📚 Year 9 CCEA Further Mathematics Case Study Practice | 九年级 CCEA 进阶数学案例分析实战演练
Case studies in Further Mathematics challenge you to apply your skills to real-world situations. This article presents a series of practical problems covering algebra, geometry, statistics and logical reasoning. Each case is designed to strengthen your problem-solving ability and deepen conceptual understanding for the CCEA curriculum.
进阶数学中的案例研究要求你将所学技能应用到实际情境中。本文提供一系列涵盖代数、几何、统计和逻辑推理的实践问题。每个案例旨在增强你解决问题的能力，并加深你对 CCEA 课程概念的理解。
1. Mobile Plan Cost Comparison | 案例一：手机套餐费用比较
Anna is choosing between two mobile phone contracts. Plan A charges a fixed monthly fee of £12 plus £0.05 per minute of calls. Plan B has no fixed fee but charges £0.12 per minute.
安娜正在两个手机套餐之间选择。套餐 A 每月固定费用 12 英镑，另加每分钟 0.05 英镑的通话费。套餐 B 没有固定费用，但每分钟收费 0.12 英镑。
First, write an expression for the monthly cost of each plan for x minutes of calls.
首先，写出通话 x 分钟时每个套餐的月费用表达式。
Plan A: Cost = 12 + 0.05x. Plan B: Cost = 0.12x.
套餐 A：费用 = 12 + 0.05x。套餐 B：费用 = 0.12x。
Find the number of minutes for which both plans cost the same.
求使两种套餐费用相同的通话分钟数。
0.07x, hence x = 12 ÷ 0.07 ≈ 171.4 minutes. So at about 171 minutes, the costs are equal.
设 12 + 0.05x = 0.12x，得出 12 = 0.07x，因此 x = 12 ÷ 0.07 ≈ 171.4 分钟。因此，约 171 分钟时费用相等。
If Anna talks for 200 minutes per month, which plan is cheaper?
如果安娜每月通话 200 分钟，哪个套餐更便宜？
Plan A: £12 + 0.05 × 200 = £22. Plan B: 0.12 × 200 = £24. So Plan A is cheaper by £2.
套餐 A：12 + 0.05 × 200 = 22 英镑。套餐 B：0.12 × 200 = 24 英镑。因此套餐 A 便宜 2 英镑。
The break-even point can also be found graphically by plotting both cost lines and finding their intersection.
也可以通过绘制两条费用直线并找到它们的交点，用图形方法找到盈亏平衡点。
2. Maximising Rectangular Area with Fixed Perimeter | 案例二：固定周长下最大化矩形面积
A gardener has 40 metres of fencing to enclose a rectangular vegetable plot against a wall, using the wall as one side. Find the dimensions that give the maximum area.
一位园丁有 40 米围栏，想靠墙围一个矩形菜地，用墙作为一边。求使面积最大的尺寸。
cular to the wall each be x metres. The total fencing used is y + 2x = 40, so y = 40 − 2x.
设平行于墙的一边为 y 米，垂直于墙的两边各为 x 米。使用的总围栏为 y + 2x = 40，所以 y = 40 − 2x。
The area A = x × y = x(40 − 2x) = 40x − 2x².
面积 A = x × y = x(40 − 2x) = 40x − 2x²。
e or use the symmetry: x = −b/(2a) with a = −2, b = 40. So x = −40/(2 × (−2)) = 10 metres.
这是一个二次函数。为求最大值，可通过配方或利用对称性：x = −b/(2a)，其中 a = −2, b = 40。因此 x = −40/(2 × (−2)) = 10 米。
Then
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