Linear Programming for GCSE CIE Mathematics: Key Points | GCSE CIE 数学：线性规划考点精讲

📚 Linear Programming for GCSE CIE Mathematics: Key Points | GCSE CIE 数学：线性规划考点精讲

Linear programming is a powerful method used to find the best possible outcome—such as maximum profit or minimum cost—when the problem can be modelled using linear equations and inequalities. In GCSE CIE Mathematics, you will learn to translate word problems into constraints, sketch the feasible region, and then determine the optimal solution using the vertices of that region or by sliding the objective function line.

线性规划是一种强大的方法，当问题可以用线性方程和不等式建模时，用于寻找最佳结果——例如最大利润或最小成本。在 GCSE CIE 数学中，你将学习将文字问题转化为约束条件，绘制可行域，然后利用可行域的顶点或通过平移目标函数线来确定最优解。

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

Linear programming involves optimising (maximising or minimising) a linear objective function subject to a set of linear inequalities called constraints. In practical situations, resources such as time, money, or materials are limited. The variables, often denoted x and y, represent the quantities we can control. Typical constraints include the non-negativity restrictions x ≥ 0, y ≥ 0, ensuring we do not produce negative amounts.

线性规划涉及在一组线性不等式（称为约束条件）的限制下，优化（最大化或最小化）一个线性目标函数。在实际情形中，时间、资金或材料等资源是有限的。变量通常用 x 和 y 表示，代表我们可以控制的数量。典型的约束条件包括非负限制 x ≥ 0、y ≥ 0，确保我们不会产出负数。

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2. Formulating Constraints from Word Problems | 从文字题中建立约束条件

Begin by reading the problem carefully and defining your variables. Identify each resource and its limitation. Express every limitation as a linear inequality. For example, if a machine has at most 16 hours available and each unit of product X requires 2 hours while product Y needs 1 hour, then the constraint is 2x + y ≤ 16. Write all constraints clearly and keep your variable definitions nearby for reference.

首先仔细读题并定义变量。识别每种资源及其限制。将每个限制表达为一个线性不等式。例如，如果一台机器最多有 16 小时可用，每单位产品 X 需要 2 小时，产品 Y 需要 1 小时，则约束条件为 2x + y ≤ 16。清晰地写出所有约束条件，并将变量定义放在一旁以便参考。

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

To graph an inequality like 2x + y ≤ 16, first draw the boundary line 2x + y = 16. Find the intercepts: when x = 0, y = 16; when y = 0, x = 8. Since the inequality includes ‘equal to’, draw a solid line. To determine which side to shade, test a point not on the line—usually (0,0) is easiest. Substitute into the inequality: 2(0) + 0 ≤ 16 is true, so shade the region containing (0,0). Use arrows or light shading, and label the line.

要画不等式 2x + y ≤ 16 的图像，首先画出边界线 2x + y = 16。求截距：当 x = 0 时 y = 16；当 y = 0 时 x = 8。因为不等式包含“等于”，所以要画实线。要确定涂哪一侧，选取不在线上的一点——通常 (0,0) 最简便。代入不等式：2(0) + 0 ≤ 16 成立，所以涂上包含 (0,0) 的区域。使用箭头或轻涂，并标注直线。

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4. Shading the Feasible Region | 标出可行域

The feasible region is the area that satisfies all constraints at once. When you have several inequalities, shade each one lightly or, as common in CIE exams, shade the region that does not satisfy each inequality, leaving the feasible region clear. Always check the instruction: ‘shade the unwanted region’ is a frequent request. The feasible region should be a polygon, and you must identify its vertices accurately.

可行域是同时满足所有约束条件的区域。当有多个不等式时，可以分别轻涂，或者，如 CIE 考试中常见的，涂上每个不等式不满足的区域，从而留出清晰的可行域。一定要核对指令：“涂上不需要的区域”是常见要求。可行域应是一个多边形，你必须准确地找出其顶点。

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5. The Objective Function | 目标函数

The objective function is what you aim to maximise or minimise—for instance, total profit P = 30x + 40y. It is a linear expression containing the decision variables. Once the feasible region is drawn, the optimum value will occur at one of the vertices (corner points) of the feasible region, provided the region is bounded and the coefficients are fixed.

目标函数是你要最大化或最小化的对象——例如，总利润 P = 30x + 40y。它是一个包含决策变量的线性表达式。一旦画出可行域，只要区域有界且系数固定，最优值就会出现在可行域的某一个顶点（角点）上。

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6. Finding Optimal Solutions (Vertex Method) | 求最优解（顶点法）

List all the vertices of the feasible region. These are found by solving pairs of boundary equations simultaneously. Then evaluate the objective function at each vertex. The vertex giving the highest value is the maximum; the lowest gives the minimum. In our chair example, vertices are (0,0), (8,0), (0,6) and the intersection (6,4). After substituting into P = 30x + 40y, (6,4) gives 340, which is clearly the maximum.

列出可行域的所有顶点。这些顶点通过联立求解边界方程组得到。接着计算每个顶点处的目标函数值。给出最高值的顶点即为最大值点；给出最低值的为最小值点。在我们的椅子例子中，顶点为 (0,0)、(8,0)、(0,6) 和交点 (6,4)。代入 P = 30x + 40y 后，(6,4) 给出 340，显然是最大值。

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7. Using Parallel Lines to Locate Optimal Point | 使用平行线法找最优解

Draw a line representing the objective function, for example 30x + 40y = 1200, choosing a constant that gives a convenient line. Then slide this line parallel to itself across the feasible region. For maximisation, move the line in the direction that increases the value; for minimisation, move it to decrease the value. The last vertex the line touches before leaving the feasible region is the optimal point. This method visually confirms the vertex method.

画一条代表目标函数的直线，例如 30x + 40y = 1200，选择一个常数使直线便于绘制。然后将这条直线平行滑动，使其穿过可行域。对于最大化问题，沿目标值增大的方向移动直线；对于最小化问题，沿目标值减小的方向移动。直线离开可行域前最后接触的顶点就是最优点。此方法可从图形上验证顶点法。

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8. Integer Solutions in Linear Programming | 线性规划中的整数解

Many real-world contexts require integer solutions—you cannot produce 6.3 chairs. Fortunately, in CIE exam problems the optimal vertex often has integer coordinates. If the vertex comes out as a fraction, you must test integer points near it that lie inside the feasible region. Evaluate the objective function at these integer points to find the best one. Always check the wording to see if integers are required.

许多现实情境要求整数解——你无法生产 6.3 把椅子。幸运的是，在 CIE 考题中，最优顶点的坐标通常是整数。如果顶点出现分数，你必须测试其在可行域内附近的整数点。在这些整数点处计算目标函数值，找出最佳点。务必检查题目措辞，看是否要求整数。

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9. Testing Vertices for Maximum/Minimum | 检验顶点求最大最小值

Creating a table is an excellent way to compare values systematically. For our chair production example, vertices are (0,0), (8,0), (0,6) and (6,4). Evaluate P = 30x + 40y at each one, and note the results. The table below demonstrates this clearly. Always double-check that each vertex you list satisfies all the original constraints—one misplotted line can lead to a false vertex.

创建表格是系统地比较数值的绝佳方式。对于我们的椅子生产例子，顶点为 (0,0)、(8,0)、(0,6) 和 (6,4)。在每个点计算 P = 30x + 40y，并记录结果。下表清晰地展示了这一点。务必反复检查你所列的每个顶点都满足原约束条件——画错一条线就可能导致虚假顶点。

Vertex (x, y)	P = 30x + 40y
(0, 0)	0
(8, 0)	240
(0, 6)	240
(6, 4)	340

If any vertex had been non-integer, you would extend the table to test surrounding integer points, ensuring they remain inside the feasible region.

若某个顶点为非整数，你就要扩展该表格，测试周围的整数点，并确保它们仍在可行域内。

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10. Worked Example: Maximising Profit | 例题：利润最大化

Let’s consolidate with a complete walkthrough. A workshop makes tables (x) and chairs (y). Constraints: cutting department 2x + y ≤ 16; assembly department x + 3y ≤ 18; x ≥ 0, y ≥ 0. Profit per table £30, per chair £40, so P = 30x + 40y. Draw the lines: for 2x + y = 16, intercepts (0,16) and (8,0). For x + 3y = 18, intercepts (0,6) and (18,0). The intersection is found by solving 2x + y = 16 and x + 3y = 18, giving x = 6, y = 4. The feasible quadrilateral has vertices (0,0), (8,0), (6,4) and (0

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