Linear Programming in A-Level OCR Mathematics | A-Level OCR 数学：线性规划 考点精讲

📚 Linear Programming in A-Level OCR Mathematics | A-Level OCR 数学：线性规划 考点精讲

Linear programming is a powerful mathematical technique used to determine the best possible outcome in a given model – such as maximum profit or minimum cost – subject to a set of linear constraints. In OCR A-Level Mathematics, this topic bridges pure algebra, coordinate geometry, and decision mathematics, requiring students to formulate real-world problems, sketch feasible regions, and identify optimal solutions using graphical methods or the vertex principle. Mastery of linear programming not only secures marks in the exam but also builds foundational skills for further studies in economics, engineering, and operations research.

线性规划是一种强大的数学工具，用于在给定线性约束条件下寻找最优结果——例如最大利润或最小成本。在OCR A-Level数学中，该主题融合了纯代数、坐标几何和决策数学，要求学生将实际问题建模、绘制可行域，并通过图解法或顶点原理确定最优解。掌握线性规划不仅能确保考试得分，还能为经济学、工程学和运筹学等领域的深造打下基础。

1. What is Linear Programming? | 什么是线性规划？

Linear programming (LP) is a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships. It involves two core components: an objective function (the quantity to be maximized or minimized) and a set of linear constraints (inequalities that define the limits of the situation). The word ‘linear’ indicates that both the objective function and all constraints are of degree one – no powers, products, or trigonometric functions of variables are allowed. In OCR exams, LP problems are typically two-dimensional, allowing solutions to be found graphically on the xy-plane.

线性规划（LP）是一种在数学模型（其要求由线性关系表示）中实现最佳结果的方法。它包含两个核心部分：目标函数（需要最大化或最小化的量）和一系列线性约束条件（定义情况限制的不等式）。”线性”一词意味着目标函数和所有约束条件都是一次的——不允许变量的幂、乘积或三角函数。在OCR考试中，线性规划问题通常是二维的，因此可以在xy平面上通过图解法求解。

2. Formulating a Linear Programming Problem | 问题建模

The first and often most challenging step is translating a worded scenario into mathematical inequalities and an objective function. Begin by defining your decision variables clearly – for example, let x be the number of product A produced and y be the number of product B produced. Then, identify the constraints: these often come from limited resources like machine time, labour hours, raw materials, or budget. Write each limitation as a linear inequality (e.g., 2x + 3y ≤ 240). Don’t forget the non-negativity constraints x ≥ 0, y ≥ 0 unless the context dictates otherwise. Finally, express the objective function, such as Profit P = 5x + 4y, which you will seek to maximize.

第一步，也往往是最具挑战性的一步，是将文字场景转化为数学不等式和目标函数。首先明确定义决策变量——例如，设x为产品A的生产数量，y为产品B的生产数量。然后，找出约束条件：这些通常来自有限资源，如机器时间、人工、原材料或预算。将每项限制写为线性不等式（例如：2x + 3y ≤ 240）。除非上下文另有规定，不要忘记非负约束x ≥ 0, y ≥ 0。最后，写出目标函数，如利润P = 5x + 4y，你需要使其最大化。

3. Graphing Constraints and the Feasible Region | 约束条件的图形表示与可行域

Each linear constraint divides the coordinate plane into two half-planes. To graph an inequality like 2x + 3y ≤ 240, first draw the boundary line 2x + 3y = 240 as a solid line (since the inequality includes equality). If the inequality is strict (< or >), use a dashed line. Then, test a point not on the line – the origin (0,0) is convenient – to determine which side satisfies the inequality. Shade the rejected region, leaving the allowed region unshaded, or clearly indicate the feasible side. The feasible region is the intersection of all allowed half-planes, including the non-negativity constraints, and is usually a convex polygon. In OCR exams, you are expected to accurately plot these lines on graph paper and label the axes and vertices.

每个线性约束将坐标平面分为两个半平面。要画出像2x + 3y ≤ 240这样的不等式，首先画出边界线2x + 3y = 240，用实线表示（因为不等式包含等号）。若不等式为严格不等号（< 或 >），则用虚线。然后，用线上以外的点——通常原点(0,0)最方便——测试哪一侧满足不等式。将不满足的区域涂上阴影，留下允许区域不涂，或清晰指示可行侧。可行域是所有允许半平面（包括非负约束）的交集，通常是一个凸多边形。在OCR考试中，你需要在坐标纸上精确绘制这些直线，并标注坐标轴和顶点。

4. Objective Function and Optimisation Lines | 目标函数与最优化直线

Once the feasible region is drawn, the objective function is optimised by considering its direction of increase or decrease. For a maximisation problem, set the objective function equal to some constant k to create an ‘iso-profit’ line (e.g., 5x + 4y = k). As k increases, the line moves parallelly in the direction of gradient. By sliding this line across the feasible region, the optimal point is the last vertex the line touches before leaving the region. For minimisation, you seek the first vertex entered. This graphical approach is intuitive but requires care: the optimum must occur at a vertex (or along a whole edge if the objective function is parallel to a constraint).

画出可行域后，通过考虑目标函数的增减方向来优化它。对于最大化问题，将目标函数设为某个常数k，得到一条”等利润线”（例如：5x + 4y = k）。随着k增大，该线沿梯度方向平行移动。将这条线在可行域上滑动，最优解就是它离开可行域前最后接触的顶点。对于最小化问题，则寻找它首次进入的顶点。这种图形化方法直观但需注意：最优解必定出现在某个顶点（或者，若目标函数与某约束平行，则整个边缘都是最优解）。

5. The Vertex Method for Optimal Solutions | 顶点法求最优解

The graphical method confirms that the optimum occurs at a vertex of the feasible region. Therefore, a reliable algebraic approach is the vertex method: calculate the coordinates of all vertices of the feasible region, then evaluate the objective function at each. The vertex yielding the highest value (for maximisation) or lowest value (for minimisation) is the optimal solution. Vertices are found by solving the simultaneous equations of boundary lines that intersect. For example, if lines 2x + y = 30 and x + 2y = 24 intersect, solving gives x = 12, y = 6. Evaluating P = 5x + 4y at (12,6) gives P=84. Compare this with other vertices to determine the optimum.

图形法证实最优解出现在可行域的某个顶点。因此，一种可靠的代数方法是顶点法：计算可行域所有顶点的坐标，然后将目标函数代入各个顶点求值。对于最大化问题，给出最高值的顶点就是最优解；对于最小化，则是最低值。顶点可通过求相交边界线的联立方程得到。例如，直线2x + y = 30与x + 2y = 24相交，解方程组得x = 12, y = 6。将P = 5x + 4y代入(12,6)得到P=84。将此值与其他顶点比较，即可确定最优解。

6. Integer Solutions and Discrete Variables | 整数解与离散变量

In many practical problems, decision variables must be integers – you cannot produce 2.7 bicycles or hire 1.5 workers. If the LP problem demands integer solutions, the optimal vertex might have non-integer coordinates. In that case, you must test integer points near the vertex that still lie within the feasible region. This is often done by drawing a grid of integer points inside the feasible region and checking those closest to the continuous optimum. Integer programming is specifically examined in OCR: always read the question carefully to see if integer constraints are required, and be prepared to check several integer pairs.

在许多实际问题中，决策变量必须为整数——你不可能生产2.7辆自行车或雇用1.5名工人。若LP问题要求整数解，而最优顶点的坐标可能不是整数。此时，你必须测试该顶点附近仍在可行域内的整数点。通常通过在可行域内画出整数网格，并检查接近连续最优解的点来实现。整数规划在OCR考试中有明确要求：务必仔细审题，查看是否需要整数约束，并准备好检查多个整数对。

7. Common Constraints and Special Cases | 常见约束与特殊情况

Constraints often take forms beyond simple linear inequalities. For example, a company may need to produce at least twice as many of product A as product B, leading to x ≥ 2y. Or there might be a minimum production requirement x + y ≥ 50. Be aware of multiple-choice constraints like ‘no more than 40 units in total’ (x + y ≤ 40) versus ‘the sum of the numbers is at least 20’ (x + y ≥ 20). Special cases include an unbounded feasible region (extending infinitely in some direction) – the objective function may not have a maximum if it increases in that unbounded direction. An infeasible region occurs when constraints contradict, yielding no common intersection. Both cases can appear in problem analysis questions.

约束条件有时不止简单的线性不等式。例如，一家公司可能需要生产的产品A数量至少是产品B的两倍，即x ≥ 2y。或者可能要求最低总产量x + y ≥ 50。注意多项选择类约束，如“总数不超过40”（x + y ≤ 40）对比“总和至少20”（x + y ≥ 20）。特殊情况包括无界可行域（在某方向无限延伸）——如果目标函数沿该无界方向增加，则可能不存在最大值。不可行域出现在约束条件相互矛盾，没有公共交集时。这两种情况都可能出现在问题分析题中。

8. Sensitivity Analysis and Shadow Prices | 灵敏度分析与影子价格

In OCR A-Level, a basic form of sensitivity analysis may be examined: understanding how changes in a resource’s availability affect the optimum. The ‘shadow price’ (or dual value) of a constraint is the amount by which the objective function would improve if the constraint’s right-hand side were increased by one unit. It is found by identifying which constraints are binding (active) at the optimum – those that the solution exactly meets as equalities. If a constraint is non-binding, its shadow price is zero. Graphically, shifting a binding constraint’s line outward slightly moves the optimal vertex; recalculating gives the incremental improvement. This concept links to economic interpretation and is assessed in comprehension-style questions.

OCR A-Level可能考查一种基本的灵敏度分析形式：理解资源可用量的变化如何影响最优解。一个约束条件的“影子价格”（或称对偶值）是指若将该约束的右侧值增加一个单位，目标函数能改善的数量。通过识别哪些约束在最优解处是紧的（即起作用的，解恰好满足等式）来求出。若某约束是非紧的，其影子价格为零。从图形上看，将紧约束的直线稍向外平移会使最优顶点移动；重新计算即可得到增量改进。此概念与经济解释相关，并在理解类题目中进行考查。

9. Worked Example: Maximising Profit | 例题解析：利润最大化

A factory produces two types of gadgets, X and Y. Each X requires 2 hours of assembly and 1 hour of painting. Each Y requires 1 hour of assembly and 3 hours of painting. There are 100 assembly hours and 90 painting hours available. The profit per unit is £30 for X and £40 for Y. Formulate and solve.

某工厂生产两种小器具X和Y。每件X需2小时组装和1小时喷漆。每件Y需1小时组装和3小时喷漆。可用组装时间为100小时，喷漆时间为90小时。每件X利润为£30，Y为£40。试建模并求解。

Let x = number of X, y = number of Y. Constraints: 2x + y ≤ 100 (assembly), x + 3y ≤ 90 (painting), x ≥ 0, y ≥ 0. Objective: Maximise P = 30x + 40y.

设x为X的数量，y为Y的数量。约束：2x + y ≤ 100（组装），x + 3y ≤ 90（喷漆），x ≥ 0, y ≥ 0。目标：最大化P = 30x + 40y。

Graph the lines: 2x + y = 100 passes through (50,0) and (0,100). x + 3y = 90 passes through (90,0) and (0,30). The feasible region is a quadrilateral with vertices at (0,0), (50,0), (0,30), and the intersection of the two lines. Solve: 2x + y = 100 and x + 3y = 90. From first, y = 100 − 2x. Sub in second: x + 3(100 − 2x) = 90 → x + 300 − 6x = 90 → −5x = −210 → x = 42. Then y = 100 − 84 = 16. Intersection is (42,16).

作图：直线2x + y = 100过(50,0)和(0,100)。x + 3y = 90过(90,0)和(0,30)。可行域是一个四边形，顶点为(0,0), (50,0), (0,30)以及两条直线的交点。求解：2x + y = 100和x + 3y = 90。由第一式得y = 100 − 2x。代入第二式：x + 3(100 − 2x) = 90 → x + 300 − 6x = 90 → −5x = −210 → x = 42。则y = 100 − 84 = 16。交点为(42,16)。

Evaluate P at each vertex: (0,0): P=0; (50,0): 30×50=1500; (0,30): 40×30=1200; (42,16): 30×42 + 40×16 = 1260 + 640 = 1900. Maximum profit is £1900, making 42 X and 16 Y. Since both are integers, no further integer adjustment is needed.

计算各顶点的P值：(0,0): P=0；(50,0): 30×50=1500；(0,30): 40×30=1200；(42,16): 30×42 + 40×16 = 1260 + 640 = 1900。最大利润为£1900，生产42件X和16件Y。因为已是整数，无需进一步整数调整。

10. Common Pitfalls and How to Avoid Them | 常见错误与避免方法

Students frequently lose marks by not defining variables explicitly, omitting units from the objective, or mislabelling axes on graphs. Another pitfall is shading the wrong side of the inequality – always test the origin unless the line passes through it. When finding intersection points, solve the equations accurately; a minor slip can change the optimum. For integer programming, simply rounding the continuous optimum to the nearest integer is often wrong because the rounded point may fall outside the feasible region. Always check that integer candidates satisfy all constraints. Finally, answer the question in context: write a concluding sentence in words stating the quantities and the value of the objective.

学生常犯的错误包括未明确定义变量、目标函数遗漏单位、图表坐标轴标注不清。另一误区是给不等式涂错了阴影侧——除非直线穿过原点，否则总是测试原点来判断。求交点时务必准确解方程；一个小失误就可能改变最优解。对于整数规划，简单地将连续最优解四舍五入到最近的整数往往是错误的，因为舍入后的点可能落在可行域之外。必须确保整数候选点满足所有约束。最后，在上下文中回答问题：用文字写出总结句，说明数量及目标函数的值。

11. Exam Strategy and Revision Tips | 应试策略与复习提示

Linear programming questions in OCR exams typically allocate 8–15 marks and are structured in parts: defining variables, writing constraints, drawing the graph, finding the optimal solution, and interpreting the result. Practice completing a full LP problem on graph paper under timed conditions. Memorise the key steps: define, constrain, graph, vertices, evaluate, conclude. For revision, use past paper questions and focus on problems with multiple constraints and integer requirements. When checking answers, verify that the optimum vertex indeed lies at the intersection of the binding constraints, and confirm that the objective value makes sense within the context. Mastery comes with repeated, structured practice.

OCR考试中线性规划题目通常为8–15分，且结构层次分明：定义变量、写出约束条件、绘制图形、求最优解、解释结果。练习在计时条件下于坐标纸上完成完整的LP题目。熟记关键步骤：定义、约束、作图、求顶点、求值、总结。复习时，使用往年真题，重点关注有多重约束和整数要求的题目。检查答案时，核实最优顶点确实位于紧约束的交点上，并确认目标值在题设情景中合理。通过有结构的反复练习即可精通。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）