📚 IGCSE CIE Mathematics: Linear Programming – Key Concepts Explained | IGCSE CIE 数学：线性规划考点精讲
Linear programming is a method used to find the maximum or minimum value of a quantity, subject to certain constraints that are expressed as linear inequalities. In IGCSE CIE Mathematics (Extended), you are expected to formulate constraints, graph them on a coordinate plane, identify the feasible region, and then use an objective function to determine the optimal solution. This topic blends algebraic manipulation with graphical interpretation and is a favourite for examiners testing your ability to link different areas of the syllabus.
线性规划是一种在给定若干线性不等式约束条件下，求解某个量最大值或最小值的方法。在 IGCSE CIE 数学（扩展）课程中，要求考生能够将实际问题转化为约束条件，在坐标平面上绘出图像，确定可行区域，然后利用目标函数找到最优解。这一专题融合了代数运算与图形解释，是考官用来考查你综合运用不同知识点的经典题型。
1. What is Linear Programming? | 什么是线性规划？
Linear programming deals with optimising (maximising or minimising) a linear expression, called the objective function, under a set of linear inequalities known as constraints. The variables are usually x and y, representing real-world quantities such as number of products, hours of work, or cost.
线性规划处理的是在一组被称为约束条件的线性不等式下，优化（最大化或最小化）一个线性表达式，即目标函数。变量通常为 x 和 y，代表现实世界中的数量，如产品数量、工作时长或成本。
In IGCSE, all constraints and the objective function are linear, meaning they form straight lines when plotted. The solution is always found at the boundary of the feasible region, typically at one of its vertices, unless the problem asks for integer solutions or the objective function is parallel to a constraint line.
在 IGCSE 考试中，所有约束条件和目标函数都是线性的，这意味着它们的图像都是直线。解总是位于可行区域的边界上，通常在其某个顶点处取得，除非题目要求整数解或目标函数与某条约束线平行。
You will see questions like: ‘A factory produces x chairs and y tables. The profit per chair is £20 and per table is £30. Given restrictions on labour and materials, find the maximum profit.’ This is a classic linear programming scenario.
你会遇到这样的问题：”某工厂生产 x 把椅子和 y 张桌子。每把椅子利润为 20 英镑，每张桌子利润为 30 英镑。在劳动力和材料的限制下，求最大利润。” 这正是线性规划的经典场景。
2. Formulating Constraints from Word Problems | 从文字题中建立约束条件
The first step is to translate the written statements into linear inequalities. Look for phrases like ‘at least’, ‘no more than’, ‘minimum’, ‘maximum’, ‘limited to’, ‘must not exceed’. Each constraint will be an inequality involving x and y.
第一步是将文字描述转化为线性不等式。注意”至少”、”不超过”、”最少”、”最多”、”限制为”、”不得超过”等关键词。每个约束条件都将是一个关于 x 和 y 的不等式。
For example, ‘the total number of chairs and tables produced cannot exceed 100’ becomes x + y ≤ 100. ‘The number of chairs must be at least twice the number of tables’ becomes x ≥ 2y. Always define your variables clearly at the start.
例如，”生产椅子和桌子的总数不能超过 100″可转化为 x + y ≤ 100。”椅子的数量必须至少是桌子数量的两倍”转化为 x ≥ 2y。务必在开始时就清晰地定义变量。
Non-negativity constraints: x ≥ 0, y ≥ 0 are almost always implied because you cannot produce a negative number of items. However, sometimes the context restricts x and y further, e.g., only whole numbers are allowed.
非负约束：x ≥ 0, y ≥ 0 几乎总是默认成立的，因为不可能生产负数的物品。但有时题目的语境会对 x 和 y 有进一步限制，例如只允许取整数。
3. Graphing Linear Inequalities | 绘制线性不等式图像
Each constraint is drawn as a straight line. To graph an inequality like 2x + y ≤ 10, first draw the boundary line 2x + y = 10. Use a solid line if the inequality includes equality (≤ or ≥) and a dashed line if it is strict (< or >).
每个约束条件都绘制为一条直线。要绘制如 2x + y ≤ 10 的不等式，首先画出边界线 2x + y = 10。如果不等式包含等号（≤ 或 ≥），则使用实线；若是严格不等式（< 或 >），则使用虚线。
Then determine which side of the line satisfies the inequality. Pick a test point not on the line, usually (0,0) if it is not on the line. Substitute into the inequality: if it holds, shade the side containing the test point; if not, shade the opposite side.
接着判断直线的哪一侧满足不等式。选取一个不在直线上的测试点，若 (0,0) 不在直线上则常用它。代入不等式：若成立，则测试点所在侧为解集，加以阴影标识；若不成立，则对侧为解集。
When multiple inequalities are graphed, the feasible region is the area where all shaded regions overlap. Label this region clearly, often with the letter R. Use a ruler and neat, accurate graphs – examiners will check that your lines intersect at correct points.
当绘制多个不等式时，可行区域就是所有阴影区域重叠的部分。用字母 R 清晰地标出该区域。作图时务必使用直尺，保证精度——考官会检查各直线是否在正确的点相交。
4. Identifying the Feasible Region & Vertices | 确定可行区域及其顶点
The feasible region is the set of all points that satisfy all constraints simultaneously. It is usually a polygon, though it can be unbounded in some problems. In IGCSE, it is almost always a bounded polygon, making optimisation straightforward.
可行区域是指同时满足所有约束条件的所有点的集合。它通常是一个多边形，尽管在某些问题中可能是无界的。在 IGCSE 中，几乎总是一个有界多边形，这使得优化过程变得简单。
Find the coordinates of the vertices (corners) of the feasible region by solving the equations of the intersecting lines simultaneously. These vertices are crucial because the optimal value of the objective function will occur at one of these points, provided the solution is not an integer point away from the boundary.
通过联立相邻边界线的方程组，求出可行区域各顶点（角点）的坐标。这些顶点至关重要，因为目标函数的最优值将出现在其中某个顶点处（除非最优解是偏离边界的一个整数点）。
Double-check that each vertex satisfies all constraints, including the non-negativity ones. Even a small arithmetic slip in solving simultaneous equations can cost you marks, so verify by substitution.
务必核对每个顶点是否满足所有约束条件，包括非负约束。解联立方程时哪怕一个小的计算错误都可能导致失分，因此要通过代入法验证。
5. The Objective Function | 目标函数
The objective function is the expression you want to maximise or minimise, such as profit P = 3x + 2y or cost C = 5x + 4y. It is always linear in x and y. The goal is to find the highest or lowest possible value of this function within the feasible region.
目标函数是你希望最大化或最小化的表达式，例如利润 P = 3x + 2y 或成本 C = 5x + 4y。它总是关于 x 和 y 的线性表达式。目标是在可行区域内找到该函数可能取得的最大值或最小值。
In graphical terms, the objective function can be represented by a family of parallel lines, each with the same slope. For maximising P, you slide a line with slope –(coefficient of x)/(coefficient of y) as far as possible in the direction of increase without leaving the feasible region.
从图形上看，目标函数可以用一族具有相同斜率的平行线来表示。要使 P 最大化，可让一条斜率为 –(x系数)/(y系数) 的直线尽可能向增大的方向平移，直至它刚好要离开可行区域为止。
At IGCSE, you are more often asked to evaluate the objective function at each vertex and pick the optimum. This ‘vertex method’ is safe and reliable as long as the region is bounded and the optimum is at a vertex.
在 IGCSE 考试中，更常见的做法是计算目标函数在每个顶点的值，然后选出最优值。这种”顶点法”只要可行区域有界且最优解位于顶点处，就是安全可靠的。
6. The Vertex Method – Step by Step | 顶点法——逐步详解
Step 1: List all vertices of the feasible region. Step 2: Substitute each into the objective function. Step 3: Identify the maximum or minimum value as required. Step 4: Write the conclusion as a sentence, stating the value and the corresponding x and y.
第一步：列出可行区域的所有顶点。第二步：将每个顶点代入目标函数。第三步：按要求确定最大值或最小值。第四步：用一句话写出结论，明确指出该值以及对应的 x 和 y。
Example: Maximise P = 20x + 30y subject to x + y ≤ 50, x ≤ 24, y ≤ 30, x ≥ 0, y ≥ 0. Vertices might be (0,0), (24,0), (24,26), (20,30), (0,30). Evaluations: (0,0)→0, (24,0)→480, (24,26)→20×24+30×26=480+780=1260, (20,30)→400+900=1300, (0,30)→900. Maximum P = 1300 at (20,30).
示例：在约束 x + y ≤ 50, x ≤ 24, y ≤ 30, x ≥ 0, y ≥ 0 下最大化 P = 20x + 30y。顶点可能为 (0,0)、(24,0)、(24,26)、(20,30)、(0,30)。计算得：(0,0)→0，(24,0)→480，(24,26)→1260，(20,30)→1300，(0,30)→900。故 P 的最大值为 1300，在 (20,30) 处取得。
Notice that the maximum was not at the extreme x or y alone, but at a point balancing both constraints. This illustrates why checking all vertices is essential.
注意到最大值并不在 x 或 y 的极端值处，而是在一个平衡了两个约束的点上。这说明了为什么检查所有顶点是必不可少的。
7. Parallel Objective Function and Multiple Solutions | 目标函数与约束平行及多解情况
If the objective function line is parallel to one of the constraint lines that forms an edge of the feasible region, there may be infinitely many solutions along that edge. In such a case, the maximum or minimum occurs at both endpoints of the segment, and at any point on the line segment joining them.
如果目标函数的直线与构成可行区域一条边的约束线平行，则可能在该边上有无穷多组解。此时，最大值或最小值会在这条线段的两个端点以及连接它们的线段上任何一点处同时取得。
For instance, if P = 2x + 2y and a constraint is x + y = 10 that forms part of the boundary, then any point on that segment gives the same P value. Examiners sometimes design questions to test if you notice this parallelism.
例如，若 P = 2x + 2y，而约束 x + y = 10 构成了边界的一部分，则该线段上的任意点都会得到相同的 P 值。考官有时会故意设计此类题目，以检验你是否注意到这种平行关系。
When you spot that the objective function’s gradient matches a constraint gradient, state clearly that there are multiple solutions, or that all points on the segment give the optimal value. The mark scheme usually awards marks for recognising this.
当你发现目标函数的梯度与某约束的梯度相同时，要明确说明存在多组解，或者指出线段上所有点都能取得最优值。评分标准通常会为识别出这一点而给分。
8. Integer Solutions and Mixed-Integer Constraints | 整数解与混合整数约束
In real-world problems, x and y often represent items that cannot be fractional, so we need integer solutions. After finding the optimal vertex, if its coordinates are not integers, you may need to test nearby integer points within the feasible region to find the best integer solution.
在现实问题中，x 和 y 通常代表不可分割的物体，因此需要整数解。在找到最优顶点后，若其坐标并非整数，你可能需要测试可行区域内附近整数点，以找到最佳整数解。
Sometimes the optimum integer solution is not the nearest integer point to the continuous optimum. You must draw or visualise a grid and carefully check points that satisfy all constraints and give the highest (or lowest) objective value.
有时最优整数解并非最靠近连续最优解的整数点。你必须画出或想象坐标网格，仔细检查那些同时满足所有约束且使目标函数值最大（或最小）的整数点。
Be systematic: list feasible integer x-values, then find corresponding possible y-values from constraints. Substitute each candidate into the objective function. This method guarantees you do not miss a hidden optimum.
要有条理：列出可行整数 x 值，然后根据约束求出相应的可能 y 值。将每个候选点代入目标函数。这种方法能保证你不会遗漏隐藏的最优解。
9. Drawing the Search Line and Using Ruler Method | 画出搜索线及平移法
Instead of (or in addition to) evaluating vertices, you can draw a line representing the objective function equal to some convenient value, for example P = 0 or P = 100. Then move a ruler parallel to this line across the feasible region in the direction that increases (or decreases) P until it just touches the region at the last point.
除了（或作为替代）顶点求值法，你可以画一条代表目标函数等于某个方便取值的直线，例如 P = 0 或 P = 100。然后用直尺平行于该线，沿使 P 增大（或减小）的方向平移，直至这条线刚好在最后一点与可行区域相切。
The last point the line touches before leaving the region is the optimal solution. This graphical approach helps visualise why the vertex gives the optimum and is particularly useful when the objective function’s slope is close to a constraint’s slope.
直线离开区域前最后接触的点就是最优解所在。这种图解方法有助于直观理解为什么最优解出现在顶点处，并且在目标函数斜率接近某条约束的斜率时尤为有用。
Use a clear ruler and mark the direction of increase with an arrow. Label the optimal point clearly with its coordinates. This method may be explicitly required in some questions, so practise both vertex and ruler techniques.
使用清晰的直尺，并用箭头标出增大方向。在最优解处清晰标注坐标。某些题目可能明确要求采用这种方法，因此顶点法和平移法都要熟练。
10. Common Mistakes to Avoid | 常见错误辨析
Mistake 1: Forgetting to include x ≥ 0, y ≥ 0 as constraints. Unless the problem implies otherwise, these are essential. Mistake 2: Misinterpreting the inequality direction, especially with phrases like ‘at most’ vs ‘at least’.
错误一：忘记将 x ≥ 0, y ≥ 0 列为约束条件。除非题目另有暗示，这些条件都是必要的。错误二：对不等号方向理解错误，尤其是在遇到”最多”和”至少”这类表述时。
Mistake 3: Using a dashed line where a solid line is needed (or vice versa). Remember: ≤ or ≥ → solid; < or > → dashed. Mistake 4: Shading the wrong side of a line. Always test with (0,0) if possible.
错误三：应该用实线的地方画了虚线（或反之）。记住：≤ 或 ≥ → 实线；< 或 > → 虚线。错误四：混淆了直线的解集侧。务必尽可能用 (0,0) 进行测试。
Mistake 5: Not reading the question carefully – some ask for the maximum value of the objective function, others ask for the values of x and y that give it. Mistake 6: Rounding off coordinates incorrectly when seeking integer solutions.
错误五：没有仔细审题——有些题目要求给出目标函数的最大值，有些则要求给出取得该值时的 x 和 y。错误六：在求整数解时对坐标进行了错误的取整。
11. Exam-Style Question Walkthrough | 考试题型演练
Question: A garden centre sells x rose bushes and y fruit trees. Space allows at most 60 plants in total. The number of rose bushes must be at least 10 and at most three times the number of fruit trees. Write down the constraints and find the maximum number of plants in total if the centre earns a profit of £5 per rose bush and £8 per fruit tree.
题目：某园艺中心出售 x 株玫瑰和 y 棵果树。场地最多可容纳共 60 棵植物。玫瑰的数量必须至少为 10，且最多为果树数量的三倍。写出约束条件，并求若每株玫瑰利润为 5 英镑、每棵果树利润为 8 英镑时，总利润的最大值。
Constraints: x + y ≤ 60; x ≥ 10; x ≤ 3y; x ≥ 0, y ≥ 0. Objective function: P = 5x + 8y. Plot lines: x + y = 60, x = 10, x = 3y. Feasible region vertices: Intersection of x=10 & x+y=60 → (10,50); x=10 & x=3y → (10, 10/3) but y must be integer? Not yet – but region boundary includes (10, 3.33); intersection of x=3y & x+y=60 → 3y+y=60 → y=15, x=45 → (45,15). Check (10,50) gives P=5×10+8×50=50+400=450; (45,15) gives 5×45+8×15=225+120=345. However, note x ≥ 10 and x ≤ 3y: at y=10/3, x=10 is valid but profit around 136.7, but lower. So maximum profit is 450 at (10,50).
约束条件：x + y ≤ 60；x ≥ 10；x ≤ 3y；x ≥ 0, y ≥ 0。目标函数：P = 5x + 8y。绘制直线：x + y = 60, x = 10, x = 3y。可行区域顶点：x=10 与 x+y=60 的交点→ (10,50)；x=10 与 x=3y 的交点→ (10, 10/3) 即 y≈3.33；x=3y 与 x+y=60 的交点→ y=15, x=45 即 (45,15)。计算各顶点 P 值：(10,50)→450，(45,15)→345。最大值 450 在 (10,50) 处取得。
The solution shows the importance of sketching accurately. Even though (45,15) is further along the x+y=60 line, the profit coefficients favour y, pushing the optimum toward the high-y vertex.
该解法显示了准确画图的重要性。尽管 (45,15) 在 x+y=60 直线上更靠右，但因利润系数有利于 y，最优解移向了 y 值较高的顶点。
12. Tips for Scoring Full Marks | 获取高分的技巧
Always define variables first: ‘Let x be … and y be …’. Use graph paper if provided; if not, draw your axes with a ruler and label scales evenly. Write the equation of each line next to it and shade the rejected region, leaving the feasible region unshaded – this is often clearer.
务必首先定义变量：”设 x 为……，y 为……”。若提供坐标纸则使用坐标纸；若没有，则用直尺画出坐标轴并均匀标记刻度。在每条直线旁写出其方程，并遮盖（阴影）不可行侧，从而留下清晰的无阴影可行区域——这通常更清晰。
Explicitly find and state the coordinates of all relevant intersection points. Show your method for solving simultaneous equations. After obtaining the optimum, write a final answer in the context of the problem, e.g., ‘The maximum profit is £450 when 10 rose bushes and 50 fruit trees are sold.’
明确求出并写出所有相关交点的坐标。展示解联立方程的过程。得出最优值后，在题目情境中写出最终答案，例如：”当售出 10 株玫瑰和 50 棵果树时，可获得最大利润 450 英镑。”
Check if the question requires integer values; if the optimal point has non-integer coordinates, investigate nearby lattice points. Finally, review that your inequalities match the wording and that your shading corresponds correctly.
确认题目是否要求整数解；若最优解包含非整数坐标，则需考察附近的格点。最后，复核不等式是否与文字描述相符，以及阴影区域是否正确对应。
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