📚 Case Study in Action: Practical Problem-Solving for OCR GCSE Further Maths | 案例分析实战演练：OCR GCSE 进阶数学解题实践
Case study questions in OCR GCSE Further Mathematics are designed to see how well you can link different topic areas — such as calculus, matrices, trigonometry and algebra — in one connected scenario. This article walks you through a complete project-style problem, with step‑by‑step reasoning, so you can practise exactly the skills the exam rewards.
OCR GCSE 进阶数学中的案例分析题旨在考查你是否能将微积分、矩阵、三角学与代数等不同专题串联运用到一个连贯的情景中。本文将通过一个完整的项目式问题，逐步推导，帮你精准训练考试所看重的解题能力。
1. Setting the Scene: The Golden Garden Project | 情景导入：金色花园项目
A local community decides to transform a strip of land along a straight river into a public garden. They have 100 metres of fencing to enclose a rectangular lawn against the river, so three sides need fencing. After completing the lawn, they will design a decorative emblem on the grass using matrix transformations and install a curved metal arch whose radius satisfies a cubic equation.
当地社区决定将河边一块狭长地带改造成公共花园。他们有 100 米围栏，准备沿河岸圈出一块靠墙的矩形草坪——因此只需围三边。草坪建成后，他们将利用矩阵变换在草地上设计一个装饰标志，并安装一个金属拱门，其半径需满足一个三次方程。
2. Part A: Maximising the Rectangular Lawn Area | 第一部分：最大化矩形草坪面积
Let the width of the lawn perpendicular to the river be x metres and the length parallel to the river be y metres. The fencing is used for two widths and one length, giving the constraint 2x + y = 100. The area A square metres is given by A = x y. Substitute y = 100 − 2x to express area as a quadratic function of x:
A(x) = x(100 − 2x) = 100x − 2x²
设草坪垂直于河岸的宽度为 x 米，平行于河岸的长度为 y 米。围栏用于两条宽和一条长，故约束条件为 2x + y = 100。面积 A（平方米）为 A = x y。代入 y = 100 − 2x，将面积表示为 x 的二次函数：
A(x) = x(100 − 2x) = 100x − 2x²
To find the value of x that maximises A, we differentiate with respect to x. The derivative is:
dA/dx = 100 − 4x
Setting dA/dx = 0 gives the stationary point:
100 − 4x = 0 ⇒ x = 25
When x = 25 m, the length y = 100 − 2(25) = 50 m, so the maximum area is 25 × 50 = 1250 m².
为了找到使 A 最大的 x 值，我们对 x 求导。导数为：
dA/dx = 100 − 4x
令 dA/dx = 0 得到驻点：
100 − 4x = 0 ⇒ x = 25
当 x = 25 m 时，长度 y = 100 − 2(25) = 50 m，因此最大面积为 25 × 50 = 1250 m²。
3. Part B: Verifying the Maximum with the Second Derivative | 第二部分：用二阶导数验证最大值
To confirm that the stationary point yields a maximum, we compute the second derivative. Differentiating dA/dx gives:
d²A/dx² = −4
Since d²A/dx² is negative for all x, the stationary point at x = 25 is a local (and global) maximum. The negative second derivative indicates the curve is concave down, so the area function has a peak at this critical value.
为确认该驻点对应最大值，我们计算二阶导数。对 dA/dx 求导得：
d²A/dx² = −4
由于 d²A/dx² 对所有 x 均为负值，故 x = 25 处的驻点是局部（也是全局）最大值。二阶导数为负表明曲线向下凹，面积函数在此临界值处达到峰值。
4. Part C: Matrix Transformation of a Unit Square | 第三部分：单位正方形的矩阵变换
The designers want a diamond‑shaped emblem on the lawn. They start with a unit square with vertices O(0,0), A(1,0), B(1,1) and C(0,1) and apply the transformation matrix
M = [ 2 1 ]
[ 1 3 ]
设计者想在草坪上绘制一个菱形标志。他们从一个单位正方形出发，顶点为 O(0,0)、A(1,0)、B(1,1) 和 C(0,1)，并施加变换矩阵
M = [ 2 1 ]
[ 1 3 ]
Multiply each vertex by M to obtain the image points. For example, OA’ = M × (1,0)ᵀ = (2,1). Computing all transformed coordinates:
- O'(0,0)
- A'(2,1)
- B'(2+1, 1+3) = (3,4)
- C'(1,3)
These four points form a parallelogram. The determinant of M is (2×3) − (1×1) = 5, which tells us the area scale factor of the transformation.
将每个顶点与 M 相乘得到像点。例如，OA’ = M × (1,0)ᵀ = (2,1)。计算全部变换坐标：
- O'(0,0)
- A'(2,1)
- B'(2+1, 1+3) = (3,4)
- C'(1,3)
这四个点构成一个平行四边形。M 的行列式为 (2×3) − (1×1) = 5，它告诉我们变换的面积缩放因子。
5. Part D: Area of the Transformed Emblem and the Determinant | 第四部分：变换后标志的面积与行列式
The original unit square has area 1 square unit. Under transformation by M, the area is multiplied by the absolute value of the determinant, so the emblem’s area is 5 square units. This matches the area of the parallelogram computed by vector cross product: base vector (2,1) and side vector (1,3) give area |2×3 − 1×1| = 5.
原始单位正方形的面积为 1 平方单位。经过 M 变换后，面积乘以行列式的绝对值，因此标志的面积为 5 平方单位。这与通过向量叉积计算得到的平行四边形面积一致：以 (2,1) 为底边向量、(1,3) 为侧边向量，面积 |2×3 − 1×1| = 5。
If the lawn is mapped using a scale model, knowing the determinant helps designers correctly size the emblem relative to the grass area.
如果草坪按比例模型绘制，了解行列式有助于设计者正确设定标志相对于草地的大小。
6. Part E: Solving a Cubic Equation for the Arch Radius | 第五部分：解三次方程求拱门半径
The garden arch is formed by bending a metal rod. Its curvature radius r (in metres) must satisfy the cubic equation
2r³ − 5r² − 4r + 3 = 0
To apply the factor theorem, we test integer factors of the constant term 3. Trying r = 1 gives 2(1)³ − 5(1)² − 4(1) + 3 = −4, not zero. Trying r = −1 gives −2 − 5 + 4 + 3 = 0, so r = −1 is a root. Since radius must be positive, we need to find other roots. Factorise (r + 1) out by polynomial long division:
(2r³ − 5r² − 4r + 3) ÷ (r + 1) = 2r² − 7r + 3
花园拱门由一根金属杆弯制而成。其曲率半径 r（米）需满足三次方程
2r³ − 5r² − 4r + 3 = 0
为使用因式定理，我们检验常数项 3 的整数因数。代入 r = 1 得 2(1)³ − 5(1)² − 4(1) + 3 = −4，不为零。代入 r = −1 得 −2 − 5 + 4 + 3 = 0，故 r = −1 是一个根。由于半径必须为正，我们需找出其他根。利用多项式长除法提取因式 (r + 1)：
(2r³ − 5r² − 4r + 3) ÷ (r + 1) = 2r² − 7r + 3
Now solve the quadratic 2r² − 7r + 3 = 0. Factorising: (2r − 1)(r − 3) = 0, giving r = ½ or r = 3. The feasible positive radii are 0.5 m and 3 m. The larger radius 3 m is chosen for the archway to give a grand appearance.
现求解二次方程 2r² − 7r + 3 = 0。因式分解得 (2r − 1)(r − 3) = 0，解得 r = ½ 或 r = 3。合理的正半径为 0.5 m 和 3 m。最终选择了较大的 3 m 作为拱门半径，以呈现宏伟效果。
7. Part F: Trigonometric Calculation for a Diagonal Support | 第六部分：对角支撑的三角计算
To strengthen the rectangular lawn’s corner posts, a diagonal cable is fixed from the corner of the lawn to the opposite top of a post. The lawn dimensions from Part A are width 25 m and length 50 m. The diagonal distance d across the rectangle is found using Pythagoras:
d = √(25² + 50²) = √(625 + 2500) = √3125 = 25√5 ≈ 55.9 m
为加固草坪角柱，要从草坪一角拉一条对角缆绳到对角柱顶。由第一部分知草坪宽 25 m、长 50 m。用力学家定理求对角线长 d：
d = √(25² + 50²) = √(625 + 2500) = √3125 = 25√5 ≈ 55.9 m
Now calculate the angle θ between the diagonal and the 50 m side. Using tangent:
tan θ = opposite/adjacent = 25/50 = 0.5 ⇒ θ = tan⁻¹(0.5) ≈ 26.6°
This angle is used to cut the cable ends appropriately so they fit flush with the posts.
现在计算对角线与 50 m 长边之间的夹角 θ。利用正切：
tan θ = 对边/邻边 = 25/50 = 0.5 ⇒ θ = tan⁻¹(0.5) ≈ 26.6°
该角度用于在缆绳端部进行适当切割，使其与立柱紧密贴合。
8. Part G: Combining Skills – An Exam‑Style Problem | 第七部分：综合应用——考试型问题
OCR often presents a final part that links several concepts. Here we ask: “Using the maximum lawn area of 1250 m², determine the width w of a uniform internal flower border (running all around inside the rectangle) that reduces the remaining grass area by 20%. Show that w satisfies a quadratic equation and solve it.”
OCR 经常会在最后一就问中串联多个概念。我们设问：“利用最大草坪面积 1250 m²，求内部均匀花坛边界（围绕矩形内侧一周）的宽度 w，使得剩余草地面积减少 20%。证明 w 满足一个二次方程并求解。”
Inner rectangle dimensions are (50 − 2w) by (25 − 2w). The new grass area is to be 80% of 1250, i.e. 1000 m². Hence:
(50 − 2w)(25 − 2w) = 1000
Expanding: 1250 − 100w − 50w + 4w² = 1000 → 4w² − 150w + 1250 = 1000 → 4w² − 150w + 250 = 0. Divide by 2: 2w² − 75w + 125 = 0. Use the quadratic formula:
w = [75 ± √(75² − 4×2×125)] / (2×2) = [75 ± √(5625 − 1000)] / 4 = [75 ± √4625] / 4
√4625 = √(25×185) = 5√185 ≈ 68.0, so w ≈ (75 − 68.0)/4 ≈ 1.75 m (taking the sensible root smaller than half the width). The other root is too large. Thus the border width is about 1.75 m.
内部矩形尺寸为 (50 − 2w) × (25 − 2w)。新的草地面积需为 1250 的 80%，即 1000 m²。因此：
(50 − 2w)(25 − 2w) = 1000
展开：1250 − 100w − 50w + 4w² = 1000 → 4w² − 150w + 1250 = 1000 → 4w² − 150w + 250 = 0。除以 2：2w² − 75w + 125 = 0。使用二次公式：
w = [75 ± √(75² − 4×2×125)] / (2×2) = [75 ± √(5625 − 1000)] / 4 = [75 ± √4625] / 4
√4625 = √(25×185) = 5√185 ≈ 68.0，故 w ≈ (75 − 68.0)/4 ≈ 1.75 m（取小于宽度一半的合理根）。另一根过大。因此花坛边界宽度约为 1.75 m。
9. Reviewing the Whole Case: Key Takeaways | 总结回顾：关键要点
This case study demonstrates how differentiation (Part A–B), matrix transformations (Part C–D), polynomial equations (Part E), trigonometry (Part F) and algebraic modelling (Part G) interweave in a realistic project. Always identify the constraints first, write functions clearly, and check that answers are practical within the context.
本案例展示了微分（第一、二部分）、矩阵变换（第三、四部分）、多项式方程（第五部分）、三角学（第六部分）以及代数建模（第七部分）如何在一个真实项目中交织运用。解题策略永远是先明确约束条件，清晰写出函数，并检验答案在情境中是否合理。
When tackling similar exam case studies, read the whole scenario before jumping in, highlight
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