📚 Linear Programming in IGCSE Math: Key Concepts & Exam Tips | IGCSE 数学：线性规划 考点精讲
Linear programming is an optimisation technique used to find the maximum or minimum value of a linear function subject to a set of linear inequalities. In IGCSE Mathematics, this topic tests your ability to graph constraints, identify a feasible region, and systematically determine the best possible outcome. Understanding how to interpret real-world problems as mathematical models is the core skill here.
线性规划是一种优化方法，用于在满足一组线性不等式的条件下，求一个线性函数的最大值或最小值。在IGCSE数学中，这个主题考察你绘制约束条件图像、识别可行域、并系统性地找到最优解的能力。把实际问题转化为数学模型并加以理解，是这里的核心技能。
1. Introduction to Linear Programming | 线性规划简介
Linear programming involves two or more variables, usually x and y, and a set of constraints expressed as linear inequalities. The objective is to maximise or minimise a linear expression called the objective function, often representing profit, cost, or output. The solution must lie within the feasible region, which is the intersection of all constraint graphs.
线性规划涉及两个或多个变量（通常是 x 和 y），以及一组用线性不等式表示的约束条件。目标是最大化或最小化一个线性表达式，叫作目标函数，通常代表利润、成本或产量。解必须落在可行域内，即所有约束条件图像的交集区域。
The constraints are typically given as inequalities like ax + by ≤ c, ax + by ≥ c, x ≥ 0, or y ≥ 0. The last two are non‑negativity constraints, reflecting that variables often cannot take negative values in real‑life scenarios. A clear sketch of the region is essential before any optimisation step.
约束条件通常以不等式形式出现，如 ax + by ≤ c、ax + by ≥ c、x ≥ 0 或 y ≥ 0。后两个是非负限制，反映实际场景中变量通常不能取负值。在进行任何优化步骤之前，清晰地绘制出该区域是至关重要的。
2. Graphing Linear Inequalities | 绘制线性不等式
To graph an inequality such as 2x + y ≤ 10, first draw the boundary line 2x + y = 10. Use a solid line for ≤ or ≥, and a dashed line for strict inequalities < or >. Then determine which side of the line satisfies the inequality by testing a point, usually (0,0) if it does not lie on the line.
要绘制一个不等式，如 2x + y ≤ 10，首先画出边界线 2x + y = 10。对于 ≤ 或 ≥ 使用实线，对于严格的不等式 < 或 > 使用虚线。然后通过测试一个点来判断线的哪一侧满足不等式，通常使用 (0,0)，如果它不在边界线上的话。
For example, substituting (0,0) into 2x + y ≤ 10 gives 0 ≤ 10, which is true, so shade the side containing the origin. If the test point does not satisfy the inequality, shade the opposite side. Always label the shaded region or use arrows to indicate the feasible side.
例如，将 (0,0) 代入 2x + y ≤ 10 得到 0 ≤ 10，成立，因此涂色包含原点的那一侧。如果测试点不满足不等式，就涂色另一侧。务必标注出涂色区域或用箭头标示可行侧。
The intersection of all shaded regions forms the feasible region. In IGCSE, you are often asked to clearly mark this region with a letter R or leave it unshaded while shading the unwanted areas. Read the question carefully to follow the required convention.
所有涂色区域的交集构成了可行域。在IGCSE考试中，题目通常要求你清晰地用字母 R 标出该区域，或者采用“涂掉不可行区域、保留可行区域不涂色”的方式。仔细审题，遵循题目规定的做法。
3. Identifying the Feasible Region | 识别可行域
The feasible region is the set of all points that satisfy every constraint simultaneously. It is a convex polygon (or sometimes an unbounded area) bounded by the constraint lines. The vertices of this polygon are called corner points, and they play a vital role in finding the optimal solution.
可行域是指同时满足所有约束条件的所有点的集合。它通常是一个凸多边形（有时是无界区域），由约束线围成。该多边形的顶点叫作极点，在寻找最优解的过程中起着至关重要的作用。
If the feasible region is bounded, the objective function will attain both a maximum and a minimum at one of the vertices. If the region is unbounded, a maximum or minimum might still exist, but we must check that the objective function is not tending to infinity. In IGCSE exams, the region is almost always a closed polygon.
如果可行域是有界的，目标函数必在其某一个顶点处同时取得最大值和最小值。如果区域是无界的，最大值或最小值仍可能存在，但我们必须验证目标函数不会趋向无穷。在IGCSE考试中，区域几乎总是封闭的多边形。
4. Corner Points and Optimal Solutions | 顶点与最优解
According to the corner-point principle, the maximum or minimum of a linear objective function over a closed convex polygon will occur at a corner point. This means we only need to evaluate the objective function at the vertices, rather than checking every possible point inside the region.
根据顶点原理，线性目标函数在封闭凸多边形上的最大值或最小值必出现在一个顶点处。这表示我们只需在顶点处计算目标函数值，而无需检查区域内部的每一个点。
To find the coordinates of the vertices, solve the simultaneous equations for each pair of boundary lines that intersect at a corner. Be sure to check that the intersection point satisfies all other inequalities; if it does not, it is not a true vertex of the feasible region.
要找到顶点的坐标，需求解在顶点处相交的每一对边界线的联立方程。务必检验该交点是否满足所有其他不等式；如果不满足，它就不是可行域真正的顶点。
Once all corner points are identified, substitute each into the objective function and compare the values. The largest value gives the maximum; the smallest gives the minimum. This method is reliable and expected in IGCSE solutions.
一旦确定了所有顶点，就将每个点代入目标函数并比较函数值。最大值对应最大的函数值，最小值对应最小的函数值。这种方法可靠且是IGCSE答题所期望的。
5. The Objective Function | 目标函数
The objective function is the expression to be maximised or minimised, for example P = 3x + 2y. On a graph, lines of constant objective value are called isolines or level lines. They are parallel lines with slope determined by the coefficients of x and y.
目标函数是需要被最大化或最小化的表达式，比如 P = 3x + 2y。在图形上，目标函数值相等的直线称为等值线或水平线。它们是互相平行的直线，其斜率由 x 和 y 的系数决定。
To visualise optimisation, you can draw a ruler line with the gradient of -a/b (from P = ax + by) and slide it parallel to itself across the feasible region. The last point it touches when moving in the direction of increasing P yields the maximum; the first point gives the minimum.
为了直观地理解优化过程，你可以画出斜率为 -a/b 的参照线（来自 P = ax + by），并将它平行地推移过可行域。沿着使 P 增大的方向移动时，最后接触的那个点即为最大值点；最早接触的点即为最小值点。
In IGCSE, the equation of the objective function is usually given as part of the question. You may be asked to write down the value, find the coordinates that give the optimum, or interpret the result in context. Clear working is essential to gain full marks.
在IGCSE中，目标函数的方程通常是作为题目的一部分给出的。你可能需要写出最优值、找出最优解对应的坐标，或者结合具体情境解释结果。清晰的解题步骤对获得满分至关重要。
6. Maximisation Problems | 最大化问题
A typical maximisation problem asks for the largest possible value of profit, revenue, or production level subject to resource constraints. The constraints might represent limited labour hours, raw materials, or machine capacity. All inequalities must be set up correctly from the problem statement.
典型的最大化问题要求你在资源限制下，求出利润、收入或产量的最大可能值。这些限制可能代表有限的劳动时间、原材料或机器产能。必须根据问题描述正确地建立所有不等式。
For instance, a factory produces two items, X and Y. Each X requires 2 hours of labour and each Y requires 3 hours. If total labour is at most 48 hours, the constraint is 2x + 3y ≤ 48. Similarly, other constraints could involve material limits or demand requirements. The objective could be Profit = 5x + 7y.
例如，一家工厂生产两种产品 X 和 Y。每件 X 需要 2 小时劳动力，每件 Y 需要 3 小时。如果总劳动时间最多为 48 小时，则约束条件为 2x + 3y ≤ 48。类似地，其他约束可能涉及材料限制或需求要求。目标函数可以是 利润 = 5x + 7y。
After graphing all constraints and identifying the feasible region, find the corner points. Substitute each into the profit function and choose the maximum. Don’t forget to state the answer with correct units and in the context of the question, such as ‘Produce 8 units of X and 12 units of Y for a maximum profit of £124.’
在绘制完所有约束条件并确定可行域之后，找出所有顶点。将每个点代入利润函数，选择最大值。别忘了在回答中带上正确的单位并结合题意，例如“生产 8 件 X 和 12 件 Y，可获得最大利润 124 英镑”。
7. Minimisation Problems | 最小化问题
Minimisation problems often involve reducing costs, waste, or travel distance. The logic is identical: define variables, write constraints as inequalities, graph the feasible region, and test corner points. The only difference is that you look for the smallest value of the objective function.
最小化问题通常涉及降低成本、损耗或行程距离。其逻辑完全相同：定义变量、将约束表示为不等式、绘制可行域、检验顶点。唯一的区别在于你要寻找目标函数的最小值。
Consider a diet problem where a person needs at least 60 g of protein and 40 g of fibre. Two foods provide these nutrients, and costs differ. Constraints will be of the type ‘at least’, so inequalities like x + 2y ≥ 60. The feasible region may be unbounded, but the minimum cost still occurs at a vertex if it exists.
考虑一个饮食问题：某人需要至少 60 克蛋白质和 40 克纤维。两种食物提供这些营养，且成本不同。约束条件会是“至少”类型，因此出现 x + 2y ≥ 60 这样的不等式。可行域可能是无界的，但如果最小值存在，它仍出现在某个顶点。
When a feasible region is unbounded, check that the objective function does not decrease indefinitely as you move outward. Often a combination of constraints prevents this, and a single vertex gives the minimum cost. Always verify with the corner-point method.
当可行域无界时，要检验当你向外移动时目标函数是否不会无限减小。通常，多个约束条件共同作用可防止这种情况，某一个单独的顶点会带来最小成本。始终用顶点法加以验证。
8. Integer Solutions and Integer Programming | 整数解与整数规划
In many real-life contexts, variables must be whole numbers. For example, you cannot produce 3.7 chairs or sell 5.2 tickets. If the optimal corner point has non‑integer coordinates, you must look for the best integer point within or on the boundary of the feasible region.
在许多实际情境中，变量必须是整数。例如，你不能生产 3.7 把椅子或者卖出 5.2 张票。如果最优顶点带有非整数坐标，你就必须在可行域内部或边界上寻找最佳的整数点。
Simply rounding the coordinates of the corner point might not give the true optimum because the rounded point could lie outside the feasible region or yield a sub‑optimal value. Instead, list integer points near the vertex, check feasibility, and evaluate the objective function to find the best one.
简单地对顶点坐标进行四舍五入可能无法得到真正的最优值，因为舍入后的点可能落在可行域之外，或者产生一个次优值。正确的做法是：列出顶点附近的整数点，检验其可行性，并计算目标函数值以找出最佳点。
IGCSE questions occasionally specify that solutions must be integers, or they ask for the number of whole-number combinations. This adds a small extra step but reinforces the understanding that mathematical solutions must be interpreted within real‑world constraints.
IGCSE 考题偶尔会明确要求解必须是整数，或者会询问整数组合的个数。这增加了一个额外的小步骤，但强化了一个认识：数学解必须结合实际约束条件来解释。
9. Writing Inequalities from Word Problems | 根据应用题写不等式
Translating a word problem into inequalities is a crucial skill. Look for keywords: ‘at most’ means ≤, ‘at least’ means ≥, ‘no more than’ means ≤, ‘must exceed’ means >, ‘cannot be less than’ means ≥. Phrases like ‘a maximum of’, ‘up to’, and ‘limited to’ all indicate upper bounds.
将应用题转化为不等式是关键技能。寻找关键词：“最多”对应 ≤，“至少”对应 ≥，“不超过”对应 ≤，“必须超过”对应 >，“不能少于”对应 ≥。“最大值为”、“高达”、“限于”等短语都表示上界。
Define the variables explicitly, e.g., ‘Let x be the number of standard packages and y be the number of deluxe packages.’ Then write each constraint as an inequality. Do not forget non‑negativity constraints x ≥ 0 and y ≥ 0 unless the context naturally implies them.
明确地定义变量，例如“设 x 为标准套餐的数量，y 为豪华套餐的数量”。然后将每条约束写成不等式。除非题意本身已暗含，否则不要忘记非负约束 x ≥ 0 和 y ≥ 0。
When a problem involves time, weight, or volume, consistency of units is vital. If labour is given in hours and production time in minutes, convert both to the same unit before writing inequalities. This prevents scaling errors that can lead to an incorrect feasible region.
当问题涉及时间、重量或体积时，单位的一致性至关重要。如果劳动时间以小时给出，而生产时间以分钟给出，在写不等式之前将两者转换为同一单位。这可以防止因缩放错误而导致的可行域不正确。
10. Common Exam Pitfalls | 常见考试陷阱
One common mistake is shading the wrong side of a boundary line. Always test a point, and do not rely on guessing the direction of the inequality arrow. Another error is misidentifying the feasible region by failing to consider all inequalities, especially the x ≥ 0 and y ≥ 0 constraints.
一个常见错误是涂色时搞错了边界线的两侧。务必测试一个点，不要凭猜测来判断不等号指向的方向。另一类错误是遗漏了某些不等式，特别是 x ≥ 0 和 y ≥ 0，从而导致识别出的可行域有误。
Using a solid line for a strict inequality or a dashed line for ≤/≥ will lose marks in the graphing part. Also, be careful when reading the scale of the graph. Slight inaccuracies in plotting can shift a vertex enough to change the objective value, so use precise coordinates.
对严格不等式使用了实线，或对 ≤/≥ 使用了虚线，都会在绘图部分失分。此外，阅读图形刻度时也要细心。绘图时的微小不精确可能使顶点位置偏移，从而改变目标函数值，因此要使用精确的坐标。
Many students forget to interpret the final answer in the given context. If the question asks ‘how many of each type should be produced?’, you must state the integer result, not just the value of the objective function. Always answer the specific question asked.
很多学生忘记结合给定情境解释最终的答案。如果题目问“每种应生产多少件？”，你必须给出整数结果，而不只是目标函数的值。一定要回答所问的具体问题。
11. Step-by-Step Problem-Solving Strategy | 逐步解题策略
Follow a structured approach to linear programming questions:
解答线性规划问题时，请遵循以下结构化步骤：
- Step 1: Identify the decision variables and define them clearly. / 步骤1：确定决策变量并清晰地定义它们。
- Step 2: Formulate the constraints as linear inequalities. Include non‑negativity restrictions if appropriate. / 步骤2：将约束条件表述为线性不等式。在适当情况下包含非负限制。
- Step 3: Graph the inequalities on a coordinate plane. Use a suitable scale and clearly label the lines. / 步骤3：在坐标平面上绘制不等式。使用合适的刻度并清楚地标记直线。
- Step 4: Identify and shade (or unshade) the feasible region exactly as instructed. / 步骤4：严格按照题目指令识别并涂色（或保留未涂色）可行域。
- Step 5: Find the coordinates of all corner points of the feasible region by solving simultaneous equations. / 步骤5：通过解联立方程求出可行域所有顶点的坐标。
- Step 6: Write down the objective function. / 步骤6：写出目标函数。
- Step 7: Evaluate the objective function at each corner point and tabulate the results. / 步骤7：在每个顶点处计算目标函数值，并列表展示结果。
- Step 8: Select the optimum value and state the corresponding variable values. / 步骤8：选出最优值并说明对应的变量取值。
- Step 9: Interpret the answer in the context of the problem, paying attention to integer requirements. / 步骤9：结合问题情境解释答案，注意整数要求。
Keeping a logical sequence of working not only helps avoid mistakes but also ensures examiners can follow your reasoning and award method marks even if a plotting slip occurs.
保持逻辑清晰的工作顺序不仅有助于避免错误，也能让考官看懂你的推理过程，即使绘图有轻微失误，依然能给予方法分。
12. Worked Example | 实例精练
A company produces two types of souvenirs, A and B. Each type A requires 1 hour of machining and 2 hours of assembly. Each type B requires 3 hours of machining and 1 hour of assembly. The machining department has a maximum of 36 hours available, while the assembly department has a maximum of 32 hours. The profit is £8 per type A and £10 per type B. Find the number of each type that should be produced to maximise profit.
一家公司生产两种纪念品 A 和 B。每件 A 需要 1 小时机加工和 2 小时组装。每件 B 需要 3 小时机加工和 1 小时组装。机加工部门最多可用 36 小时，组装部门最多可用 32 小时。每件 A 获利 8 英镑，每件 B 获利 10 英镑。求为最大化利润，每种应生产多少件。
Solution: Let x = number of type A, y = number of type B. Constraints:
Machine: 1x + 3y ≤ 36
Assembly: 2x + 1y ≤ 32
Non‑negativity: x ≥ 0, y ≥ 0.
解：设 x = A 型数量，y = B 型数量。约束条件：
机加工：1x + 3y ≤ 36
组装：2x + 1y ≤ 32
非负：x ≥ 0, y ≥ 0。
Corner points from intersections: (0,0), (0,12) from 1x + 3y = 36 and x=0, (16,0) from 2x + y = 32 and y=0, and intersection of the two lines: solving
x + 3y = 36
2x + y = 32
gives x = 12, y = 8.
顶点由各直线交点得出：(0,0), (0,12) 来自 x=0 与 x + 3y = 36, (16,0) 来自 y=0 与 2x + y = 32, 以及两直线交点：解方程组
x + 3y = 36
2x + y = 32
得到 x = 12, y = 8。
Objective function: Profit P = 8x + 10y.
目标函数：利润 P = 8x + 10y。
| Corner (x,y) | P = 8x + 10y |
|---|---|
| (0,0) | 0 |
| (0,12) | 120 |
| (16,0) | 128 |
| (12,8) | 8(12)+10(8)=96+80=176 |
The maximum profit of £176 is achieved at x = 12, y = 8. Since both values are integers, no further adjustment is needed. Answer: Produce 12 of type A and 8 of type B for a maximum profit of £176.
最大利润为 176 英镑，在 x = 12, y = 8 时取得。两者均为整数，无需进一步调整。答案：生产 12 件 A 型和 8 件 B 型，可获得最大利润 176 英镑。
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