Linear Programming | 线性规划考点精讲

📚 Linear Programming | 线性规划考点精讲

Linear programming is a powerful mathematical method used to find the best outcome – such as maximum profit or minimum cost – in a given model whose requirements are represented by linear relationships. In IGCSE AQA Mathematics, it involves interpreting real‑world constraints as linear inequalities, graphing them to form a feasible region, and then using an objective function to locate the optimal solution. This article breaks down every essential concept with clear explanations and exam tips.

线性规划是一种强大的数学方法，用于在由线性关系描述的模型中寻找最优结果，例如最大利润或最小成本。在 IGCSE AQA 数学中，你需要将现实中的约束条件转化为线性不等式，画出它们的图像形成可行区域，再借助目标函数找到最优解。本文通过清晰的中英讲解和考试技巧，逐一剖析所有核心考点。

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

Linear programming deals with the problem of optimising (maximising or minimising) a linear function subject to a set of linear inequalities. The variables, usually x and y, represent quantities that cannot be negative. The set of inequalities defines a feasible region on the coordinate plane, and the objective function’s highest or lowest value within that region gives the optimal decision.

线性规划处理的是在一组线性不等式的约束下，优化（最大化或最小化）一个线性函数的问题。变量通常用 x 和 y 表示，代表非负的量。这些不等式共同在坐标平面上围成一个可行区域，目标函数在该区域内的最大值或最小值就对应着最优决策。

2. Translating Words into Inequalities | 把文字转化为不等式

The first step in any linear programming problem is to express the given constraints as linear inequalities. Read each sentence carefully. Words like ‘at least’, ‘no more than’, ‘at most’, ‘minimum’ and ‘maximum’ signal inequality signs. For example, ‘the number of type‑A products cannot exceed twice the number of type‑B products’ becomes x ≤ 2y. ‘Total resources available are 100 units’ leads to an inequality like 3x + 4y ≤ 100.

任何线性规划问题的第一步都是把题目给出的约束条件用线性不等式表达出来。要仔细阅读每一句话。像“至少”、“不超过”、“最多”、“最小”、“最大”这样的词都提示着不等号的方向。例如，“A 类产品的数量不能超过 B 类产品数量的两倍”可以写成 x ≤ 2y；“可用资源总共为 100 单位”则会导出 3x + 4y ≤ 100 这样的不等式。

Always identify the variables first and state clearly what x and y represent, e.g. ‘Let x be the number of cakes and y be the number of biscuits.’ Remember to include the non‑negativity constraints x ≥ 0 and y ≥ 0 unless the context says otherwise.

首先要确定变量，并清楚地说明 x 和 y 代表什么，比如“设 x 为蛋糕的数量，y 为饼干的数量”。除非题目另有说明，否则一定要加上非负约束 x ≥ 0 和 y ≥ 0。

3. Graphing Linear Inequalities | 绘制线性不等式图像

To graph a linear inequality like 2x + y ≤ 10, first draw the boundary line 2x + y = 10 as a solid line (for ≤ or ≥) or a dashed line (for < or >). Use standard methods to plot the line: find where it crosses the axes – when x = 0, y = 10; when y = 0, x = 5. Join these intercepts.

要画出 2x + y ≤ 10 这样的线性不等式图像，首先画出边界直线 2x + y = 10。如果是 ≤ 或 ≥ 就画成实线，如果是 < 或 > 则画成虚线。使用常规方法画出直线：找到它与坐标轴的交点——当 x = 0 时 y = 10，当 y = 0 时 x = 5，然后连接这两个截点。

After drawing the line, decide which side of it satisfies the inequality. Pick a test point not on the line, such as (0,0). Substitute into 2x + y ≤ 10: 2(0)+0 = 0 ≤ 10, which is true, so shade the side containing (0,0). If the test point fails, shade the opposite side.

画好直线后，需要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，例如 (0,0)。代入 2x + y ≤ 10：2(0)+0 = 0 ≤ 10，不等式成立，因此涂上包含 (0,0) 的一侧。如果测试点不满足，则涂上另一侧。

4. Identifying the Feasible Region | 确定可行区域

When several inequalities are graphed together, the feasible region is the area where all shaded regions overlap. It is often a polygon, bounded or unbounded. On the graph, clearly label this region. The vertices (corner points) of this polygon are particularly important because the optimal solution of a linear programme always occurs at one of these vertices, provided the region is bounded and the objective function is linear.

当多个不等式画在同一坐标系中时，可行区域就是所有阴影部分重叠的区域。它通常是一个多边形，可以是有界的，也可以是无界的。在图上要清楚地标出这个区域。该多边形的顶点（角点）特别重要，因为只要区域有界且目标函数是线性的，线性规划的最优解一定出现在这些顶点中的某一个处。

Use different shading patterns or clearly hatch out the unwanted sides so the feasible region stands out. Examiners expect to see a neat graph with all lines labelled and the feasible region marked, for example with the letter ‘R’.

可以使用不同的阴影方向或把不满足条件的部分划掉，使可行区域一目了然。考官希望看到整洁的图形，所有直线都已标注，可行区域用字母（如 ‘R’）标出。

5. The Objective Function | 目标函数

The objective function is the expression you are trying to maximise or minimise. It is usually given in the form P = ax + by, where a and b are constants. For instance, a company wants to maximise profit: P = 5x + 3y, where x and y are the numbers of two products.

目标函数就是你要寻求最大值或最小值的那个表达式。它通常形如 P = ax + by，a 和 b 为常数。例如，某公司要最大化利润：P = 5x + 3y，其中 x 和 y 是两种产品的产量。

To find the optimal value, you cannot simply test every point inside the feasible region – there are infinitely many. Instead, use the property that the optimal solution lies at a vertex of the feasible region. The objective function line can be visualised by drawing the line ax + by = k for some value k, then sliding it parallel to itself until it just leaves the feasible region. The last point it touches gives the optimal solution.

要找到最优值，你无法把可行区域内的所有点都测试一遍——太多了。正确的方法是运用“最优解在可行区域顶点处”这一性质。你也可以通过画出目标函数的等值线 ax + by = k，然后把它向增大或减小的方向平行移动，直到刚好离开可行区域。最后接触的那个点就是最优解点。

6. Finding the Optimal Point – Vertex Method | 寻找最优解点——顶点法

The vertex method is the most reliable technique for IGCSE. Follow these steps:

顶点法是 IGCSE 考试中最可靠的方法，按以下步骤操作：

1. Accurately graph all inequalities and identify the feasible region.
1. 准确画出所有不等式的图像，确定可行区域。

2. Find the coordinates of every vertex of the feasible region. These usually require solving pairs of simultaneous equations from the boundary lines.
2. 求出可行区域每一个顶点的坐标。通常需要联立两条边界直线的方程来求解。

3. Substitute each vertex’s coordinates into the objective function and note the value.
3. 把每个顶点的坐标代入目标函数，算出对应的函数值。

4. The vertex giving the highest value is the maximum; the one giving the lowest is the minimum.
4. 得到最大值的顶点即为最大值点；得到最小值的顶点即为最小值点。

Always check that the vertex actually satisfies all constraints (it will, by definition). In an exam, show your substitution clearly and state the final answer in the context of the problem – e.g. ‘Produce 4 cakes and 3 biscuits for a maximum profit of £29.’

一定要检查顶点是否真的满足所有约束条件（按定义它满足）。考试时要清楚写出代入过程，并在题目情境下陈述最终答案，比如“生产 4 个蛋糕和 3 块饼干，可获得最大利润 29 英镑”。

7. Cases with Integer Solutions | 整数解的情形

Sometimes the variables represent items that cannot be fractional, like people or cars. The optimal vertex may have non‑integer coordinates, e.g. (3.5, 2.8). In such cases, you must search the integer points near the vertex that lie inside the feasible region to find the best whole‑number solution.

有时变量代表不能分割的实物，比如人数或车辆数。此时最优顶点的坐标可能不是整数，例如 (3.5, 2.8)。这种情况下，你必须在可行区域内、靠近该顶点的整数点中寻找最好的整数解。

Test a small grid of integer coordinates around the fractional vertex, ensuring they satisfy all constraints, and compare the objective values. Remember that the optimal integer solution may not be immediately adjacent to the continuous optimum, so check at least two or three points in each direction.

围绕分数顶点的坐标取一小片整数格点，检查它们是否满足所有约束，并比较目标函数的值。注意，最优整数解不一定紧挨着连续最优解，所以要在各个方向上至少测试两三个点。

8. Interpreting the Solution | 解释解的含义

After finding the optimal values of x and y, you must relate them back to the original real‑world situation. State clearly what the values mean, e.g. ‘The school should hire 5 coaches and 12 minibuses.’ Also give the maximum or minimum value of the objective function in appropriate units, such as ‘The minimum total cost is £850.’

求出 x 和 y 的最优值后，必须将其与现实情境联系起来。清楚地说明这些值代表什么，例如“学校应租用 5 辆大巴和 12 辆小巴”。同时用合适的单位给出目标函数的最大值或最小值，如“最低总费用为 850 英镑”。

If the question asks for ‘how many of each’ or ‘the least cost’, make sure your final answer is a sentence, not just a pair of coordinates. This shows the examiner you understand the practical application.

如果问题问的是“各需多少”或“最少花费是多少”，确保最终答案写成一个完整的句子，而不仅仅是一对坐标。这能让考官看到你理解了实际应用。

9. Common Errors and How to Avoid Them | 常见错误及避免方法

1. Forgetting x ≥ 0 and y ≥ 0: Many students omit these non‑negativity constraints, which can seriously affect the feasible region. Always include them unless the problem explicitly allows negative quantities.

1. 忘记 x ≥ 0 和 y ≥ 0：很多同学会漏掉这些非负约束，这会严重影响可行区域的形状。除非题目明确允许负值，否则一定要加上。

2. Using a dashed line for ≤ or ≥: A boundary line with ≤ or ≥ should be solid to show that points on the line are included. Using a dashed line may lose marks.

2. 对于 ≤ 或 ≥ 用了虚线：带 ≤ 或 ≥ 的边界线应该画成实线，表示直线上的点也满足不等式。画成虚线会丢分。

3. Incorrect shading: Shade the unwanted side or clearly mark the feasible region. If you shade the wanted side for one inequality differently from another, the overlap becomes messy. Use a systematic approach – cross out the side that does NOT satisfy each inequality.

3. 阴影标识错误：最好涂掉不满足的一侧，或者清晰地标出可行区域。如果每条不等式的阴影方式不同，重叠部分会变得杂乱。可以系统性地划掉不满足每一不等式的一侧。

4. Poor scaling: If axes are not scaled evenly, lines may not intersect correctly. Use graph paper or draw a neat grid and label the axes carefully.

4. 坐标比例不当：如果坐标轴刻度不均匀，直线的交点可能不准确。应使用坐标纸或画出整齐的网格，并仔细标注坐标轴。

10. Exam Question Walkthrough | 典型考题思路分析

An IGCSE AQA question typically gives a scenario: ‘A factory produces chairs (x) and tables (y). Each chair requires 2 hours on machine A and 1 hour on machine B. Each table requires 1 hour on machine A and 3 hours on machine B. Machine A is available for 10 hours and machine B for 12 hours. The profit is £5 per chair and £8 per table. Determine how many of each should be produced to maximise profit.’

一道典型的 IGCSE AQA 考题会给出这样的情境：“某工厂生产椅子 (x) 和桌子 (y)。每把椅子需在机器 A 上加工 2 小时，在机器 B 上加工 1 小时；每张桌子需在机器 A 上加工 1 小时，在机器 B 上加工 3 小时。机器 A 可用 10 小时，机器 B 可用 12 小时。每把椅子利润为 5 英镑，每张桌子利润为 8 英镑。为使利润最大，应各生产多少？”

Step 1 – Inequalities: Constraint A: 2x + 1y ≤ 10. Constraint B: 1x + 3y ≤ 12. Non‑negativity: x ≥ 0, y ≥ 0.

步骤 1——不等式：机器 A 约束：2x + 1y ≤ 10；机器 B 约束：1x + 3y ≤ 12；非负约束：x ≥ 0，y ≥ 0。

Step 2 – Graph: Draw lines 2x + y = 10 (intercepts 5 and 10) and x + 3y = 12 (intercepts 12 and 4). Shade the feasible quadrilateral.

步骤 2——画图：画出直线 2x + y = 10（截距 5 和 10）和 x + 3y = 12（截距 12 和 4），标示出可行四边形区域。

Step 3 – Vertices: The vertices are (0,0), (5,0), (0,4), and the intersection of the two lines. Solve: 2x + y = 10 and x + 3y = 12 → from first y = 10 – 2x → substitute: x + 3(10 – 2x) = 12 → x + 30 – 6x = 12 → -5x = -18 → x = 3.6, y = 10 – 7.2 = 2.8. So (3.6, 2.8).

步骤 3——顶点：顶点为 (0,0)，(5,0)，(0,4)，以及两条直线的交点。解方程组：2x + y = 10 和 x + 3y = 12 → 由第一式得 y = 10 – 2x → 代入：x + 3(10 – 2x) = 12 → x + 30 – 6x = 12 → -5x = -18 → x = 3.6，y = 10 – 7.2 = 2.8，故交点为 (3.6, 2.8)。

Step 4 – Objective: Profit P = 5x + 8y. Evaluate at vertices: (0,0) → 0; (5,0) → 25; (0,4) → 32; (3.6,2.8) → 5(3.6)+8(2.8)=18+22.4=40.4. Since we need integer numbers of items, test nearby integer points: (3,3) gives 15+24=39, in feasible region? Check: 2(3)+3=9≤10; 3+3(3)=12≤12 → yes. (4,2): 8+2=10≤10; 4+6=10≤12 → yes, profit=20+16=36. So maximum integer profit is 39 at (3,3). Answer: 3 chairs and 3 tables, profit £39.

步骤 4——目标函数：利润 P = 5x + 8y。在顶点的取值：(0,0) → 0；(5,0) → 25；(0,4) → 32；(3.6,2.8) → 5(3.6)+8(2.8)=18+22.4=40.4。因为产品数量必须为整数，测试邻近的整数点：(3,3) 利润 15+24=39，检查约束：2(3)+3=9≤10；3+3(3)=12≤12，满足。(4,2)：8+2=10≤10；4+6=10≤12，利润=20+16=36。最大整数利润为 39，在 (3,3) 取得。答案：生产 3 把椅子和 3 张桌子，利润 39 英镑。

11. Drawing Tips for the Exam | 考试绘图技巧

Use a sharp pencil and a ruler. Draw axes with arrows and label them x and y. Mark a consistent scale – 1 cm to 1 unit or as appropriate for the numbers. Plot points with a cross or dot, and draw straight lines carefully. After shading the feasible region, outline it heavily or colour it lightly so it is obvious to the examiner.

用削尖的铅笔和直尺作图。坐标轴要画上箭头并标出 x 和 y。刻度要保持一致——可以规定 1 cm 代表 1 个单位，或根据数据选择合适的比例。用十字或圆点标出坐标点，仔细画出直线。涂出可行区域后，可以用粗线勾勒轮廓或轻轻上色，让考官一目了然。

Write the equations of the boundary lines next to them. If you make a mistake, erase cleanly. Neatness saves time when locating vertices. If the vertices do not have whole‑number coordinates, you may be allowed to read them from the graph in some foundation‑tier papers, but on higher tier you are expected to solve simultaneous equations exactly.

在每条边界线旁边写出它的方程。如果画错了，要擦干净再画。整洁的图形有助于更快地找到顶点。如果顶点的坐标不是整数，部分基础卷可能允许直接从图上读取，但在较高层级的考试中，需要你通过解方程组精确求得。

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Linear Programming | 线性规划 考点精讲