📚 Linear Programming in AS Mathematics | AS 数学：线性规划 考点精讲
Linear programming is a powerful mathematical technique used to find the best possible outcome in a given situation, such as maximising profit or minimising cost, subject to a set of limitations. In AS Mathematics, you learn the graphical approach to linear programming – a topic that combines algebra, inequalities, and coordinate geometry. Mastering it involves interpreting word problems, translating them into mathematical constraints, drawing the feasible region, and using the vertex method to identify the optimal solution. This article walks you through every key concept, common pitfalls, and exam-style examples to build your confidence.
线性规划是一种强大的数学方法，用来在给定条件下寻找最佳结果，例如最大化利润或最小化成本，同时受到一系列限制。在AS数学中，你将学习线性规划的图解法——这一主题融合了代数、不等式和解析几何。掌握它需要读懂文字题，将之转化为数学约束，画出可行域，并利用顶点法找到最优解。本文将带你逐一梳理每个核心概念、常见错误以及考试型例题，帮助你建立信心。
1. Introduction to Linear Programming | 线性规划简介
Linear programming deals with optimising a linear objective function while satisfying several linear inequality constraints. The word 'programming' refers to planning rather than computer coding. In AS exams, all problems are limited to two decision variables, so you can solve them by drawing graphs and shading regions.
线性规划处理的是在满足多个线性不等式约束的同时优化一个线性目标函数的问题。'规划'一词指的是计划，而不是计算机编程。在AS考试中，所有问题都限定在两个决策变量内，因此你可以通过绘制图形和阴影区域来求解。
Typical real-world scenarios include a factory deciding how many units of two products to manufacture, given limits on machine time and raw materials. The goal is to maximise profit. Other applications involve minimising transport costs or optimising blends of ingredients.
典型的现实场景包括：一家工厂在机器时间和原材料有限的条件下，决定生产两种产品各多少件，目标是最大化利润。其他应用还包括最小化运输成本或优化配料混合。
The main steps are: define the variables, write the constraints as linear inequalities, form the objective function, draw the feasible region, find the vertices, and evaluate the objective function at each vertex to pick the best value. We will unpack each step in detail.
主要步骤是：定义变量，将约束写为线性不等式，构造目标函数，画出可行域，求出顶点，并计算每个顶点的目标函数值以选择最优值。我们将逐个详细解析这些步骤。
2. Decision Variables and Objective Function | 决策变量与目标函数
S Mathematics, variables are usually non-negative, so x ≥ 0 and y ≥ 0 automatically apply.
决策变量是你可以控制的数量。你应该在一开始就清晰地定义它们。使用与问题相关的字母，例如，设 x 为生产的椅子数量，y 为生产的桌子数量。在AS数学中，变量通常为非负值，因此自动有 x ≥ 0 和 y ≥ 0。
ised or minimised. It has the form Z = ax + by, where a and b are constants. You might see 'Maximise P = 5x + 3y' or 'Minimise C = 2x + 4y' because the method then seeks the largest or smallest Z-value within the feasible region.
目标函数表达的是要最大化或最小化的量，其形式为 Z = ax + by，其中 a 和 b 是常数。你可能看到'最大化 P = 5x + 3y'或'最小化 C = 2x + 4y'。考试时，务必注意题目要求最大值还是最小值，因为方法随后是在可行域内寻找 Z 的最大或最小值。
Use centred bold styling for your objective function and constraints when presenting them in your work:
Maximise Z = 5x + 3y
3. Formulating Constraints from Word Problems | 从文字题构建约束条件
y come from resources like time, money, materials, or space. To translate a statement like ' es 2 hours of labour and each table requires 3 hours, and total labour is at most 60 hours', you write 2x + 3y ≤ 60. Remember: 'at most' means ≤, 'at least' means ≥, and 'exactly' means =.
约束是限制决策变量取值的不等式。它们来源于时间、金钱、材料或空间等资源。要将'每把椅子需要2小时人工，每张桌子需要3小时人工，总人工最多为60小时'这类陈述翻译为数学语言，你写作 2x + 3y ≤ 60。记住：'最多'意味着 ≤，'至少'意味着 ≥，'恰好'意味着 =。
Other common phrases: 'cannot exceed' → ≤, 'no more than' → ≤, 'a minimum of' → ≥, 'must be at least' → ≥. Also watch out for combined constraints like 'the number of chairs must be at least twice the number of tables' . Rearranging such inequalities to a standard form such as x − 2y ≥ 0 helps when graphing.
其他常见短语：'不能超过' → ≤，'不多于' → ≤，'至少' → ≥，'必须不少于' → ≥。还要注意组合约束，如'椅子的数量必须至少是桌子数量的两倍'，可得 x ≥ 2y。将这类不等式改写为标准形式如 x − 2y ≥ 0 在作图时更有帮助。
imes a constraint states a maximum number of items regardless of type, such as x + y ≤ 50.
始终单独列出约束，包括非负条件。一个典型列表形如：2x + y ≤ 40, x + 3y ≤ 60, x ≥ 0, y ≥ 0。有时约束会给出不分类型的总量上限，例如 x + y ≤ 50。
4. Graphing Linear Inequalities and Feasible Region | 绘制线性不等式与可行域
nequality includes equality (≤). Use a dashed line for strict inequalities such as < or >.
要绘制诸如 2x + y ≤ 40 的不等式，首先画出边界直线 2x + y = 40。求出 x 轴和 y 轴截距：当 x = 0 时 y = 40；当 y = 0 时 x = 20。用实线连接这些点，因为不等式包含等号 (≤)。对于严格不等式如 < 或 >，则使用虚线。
decide which side of the line represents the inequality. Pick a test point not on the line – hade the side containing (0,0). If false, shade the opposite side. Repeat for all constraints. The feasible region is the intersection where all shadings overlap – in other words, the unshaded region if you use the 'shade out' method favoured by many exam boards.
接下来，判断直线的哪一侧代表该不等式。选取直线上方区域外一个测试点——原点 (0,0) 通常最方便。代入不等式：2(0)+0 ≤ 40 得 0 ≤ 40，成立，因此涂阴影于包含 (0,0) 的一侧。若不成立，则涂对侧。对所有约束重复此步骤。可行域是所有阴影重叠的区域——换句话说，如果你采用许多考试局偏爱的'向外涂'方法，可行域就是未被涂色的区域。
When shading, label your lines and clearly mark the feasible region. Use a ruler for accuracy, as poorly drawn lines can lead to incorrect vertex identification. In the exam, you may be given graph paper or asked to complete a partially drawn graph.
涂阴影时，标注直线并清晰标出可行域。使用直尺以确保准确，因为线条画得不好可能导致顶点识别错误。在考试中，你可能会得到坐标纸，或被要求完成部分已绘制的图形。
5. Finding the Vertices of the Feasible Region | 求可行域顶点
The optimal solution of a linear programming problem will always occur at a vertex (corner point) of the feasible region, provided the region is bounded and the solution is unique. Therefore, you need to find the coordinates of all vertices. Some vertices come from intercepts with axes, while others are intersections of two constraint lines.
如果一个线性规划问题的最优解是唯一的且可行域有界，那么最优解总是出现在可行域的顶点（角点）上。因此，你需要找到所有顶点的坐标。有些顶点来自与坐标轴的截距，另一些则来自两条约束直线的交点。
2, then y = 6. The vertex is (2,6). Always check that the point satisfies all inequalities – if it lies on a boundary but violates another constraint, it is not a feasible vertex.
要找到交点，联立求解两条边界直线的方程。例如，直线 x + y = 8 和 2x + y = 10 相交，用第二个方程减去第一个得 x = 2，然后 y = 6。该顶点为 (2,6)。始终检查该点是否满足所有不等式——如果它位于某条边界线上但却违反了另一约束，那么它就不是一个可行顶点。
List vertices methodically, often starting from the origin and moving around the region. A systematic table can help:
| Vertex (x, y) | Obtained from |
|---|---|
| (0, 0) | Intersection of x = 0, y = 0 |
| (5, 0) | 2x + y = 10 and y = 0 |
| (2, 6) | x + y = 8 and 2x + y = 10 |
| (0, 8) | x + y = 8 and x = 0 |
系统性地列出顶点，通常从原点开始并沿区域移动。系统化的表格会有所帮助。表格如上所示。
6. The Vertex Method for Optimal Solution | 顶点法求最优解
losed and bounded. For example, using Z = 3x + 2y and vertices (0,0), (5,0), (2,6), (0,8):
一旦你得到所有可行顶点，就将每一对坐标代入目标函数。记录得到的 Z 值。如果可行域是封闭且有界的，这些值中的最大或最小值就是最优答案。例如，使用 Z = 3x + 2y 以及顶点 (0,0), (5,0), (2,6), (0,8)：
At (0,0): Z = 0; At (5,0): Z = 15; At (2,6): Z = 3(2)+2(6) = 18; At (0,8): Z = 16.
The maximum is 18 at (2,6). So the optimal strategy is to produce 2 of product A and 6 of product B to achieve the highest profit of 18 units. Always write a concluding statement in the context of the problem.
最大值为 18，在点 (2,6) 处取得。因此最优策略是生产 2 个产品 A 和 6 个产品 B，以获得最高利润 18 单位。始终结合题目情境写出结论性陈述。
If you are minimising, pick the smallest Z-value from the vertices. It is good practice to evaluate all vertices even if one seems obviously larger or smaller, as exam marks rely on complete working. Some questions may also ask you to find the optimal value rather than the point, so be careful to read the requirement.
如果求最小值，就选取顶点中最小的 Z 值。即使某个值明显更大或更小，也最好对所有顶点求值，因为考试评分依赖完整的解题过程。有些问题可能要求找出最优值而非最优解点，因此要仔细阅读题意。
7. Multiple Optimal Solutions and Special Cases | 多重最优解与特殊情况
Sometimes the objective function is parallel to one of the constraint boundaries that forms an edge of the feasible region. In that case, all points along that edge give the same optimal Z-value. This is called multiple or infinite optimal solutions. You can detect it when the slope of the objective line equals the slope of a bounding line, and two adjacent vertices yield the same Z-value.
有时目标函数与构成可行域某条边界的约束直线平行。在这种情况下，该边上的所有点都给出相同的最优 Z 值。这称为多重最优解或无限最优解。当目标直线斜率等于某条边界直线的斜率，且两个相邻顶点得到相同的 Z 值时，你可以发现这种情形。
and (0,3) and all points on the segment joining them. The answer can then be expressed as 'any point on the line segment between (0,3) and (2,2) inclusive'.
例如，最大化 Z = x + 2y，约束为 x + y ≤ 4, x + 2y ≤ 6, x, y ≥ 0。顶点为：(0,0) Z=0；(4,0) Z=4；(2,2) Z=6；(0,3) Z=6。在 (2,2) 和 (0,3) 处均取得最大值 Z=6，并且连接这两点的线段上所有点也都得到最大值。答案可表达为'线段 (0,3) 到 (2,2) 上的任意点（含端点）'。
In an exam, if you get the same Z-value at two adjacent vertices, check whether the edge linking them has the same gradient as the objective function. If it does, state that there are multiple optimal solutions along that edge. Do not assume it is a calculation error.
考试中，如果你在两个相邻顶点得到相同的 Z 值，检查连接它们的边的斜率是否与目标函数相同。如果是，就应说明沿着那条边存在多重最优解。不要误以为是计算错误。
8. Unbounded and Infeasible Problems | 无界问题与无可行解
rbitrarily large by taking x and y very large. The problem is then said to be unbounded. However, minimising the same objective over an unbounded region could still yield a finite minimum at a vertex.
并非所有线性规划问题都有唯一最优解。如果可行域无界（在某方向上无限延伸），最大化问题可能没有有限的最大值。例如，如果约束仅为 x ≥ 0, y ≥ 0 且目标为最大化 Z = x + y，你可以通过取非常大的 x 和 y 使 Z 任意大。此时问题被称为无界问题。然而，在无界区域上最小化同样的目标函数，仍可能在某个顶点处获得有限的最小值。
9. The minimum sum would be 10 if both are at least 5, which exceeds 9, so no point satisfies all three. The feasible region is empty. In such a case, you would state that no solution exists.
当约束条件相互矛盾，不存在任何一组值能同时满足所有不等式时，就会产生无可行解问题。例如，x ≥ 5, y ≥ 5 且 x + y ≤ 9。若两者均至少为 5，则最小和为 10，这超过了 9，因此没有任何点能满足全部三个条件。可行域为空。在这种情况下，你应该说明解不存在。
Your graph can reveal these situations: an unbounded feasible region will look like an open shape, while an infeasible problem will have no overlapping shaded area. Always check whether your feasible region makes sense.
你的图形可以揭示这些情况：无界可行域看起来像一个开放形状，而无可行解问题则表现为没有重叠的阴影区域。永远要检查你的可行域是否合理。
9. Integer Programming – A First Look | 整数规划初探
In many real-life problems, the decision variables must be whole numbers (integers). For instance, you cannot manufacture 2.5 tables. AS problems sometimes ask for integer solutions. The graphical method gives continuous solutions at vertices, but the optimal integer solution may not be exactly at a vertex. You must test integer points near the continuous optimum.
在许多现实问题中，决策变量必须是整数。例如，你无法生产 2.5 张桌子。AS 题目有时会要求整数解。图解法给出位于顶点的连续解，但最优整数解未必恰好在一个顶点上。你必须检验连续最优解附近的整数点。
Suppose the vertex giving maximum profit is (3.6, 2.7). You would examine integer points around it, such as (3,2), (3,3), (4,2), (4,3), while ensuring each candidate satisfies all constraints. Calculate the objective function at each feasible integer point and select the best. There is no shortcut – you must perform this check manually when instructed to find integer solutions.
假设给出最大利润的顶点是 (3.6, 2.7)。你需要检查它周围的整数点，如 (3,2), (3,3), (4,2), (4,3)，同时确保每个候选点满足所有约束。在每个可行的整数点处计算目标函数，并选出最优值。没有捷径——当题目要求找出整数解时，你必须手动完成这一检验。
Be careful: the nearest integer point to the continuous optimum is not always the best integer solution. The integer feasible region is a set of discrete points, and its optimum can lie further away.
注意：最接近连续最优解的整数点并不总是最佳整数解。整数可行域是一组离散点，其最优解可能位于更远的地方。
10. Reading and Interpreting Inequality Regions | 阅读与解释不等式区域
on and see which side of each line it lies on. This tells you whether the inequality is ≤ or ≥.
考官通常会给出一个带阴影可行域和若干边界线的图形，要求你写出定义该区域的不等式。为此，首先确定每条直线的方程（形式为 ax + by = c）。接下来，在阴影区域内明确选取一个测试点，看它位于每条直线的哪一侧。这告诉你该用 ≤ 还是 ≥。
esting the origin or another easy point. If the region is unshaded (shaded outside), ensure you still write inequalities satisfied by the unshaded area.
对于垂直线 x = k，不等式将是 x ≤ k 或 x ≥ k，取决于区域在左侧还是右侧。类似地，水平线 y = c 给出 y ≤ c 或 y ≥ c。始终通过测试原点或其他简单点来复核。如果区域未涂色（阴影涂在外面），你仍然要确保写出的是未被涂色区域所满足的不等式。
Sometimes the graph contains additional points labelled A, B, C, and you must find the coordinates or interpret their meaning, like 'minimum cost occurs at point B'. Use the scale carefully – misreading coordinates is a common error.
有时图形中还包含标注为 A、B、C 的附加点，你必须求出它们的坐标或解释其含义，例如'最小成本出现在点 B'。仔细使用比例尺——读错坐标是个常见错误。
11. Worked Example: A Complete Linear Programming Problem | 完整例题解析
Let's walk through a typical AS linear programming problem to consolidate the process.
让我们走一遍典型的 AS 线性规划问题，以巩固所学的步骤。
Problem: A company makes two types of souvenir, Basic and Deluxe. Each Basic requires 2 minutes on machine M and 3 minutes on machine N. Each Deluxe requires 4 minutes on M and 2 minutes on N. Machine M can operate for at most 120 minutes per day, and machine N for at most 90 minutes. The profit is £3 per Basic and £4 per Deluxe. The company wants to maximise profit. How many of each should be produced?
问题：一家公司生产两种纪念品：基础款和豪华款。每件基础款需要在机器 M 上加工 2 分钟，在机器 N 上加工 3 分钟。每件豪华款需要在机器 M 上加工 4 分钟，在机器 N 上加工 2 分钟。机器 M 每天最多运行 120 分钟，机器 N 每天最多运行 90 分钟。基础款每件利润为 £3，豪华款每件利润为 £4。公司希望最大化利润。每种应生产多少件？
Step 1 – Define variables: Let x be the number of Basics and y be the number of Deluxe products.
步骤 1 – 定义变量：设 x 为基础款的数量，y 为豪华款的数量。
Step 2 – Constraints: M: 2x + 4y ≤ 120; N: 3x + 2y ≤ 90; Non-negativity: x ≥ 0, y ≥ 0.
步骤 2 – 约束：M: 2x + 4y ≤ 120; N: 3x + 2y ≤ 90; 非负: x ≥ 0, y ≥ 0.
Step 3 – Objective: Maximise P = 3x + 4y.
步骤 3 – 目标：最大化 P = 3x + 4y.
Step 4 – Graph: 15. Substitute back: 2(15) + 4y = 120 → 30 + 4y = 120 → 4y = 90 → y = 22.5. Intersection is (15,22.5).
步骤 4 – 绘图：画出直线 2x+4y=120（截距 (60,0) 和 (0,30)）和 3x+2y=90（截距 (30,0) 和 (0,45)）。涂出可行域。求出 2x+4y=120 与 3x+2y=90 的交点：联立求解。将第二个方程乘以 2：6x+4y=180。减去第一个方程：(6x+4y) − (2x+4y) = 180 − 120 → 4x = 60 → x = 15。回代：2(15) + 4y = 120 → 30 + 4y = 120 → 4y = 90 → y = 22.5。交点为 (15,22.5)。
Step 5 – Vertices: feasible vertices are
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