(x - 3)² + 18 6 m. So the optimal dimensions are 3 m by 6 m, giving the largest possible area of 18 m².
这是一个形式为 -2x² + 12x 的二次式。要找到最大面积，我们进行配方：首先提取 -2：A = -2(x² - 6x) = -2[(x - 3)² - 9] = -2(x - 3)² + 18。A 的最大值为18 m²，出现在 x - 3 = 0 即 x = 3 m 时。相应的长度为 12 - 2×3 = 6 m。因此最优尺寸为3 m × 6 m，可得到最大面积18 m²。
10. Financial Planning and Fundraising | 财务规划与筹款
The total estimated cost of the garden project includes fencing (£808.50), soil (£165), paving slabs for paths (£230), the ramp materials (£95), and plants/seeds (£120). Summing these: £808.50 + £165 + £230 + £95 + £120 = £1,418.50. The school plans to fund 60% of this from its budget and raise the rest through a bake sale.
花园项目的估计总成本包括围栏（808.50英镑）、土壤（165英镑）、铺路石板（230英镑）、坡道材料（95英镑）以及植物/种子（120英镑）。总和为：808.50 + 165 + 230 + 95 + 120 = 1,418.50英镑。学校计划从其预算中支付其中的60%，剩余通过烘焙义卖筹集。
School contribution = 0.60 × 1418.50 = £851.10. Amount to raise = 1418.50 - 851.10 = £567.40. If each bake sale item sells for £1.50 and makes a profit of £0.90, the number of items needed is: 567.40 ÷ 0.90 ≈ 631 items. This project shows how mathematical reasoning drives every stage of a real-life undertaking.
学校出资部分 = 0.60 × 1418.50 = 851.10英镑。需筹集金额 = 1418.50 - 851.10 = 567.40英镑。如果每件烘焙甜品售价为1.50英镑，利润为0.90英镑，则所需售出的物品数量为：567.40 ÷ 0.90 ≈ 631件。该项目展示了数学推理如何驱动现实项目中的每一个阶段。
Published by TutorHao | SQA Year 9 Advanced Mathematics Revision Series | aleveler.com
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