📚 IGCSE OCR Maths: Linear Programming Revision Guide | IGCSE OCR 数学：线性规划 考点精讲
Linear programming is a crucial topic in the IGCSE OCR Mathematics syllabus. It allows you to model real-world situations using linear inequalities and find optimal solutions, such as maximising profit or minimising cost, subject to various constraints. Mastering this topic will not only boost your exam performance but also sharpen your analytical thinking.
线性规划是 IGCSE OCR 数学大纲中的一个关键主题。它让你能够使用线性不等式对现实情况进行建模，并在各种约束条件下找到最优解，例如最大化利润或最小化成本。掌握这个主题不仅能提高你的考试成绩，还能锻炼你的分析思维能力。
1. What is Linear Programming? | 什么是线性规划？
Linear programming is a method used to find the best outcome (maximum profit or minimum cost) in a mathematical model whose requirements are represented by linear relationships. The problem is defined by a set of linear inequalities (constraints) and a linear objective function that you aim to optimise.
线性规划是一种在要求由线性关系表示的数学模型中寻找最佳结果（最大利润或最小成本）的方法。该问题由一组线性不等式（约束条件）和一个你要优化的线性目标函数定义。
In an IGCSE exam, you will typically be given a scenario such as a factory producing two products with limited resources. You will write inequalities, draw a feasible region on a graph, and then determine the combination that gives the maximum or minimum value of the objective function.
在 IGCSE 考试中，通常会给出一个场景，比如一家工厂生产两种产品，资源有限。你将写出不等式，在图形上画出可行区域，然后确定使目标函数达到最大值或最小值的组合。
2. Formulating Inequalities from a Problem | 从问题中列出不等式
The first step is to translate the word problem into mathematical inequalities. Read the text carefully and identify the two variables (usually x and y). Then express each constraint as a linear inequality. Common constraints include limits on raw materials, time, labour, or storage space. Always remember non-negativity constraints: x ≥ 0 and y ≥ 0, as you cannot produce a negative number of items.
第一步是将文字问题转化为数学不等式。仔细阅读文本并确定两个变量（通常是 x 和 y）。然后将每个约束条件表示为线性不等式。常见的约束包括原材料、时间、劳动力或存储空间的限制。务必记住非负性约束：x ≥ 0 和 y ≥ 0，因为你不能生产负数量的物品。
For example, ‘A factory makes two types of chairs, A and B. Each chair A requires 2 hours of cutting and 1 hour of assembling. Chair B requires 1 hour of cutting and 3 hours of assembling. The cutting department is available for at most 40 hours, and the assembling department for at most 36 hours.’ If x is the number of chair A and y is the number of chair B, the constraints are: 2x + y ≤ 40, x + 3y ≤ 36, x ≥ 0, y ≥ 0.
例如，“一家工厂生产两种椅子 A 和 B。每把椅子 A 需要 2 小时切割和 1 小时组装。椅子 B 需要 1 小时切割和 3 小时组装。切割部门最多可用 40 小时，组装部门最多可用 36 小时。”如果 x 是椅子 A 的数量，y 是椅子 B 的数量，那么约束条件为：2x + y ≤ 40，x + 3y ≤ 36，x ≥ 0，y ≥ 0。
Always double-check that you have translated the limits correctly. Phrases like ‘at least’ correspond to ≥, while ‘at most’ or ‘does not exceed’ give ≤. ‘Exactly’ would give an equation, but linear programming typically uses inequalities.
务必仔细检查你是否正确转换了限制条件。诸如“至少”对应 ≥，而“至多”或“不超过”对应 ≤。“刚好”会给出等式，但线性规划通常使用不等式。
3. Graphing Linear Inequalities | 绘制线性不等式
To graph a linear inequality, first draw the corresponding boundary line as if it were an equation. For ≤ or ≥, use a solid line to show that points on the line are included. For < or >, use a dashed line to indicate the line is not part of the region.
要绘制线性不等式，首先画出对应的边界线，如同画方程一样。对于 ≤ 或 ≥，使用实线表示线上的点包含在内。对于 < 或 >，使用虚线表示该线不属于区域。
Then decide which side of the line satisfies the inequality. Choose a test point not on the line (the origin (0,0) is easiest, unless the line passes through it). Substitute the coordinates into the inequality. If the inequality holds true, shade the side containing the test point; otherwise shade the opposite side.
然后确定直线哪一侧满足不等式。选择一个不在直线上的测试点（原点 (0,0) 最简单，除非直线经过原点）。将坐标代入不等式。如果不等式成立，则给包含测试点的那一侧涂上阴影；否则给另一侧涂上阴影。
For example, for 2x + y < 10, draw the line 2x + y = 10 as a dashed line. Test (0,0): 2(0)+0=0 < 10, so shade the side containing (0,0). For y ≥ 3, draw a solid horizontal line y = 3; test (0,0): 0 ≥ 3 is false, so shade the region above the line.
例如，对于 2x + y < 10，将直线 2x + y = 10 画成虚线。测试 (0,0)：2(0)+0=0 < 10，因此给包含 (0,0) 的一侧涂上阴影。对于 y ≥ 3，画一条实水平线 y = 3；测试 (0,0)：0 ≥ 3 不成立，则给直线上方区域涂上阴影。
4. The Feasible Region | 可行区域
The feasible region is the intersection of all the shaded areas that satisfy every constraint. It represents all possible combinations of x and y that meet the conditions. Often, you must shade out the unwanted regions, leaving the feasible region unshaded (this is sometimes called the ‘rejected region’ method). Follow the specific instruction in the exam question.
可行区域是所有满足每个约束条件的阴影区域的交集。它代表了所有满足条件的 x 和 y 的可能组合。通常，你需要将不需要的区域涂上阴影，留下可行区域不涂（有时称为“拒绝区域”方法）。请遵循考题中的具体说明。
Label the feasible region clearly, usually with the letter R. It will be a polygon (often a quadrilateral or triangle) bounded by the constraint lines and the axes if x ≥ 0 and y ≥ 0 are included. If one of the constraints is missing, the region may be unbounded, but in IGCSE problems it is nearly always bounded.
清楚地标记可行区域，通常用字母 R。它将是一个由约束直线和坐标轴（如果包含 x ≥ 0 和 y ≥ 0）围成的多边形（通常是四边形或三角形）。如果缺少某个约束条件，区域可能是无界的，但在 IGCSE 问题中，它几乎总是有界的。
5. The Objective Function | 目标函数
The objective function is the expression you want to maximise or minimise. It is often given as something like P = 5x + 8y, representing profit, or C = 3x + 2y, representing cost. Every point in the feasible region gives a possible value of the objective function, but the optimum value will always occur at a vertex (corner point) of the region.
目标函数是你想要最大化或最小化的表达式。它通常以 P = 5x + 8y（代表利润）或 C = 3x + 2y（代表成本）等形式给出。可行区域中的每个点都给出目标函数的一个可能值，但最优值总是出现在区域的顶点（角点）处。
To find the maximum or minimum, you do not need to test infinitely many points. Thanks to the linearity, the optimal solution lies at one of the vertices. This is the fundamental principle of linear programming.
要找到最大值或最小值，你不需要测试无限多个点。由于线性特性，最优解位于其中一个顶点处。这是线性规划的基本原理。
6. The Method of Vertices | 顶点法
Once you have drawn the feasible region and identified all the vertices, you can evaluate the objective function at each vertex. List the coordinates of all corner points: the origin (if included), points where lines intersect the axes, and points where two constraint lines intersect.
一旦画出了可行区域并确定了所有顶点，你就可以在每个顶点处计算目标函数的值。列出所有角点的坐标：原点（如果包含在内），直线与坐标轴的交点，以及两条约束直线相交的点。
To find the intersection of two lines, solve their equations simultaneously. For example, for 2x + y = 40 and x + 3y = 36, multiply the first equation by 3: 6x + 3y = 120. Subtract the second equation: (6x+3y) – (x+3y) = 120 – 36, giving 5x = 84, so x = 16.8. Then substitute back to find y = 40 – 2(16.8) = 6.4. Thus the vertex is (16.8, 6.4).
要找到两条直线的交点，需要联立求解它们的方程。例如，对于 2x + y = 40 和 x + 3y = 36，将第一个方程乘以 3：6x + 3y = 120。减去第二个方程：(6x+3y) – (x+3y) = 120 – 36，得到 5x = 84，因此 x = 16.8。然后代回求得 y = 40 – 2(16.8) = 6.4。所以该顶点是 (16.8, 6.4)。
It can be useful to present your working in a table. You will then see which vertex gives the highest or lowest value of the objective function.
将你的计算过程呈现在表格中会很有帮助。然后你就会看到哪个顶点使目标函数取到最高值或最低值。
7. Finding the Optimal Solution: Maximisation | 求最优解：最大化
Suppose the objective function is Profit P = 5x + 8y, representing the profit from x chairs A and y chairs B. Using the constraints from the previous example, the vertices of the feasible region are: (0,0), (0,12) [from x+3y=36 when x=0], (20,0) [from 2x+y=40 when y=0], and the intersection (16.8, 6.4). However, if the region is bounded also by a line such as y = 0 and x = 0, those are included.
假设目标函数为利润 P = 5x + 8y，代表 x 把椅子 A 和 y 把椅子 B 的利润。使用前面示例中的约束条件，可行区域的顶点有：(0,0)，(0,12) [由 x+3y=36 当 x=0 时得出]，(20,0) [由 2x+y=40 当 y=0 时得出]，以及交点 (16.8, 6.4)。但由于区域也由 y=0 和 x=0 围成，这些点都包含在内。
Now evaluate P at each vertex: at (0,0) P=0; at (0,12) P=5(0)+8(12)=96; at (20,0) P=5(20)+8(0)=100; at (16.8, 6.4) P=5(16.8)+8(6.4)=84+51.2=135.2. The maximum profit is 135.2, achieved at approximately x = 16.8 and y = 6.4. In practice, you may need to round, but in pure mathematics answers are left exact if decimals are allowed.
现在在每个顶点处计算 P 值：在 (0,0) 处 P=0；在 (0,12) 处 P=5(0)+8(12)=96；在 (20,0) 处 P=5(20)+8(0)=100；在 (16.8, 6.4) 处 P=5(16.8)+8(6.4)=84+51.2=135.2。最大利润是 135.2，在大约 x=16.8 和 y=6.4 时实现。在实际应用中可能需要取整，但在纯数学中如果允许小数则答案保留精确值。
Always state the optimal value with the corresponding values of x and y. The answer should say: ‘Maximum profit is 135.2 when x = 16.8 and y = 6.4.’
始终要说明最优值和对应的 x 与 y 的值。答案应为：“当 x = 16.8 且 y = 6.4 时，最大利润为 135.2。”
8. Finding the Optimal Solution: Minimisation | 求最优解：最小化
For a minimisation problem, the approach is identical: test the objective function at all vertices and pick the smallest value. For instance, if the objective function is Cost C = 3x + 2y and the same constraints apply, C at (0,0)=0 is clearly the minimum, but that may be unrealistic if producing zero items is not allowed. So additional constraints like x + y ≥ 10 might be introduced to ensure some production.
对于最小化问题，方法完全相同：在所有顶点处测试目标函数，并选出最小值。例如，如果目标函数是成本 C = 3x + 2y，且适用同样的约束条件，那么 C 在 (0,0)=0 显然是最小值，但如果生产零件物品不被允许，这可能不切实际。因此可能会引入额外的约束条件，如 x + y ≥ 10，以确保有一定的生产量。
Suppose a new constraint x + y ≥ 10 is added. The feasible region shrinks, and the vertices change. The origin is no longer feasible. The new vertices would be the intersection of x+y=10 with the other lines. Again, calculate and test each vertex to find the minimum cost.
假设添加一个新的约束条件 x + y ≥ 10。可行区域缩小，顶点也发生变化。原点不再可行。新的顶点将是 x+y=10 与其他直线的交点。同样，计算并测试每个顶点以找到最小成本。
9. Integer Solutions and Practical Considerations | 整数解与实际考虑
Often in real-life problems, x and y represent whole items, so the optimal vertex coordinates may not be integers. In such cases, the exam may ask for the optimal integer solution. You cannot simply round the fractional coordinates; you must check integer points near the optimal vertex that still lie inside the feasible region.
在现实问题中，x 和 y 常常代表整件物品，因此最优顶点坐标可能不是整数。此时，考试可能要求找出最优整数解。你不能简单地对分数坐标进行四舍五入；你必须检查最优顶点附近的整数点，并确保它们仍在可行区域内。
Plot integer points around (16.8, 6.4) such as (16,6), (16,7), (17,6) etc. Test each one to see if it satisfies all inequalities, and then calculate the objective function. The point giving the highest P while satisfying constraints is the optimal integer solution. Be systematic and write down your working.
画出 (16.8, 6.4) 附近的整数点，如 (16,6), (16,7), (17,6) 等。逐一测试它们是否满足所有不等式，然后计算目标函数值。在满足约束条件的同时给出最高 P 值的点即为最优整数解。要有条不紊地列出计算过程。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
One common error is forgetting to include the non-negativity constraints x ≥ 0 and y ≥ 0. Even if the text does not state them explicitly, they are usually implied because you cannot have negative quantities. Missing them can make the feasible region unbounded and lead to an incorrect answer.
一个常见错误是忘记包含非负性约束 x ≥ 0 和 y ≥ 0。即使文字没有明确说明，它们通常也是隐含的，因为不能有负数数量。遗漏这些约束会使可行区域无界，导致错误答案。
Another frequent mistake is shading the wrong side of a line. Always use a test point and substitute the coordinates carefully. Also, when two lines have similar slopes, plot them accurately so that intersections are correct. Use a ruler and a sharp pencil.
另一个常见错误是在直线的错误一侧涂上阴影。务必使用测试点并仔细代入坐标。此外，当两条直线斜率相近时，要精确绘制，以确保交点准确。使用直尺和削尖的铅笔。
In the exam, read the question to determine whether you need to label the feasible region R, shade the rejected region, or shade the feasible region. Different exam papers use different conventions. Follow the instructions exactly. Finally, if a question asks for the maximum value, state it as ‘Maximum … is … when x = … and y = …’ to gain full marks.
考试中，仔细读题以确定是否需要将可行区域标记为 R、给拒绝区域涂阴影，还是给可行区域涂阴影。不同的试卷可能使用不同的惯例。严格遵循说明。最后，如果题目要求求出最大值，要陈述为“当 x = … 且 y = … 时，最大… 为 …”，以获得全部分数。
Practising with past papers is the best way to master linear programming. You will become familiar with the wording of OCR questions and the typical scenarios involving production, cost minimisation, or diet planning. Good luck!
通过历年真题进行练习是掌握线性规划的最佳方法。你将熟悉 OCR 题目的措辞以及涉及生产、成本最小化或饮食计划的典型场景。祝你成功！
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