📚 Linear Programming: Feasible Regions | 可行区域：不等式约束下的区域表示
A feasible region is the set of all points that satisfy a system of linear inequalities. In A-Level Mathematics, we often represent these regions graphically on a coordinate plane to solve optimisation problems. Understanding how to shade, bound and interpret these regions is essential for linear programming questions in the Edexcel exam.
可行区域是满足一组线性不等式的所有点的集合。在 A-Level 数学中，我们通常在坐标平面上用图形表示这些区域以解决优化问题。理解如何绘制阴影、确定边界并解读这些区域，对于 Edexcel 考试中的线性规划题目至关重要。
1. Linear Inequalities and Half-Planes | 线性不等式与半平面
A linear inequality in two variables, such as 2x + 3y ≤ 12, divides the coordinate plane into two half-planes. The boundary line is obtained by replacing the inequality sign with an equals sign: 2x + 3y = 12. One half-plane satisfies the inequality; the other does not.
二元线性不等式（例如 2x + 3y ≤ 12）将坐标平面分为两个半平面。将不等号改为等号即可得到边界线：2x + 3y = 12。一个半平面满足不等式，另一个不满足。
2x + 3y ≤ 12 ↔ boundary line: 2x + 3y = 12
If the inequality is strict (< or >), the boundary line is drawn dashed. If it is non-strict (≤ or ≥), the boundary is drawn solid. This distinction is crucial because points on a solid boundary are included in the feasible region, while points on a dashed boundary are not.
如果不等式是严格的（< 或 >），边界线画成虚线。如果是不严格的（≤ 或 ≥），边界线画成实线。这一区别至关重要，因为实线边界上的点属于可行区域，而虚线边界上的点不属于。
2. Drawing Boundary Lines Correctly | 正确绘制边界线
To draw a boundary line, find two points that lie on it. The simplest method is to find the intercepts. For 2x + 3y = 12, set x = 0 to get y = 4, giving the point (0, 4); set y = 0 to get x = 6, giving the point (6, 0). Draw the line through these two points.
要画边界线，找直线上两个点即可。最简方法是求截距。对于 2x + 3y = 12，令 x = 0 得 y = 4，即点 (0, 4)；令 y = 0 得 x = 6，即点 (6, 0)。过这两点画直线。
| x | y | Point ✚ 点 |
|---|---|---|
| 0 | 4 | (0, 4) |
| 6 | 0 | (6, 0) |
For vertical or horizontal lines, the process is simpler. The inequality x ≥ 2 is represented by a solid vertical line at x = 2, shading to the right. The inequality y < 3 is represented by a dashed horizontal line at y = 3, shading below.
对于垂直或水平直线，过程更简单。不等式 x ≥ 2 用 x = 2 处的实心垂直线表示，向右方阴影。不等式 y < 3 用 y = 3 处的虚线水平线表示，向下方阴影。
3. Testing a Point to Determine the Correct Half-Plane | 用测试点判断正确的半平面
After drawing the boundary line, select a test point not on the line. The origin (0, 0) is usually the most convenient, unless the line passes through the origin. Substitute the test point into the inequality. If the inequality is satisfied, shade the side containing the test point; otherwise, shade the opposite side.
画出边界线后，选取一个不在线上的测试点。原点 (0, 0) 通常最方便，除非直线经过原点。将测试点代入不等式。若不等式成立，则阴影位于包含测试点的一侧；否则阴影位于另一侧。
For example, test (0, 0) in 2x + 3y ≤ 12:
例如，将 (0, 0) 代入 2x + 3y ≤ 12：
2(0) + 3(0) = 0 ≤ 12 ✔
Since 0 ≤ 12 is true, the feasible region lies on the same side of the line as the origin. Shade that half-plane.
因为 0 ≤ 12 成立，所以可行区域位于直线包含原点的一侧。将该半平面涂上阴影。
4. Systems of Inequalities: The Feasible Region | 不等式组：可行区域
A feasible region is the intersection of all half-planes defined by a system of inequalities. For example, consider:
可行区域是由一组不等式定义的所有半平面的交集。例如，考虑：
x ≥ 0, y ≥ 0, x + y ≤ 6, 2x + y ≤ 8
The first two inequalities restrict the region to the first quadrant. The lines x + y = 6 and 2x + y = 8 form the upper boundaries. The feasible region is the polygon (quadrilateral) bounded by the x-axis, y-axis, and these two lines.
前两个不等式将区域限制在第一象限。直线 x + y = 6 和 2x + y = 8 构成上边界。可行区域是以 x 轴、y 轴和这两条直线为界的多边形（四边形）。
To draw the feasible region, first sketch all boundary lines on the same axes. Then determine the correct side for each inequality by testing a point. Finally, the feasible region is the overlapping area that satisfies all inequalities simultaneously.
要绘制可行区域，先在同一个坐标系中画出所有边界线。然后通过测试点判断每个不等式的正确一侧。最后，可行区域就是同时满足所有不等式的重叠区域。
5. Finding Vertices (Corner Points) | 求顶点（角点）
The vertices of a feasible region are critical for linear programming because the objective function attains its maximum or minimum at a vertex. Vertices are found at the intersection of two boundary lines. Solve the two equations simultaneously.
可行区域的顶点对线性规划至关重要，因为目标函数的最大值或最小值在顶点处取得。顶点位于两条边界线的交点处。联立求解两个方程。
For the system above, the vertices are:
对于上述方程组，顶点为：
| Boundary lines ◆ 边界线 | Solve 求解 | Vertex ✚ 顶点 |
|---|---|---|
| x = 0 and y = 0 | Trivial 直接 | (0, 0) |
| y = 0 and 2x + y = 8 | 2x = 8 → x = 4 | (4, 0) |
| x = 0 and x + y = 6 | y = 6 | (0, 6) |
| x + y = 6 and 2x + y = 8 | Subtract 相减: x = 2, y = 4 | (2, 4) |
The coordinates of the last vertex are found by subtracting the equations: (x + y) − (2x + y) = 6 − 8, which gives −x = −2, so x = 2. Substituting back gives y = 4.
最后一个顶点的坐标通过方程相减得到：(x + y) − (2x + y) = 6 − 8，即 −x = −2，所以 x = 2。代回得 y = 4。
6. Shading the Feasible Region: Which Side to Shade? | 可行区域的阴影：应该涂哪一侧？
A common source of confusion is whether to shade the region that satisfies the inequality or the region that does not. In Edexcel A-Level questions, you should always shade the feasible region itself — the set of points that satisfy all constraints.
一个常见的困惑是：应该涂满足不等式的区域还是不满足的区域？在 Edexcel A-Level 题目中，你应当始终涂可行区域本身——即满足所有约束条件的点集。
Some textbooks and older conventions shade the unwanted region (the complement). Always read the question carefully: if it says “shade the feasible region”, shade the region satisfying all inequalities. If it says “shade the region not satisfying”, shade the outside. In linear programming, the convention is to identify the feasible region clearly, usually by shading it.
有些教科书或旧惯例涂的是不需要的区域（补集）。务必仔细读题：如果题目要求 “shade the feasible region”，则涂满足所有不等式的区域。如果要求 “shade the region not satisfying”，则涂外部区域。在线性规划中，惯例是清楚标出可行区域，通常通过涂阴影来实现。
When multiple inequalities are involved, a systematic approach is essential. Draw each boundary line, test one point per inequality, and shade the correct side. The final feasible region is where all shaded areas overlap. It is often helpful to label the feasible region with the letter R.
当涉及多个不等式时，系统化方法必不可少。画出每条边界线，对每个不等式测试一个点，并涂正确的侧。最终可行区域是所有阴影区域的重叠部分。通常用字母 R 标记可行区域。
7. Interpreting Feasible Regions: Bounded and Unbounded Regions | 解读可行区域：有界区域与无界区域
A feasible region is bounded if it can be enclosed within a finite circle; otherwise it is unbounded. The region defined by x ≥ 0, y ≥ 0, x + y ≤ 6 is bounded — it is a triangle with vertices (0, 0), (6, 0) and (0, 6). The region defined by x ≥ 0, y ≥ 0, x + y ≥ 6 is unbounded — it extends infinitely in the positive x and y directions.
可行区域如果能被一个有限圆包围，则称为有界区域；否则称为无界区域。由 x ≥ 0、y ≥ 0、x + y ≤ 6 定义的区域是有界的——它是一个以 (0, 0)、(6, 0) 和 (0, 6) 为顶点的三角形。由 x ≥ 0、y ≥ 0、x + y ≥ 6 定义的区域是无界的——它沿 x 正方向和 y 正方向无限延伸。
Boundedness affects whether an objective function like P = 3x + 5y has a finite maximum. For a bounded feasible region, both a maximum and a minimum always exist at vertices. For an unbounded region, a maximum or minimum may not exist unless the objective function is constrained in the appropriate direction.
有界性影响目标函数（如 P = 3x + 5y）是否存在有限最大值。对于有界可行区域，最大值和最小值一定在顶点处存在。对于无界区域，除非目标函数在相应方向受到约束，否则最大值或最小值可能不存在。
8. Integer Points within the Feasible Region | 可行区域内的整数点
In many real-world problems, variables represent discrete quantities such as the number of products manufactured. In such cases, we seek integer solutions within the feasible region. The optimal integer point may not be a vertex of the continuous feasible region; it is often a nearby integer point.
在许多实际问题中，变量代表离散数量，例如产品生产数量。此时我们在可行区域内寻找整数解。最优整数点可能不是连续可行区域的顶点；它通常是附近的整数点。
For example, suppose the objective function P = 3x + 5y is maximised over the feasible region defined earlier. The continuous maximum occurs at vertex (2, 4), giving P = 3(2) + 5(4) = 26. If x and y must be integers, we check all integer points inside the region. The point (2, 4) is already integer, so it is also the integer-optimal solution.
例如，假设目标函数 P = 3x + 5y 在之前定义的可行区域上最大化。连续最大值出现在顶点 (2, 4)，P = 3(2) + 5(4) = 26。如果 x 和 y 必须是整数，我们需要检查区域内所有整数点。点 (2, 4) 已经是整数，因此它也是整数最优解。
If the continuous optimum were at a non-integer point, we would examine all integer points surrounding that vertex and evaluate the objective function at each to determine the best feasible integer solution. This process is often tested in decision mathematics papers.
如果连续最优解位于非整数点，我们将检查该顶点周围的所有整数点，并在每个点上评估目标函数以确定最优可行整数解。这一过程常在决策数学试卷中出现。
9. Formulating Constraints from Word Problems | 从应用题中建立约束条件
Feasible regions arise naturally from real-world constraints. For instance, a factory produces two products A and B. Each unit of A requires 2 hours of machine time and each unit of B requires 1 hour. The machine is available for at most 100 hours per week. If x is the number of units of A and y the number of units of B, this constraint is written as 2x + y ≤ 100.
可行区域自然产生于现实世界的约束条件。例如，一家工厂生产两种产品 A 和 B。每单位 A 需要 2 小时机器时间，每单位 B 需要 1 小时。机器每周最多可用 100 小时。若 x 是 A 的产量，y 是 B 的产量，该约束写作 2x + y ≤ 100。
Common constraints include:
常见约束包括：
- Non-negativity: x ≥ 0, y ≥ 0 — production quantities cannot be negative.
- 非负约束：x ≥ 0, y ≥ 0 —— 产量不能为负。
- Resource limits: e.g. 2x + 4y ≤ 80 for a labour-hour restriction.
- 资源限制：例如 2x + 4y ≤ 80 代表工时限制。
- Demand constraints: e.g. x ≤ 30 — at most 30 units of A can be sold.
- 需求约束：例如 x ≤ 30 —— A 最多只能卖出 30 单位。
- Ratios: e.g. x ≥ 2y — at least twice as many A as B must be produced.
- 比例约束：例如 x ≥ 2y —— A 的产量至少是 B 的两倍。
Translating word descriptions into algebraic inequalities is a key skill. Look for phrases such as “at least” (≥), “at most” (≤), “no more than” (≤), “cannot exceed” (≤), and “minimum of” (≥).
将文字描述转化为代数不等式是一项关键技能。注意 “at least”（至少，≥）、”at most”（至多，≤）、”no more than”（不超过，≤）、”cannot exceed”（不能超过，≤）和 “minimum of”（最低，≥）等短语。
10. Solving Linear Programming Problems Using the Feasible Region | 利用可行区域求解线性规划问题
Once the feasible region is drawn, the objective function P = ax + by can be optimised using the vertex method. Evaluate P at each vertex of the feasible region. The largest value is the maximum, the smallest is the minimum.
画出可行区域后，可以使用顶点法优化目标函数 P = ax + by。在可行区域的每个顶点处求 P 值。最大值为最大值，最小值为最小值。
Consider the objective P = 3x + 5y over the earlier feasible region with vertices (0, 0), (4, 0), (2, 4), (0, 6):
考虑目标函数 P = 3x + 5y 在之前可行区域上的优化，顶点为 (0, 0)、(4, 0)、(2, 4)、(0, 6)：
| Vertex ✚ 顶点 | P = 3x + 5y | Value ⚖ 值 |
|---|---|---|
| (0, 0) | 3(0) + 5(0) | 0 |
| (4, 0) | 3(4) + 5(0) | 12 |
| (2, 4) | 3(2) + 5(4) | 26 |
| (0, 6) | 3(0) + 5(6) | 30 |
The maximum value is 30 at the vertex (0, 6). The minimum value is 0 at (0, 0). Thus the feasible region method reduces optimisation to a finite check of vertices.
最大值为 30，在顶点 (0, 6) 处取得。最小值为 0，在 (0, 0) 处取得。因此可行区域法将优化问题化简为对顶点的有限检查。
Alternatively, the objective function can be represented as a family of parallel lines. By sliding the line P = 3x + 5y across the feasible region, the last point of contact as the line leaves the region gives the optimal solution. This graphical method is useful for verifying the vertex method.
或者，目标函数可以表示为一族平行线。通过将直线 P = 3x + 5y 在可行区域上平移，直线离开区域时的最后一个接触点即为最优解。这种图解方法可用于验证顶点法。
11. Common Mistakes in Sketching Feasible Regions | 绘制可行区域时的常见错误
Several pitfalls frequently cause students to lose marks in exams. Being aware of them helps you avoid them.
以下几个常见陷阱经常导致学生在考试中失分。了解它们有助于避开。
- Mistake 1: Using a solid line for a strict inequality. Remember: y > 2x + 1 uses a dashed line.
- 错误一：对严格不等式画了实线。记住：y > 2x + 1 应画虚线。
- Mistake 2: Shading the wrong side of a boundary line. Always test a point — never guess from the sign alone.
- 错误二：把边界线的错误一侧涂了阴影。务必测试一个点——不要仅凭符号猜测。
- Mistake 3: Forgetting non-negativity constraints x ≥ 0, y ≥ 0 in word problems.
- 错误三：在应用题中忘记非负约束 x ≥ 0、y ≥ 0。
- Mistake 4: Incorrectly finding intersections by solving the wrong pair of equations.
- 错误四：联立错误的方程组求交点。
- Mistake 5: Not labelling the feasible region with the letter R when required.
- 错误五：需要时未用字母 R 标记可行区域。
Also, always check whether the boundary line passes through the origin before choosing (0, 0) as the test point. If the line passes through the origin, choose another point such as (1, 0) or (0, 1).
此外，选择 (0, 0) 作为测试点之前，务必检查边界线是否经过原点。如果直线经过原点，则选择其他点，如 (1, 0) 或 (0, 1)。
12. Exam-Style Worked Example | 考试风格例题
Let us work through a complete exam-style question step by step.
让我们逐步完成一道完整的考试风格例题。
Question: A company makes two products, X and Y. Let x and y be the number of units of X and Y produced per week. The constraints are:
题目：某公司生产两种产品 X 和 Y。设 x 和 y 分别为每周生产 X 和 Y 的数量。约束条件为：
x ≥ 0, y ≥ 0, x + 2y ≤ 20, 3x + y ≤ 24
Step 1: Draw the boundary lines. For x + 2y = 20, intercepts are (0, 10) and (20, 0). For 3x + y = 24, intercepts are (0, 24) and (8, 0). Both lines are solid because the inequalities are non-strict.
步骤一：画边界线。x + 2y = 20 的截距为 (0, 10) 和 (20, 0)。3x + y = 24 的截距为 (0, 24) 和 (8, 0)。由于不等式是非严格的，两条线均为实线。
Step 2: Test the origin for each inequality: 0 + 2(0) = 0 ≤ 20 ✔ and 3(0) + 0 = 0 ≤ 24 ✔. So the feasible region lies in the direction of the origin for both lines. Combine with x ≥ 0 and y ≥ 0, which restrict the region to the first quadrant.
步骤二：对每个不等式测试原点：0 + 2(0) = 0 ≤ 20 ✔，且 3(0) + 0 = 0 ≤ 24 ✔。因此对于两条直线，可行区域都在包含原点的一侧。结合 x ≥ 0 和 y ≥ 0，区域限制在第一象限。
Step 3: Find the vertices. The relevant intersection is between x + 2y = 20 and 3x + y = 24. Solve simultaneously:
步骤三：求顶点。相关的交点在 x + 2y = 20 与 3x + y = 24 之间。联立求解：
x + 2y = 20 → x = 20 − 2y
3(20 − 2y) + y = 24 → 60 − 6y + y = 24 → −5y = −36 → y = 7.2
x = 20 − 2(7.2) = 5.6
The vertices are (0, 0), (8, 0), (5.6, 7.2), and (0, 10).
顶点为 (0, 0)、(8, 0)、(5.6, 7.2) 和 (0, 10)。
Step 4: If the profit function is P = 4x + 5y, evaluate at each vertex:
步骤四：若利润函数为 P = 4x + 5y，在每个顶点处求值：
| Vertex ✚ 顶点 | P = 4x + 5y | Profit ⚖ 利润 |
|---|---|---|
| (0, 0) | 0 | 0 |
| (8, 0) | 4(8) + 5(0) | 32 |
| (5.6, 7.2) | 4(5.6) + 5(7.2) | 22.4 + 36 = 58.4 |
| (0, 10) | 4(0) + 5(10) | 50 |
The maximum profit is 58.4, achieved at x = 5.6 and y = 7.2. If integer solutions are required, test the integer points near (5.6, 7.2): (5, 7), (5, 8), (6, 7), (6, 8), etc., ensuring each satisfies all constraints. Among these, the best feasible integer point gives the optimal integer profit.
最大利润为 58.4，在 x = 5.6、y = 7.2 处取得。如果要求整数解，则测试 (5.6, 7.2) 附近的整数点：(5, 7)、(5, 8)、(6, 7)、(6, 8) 等，确保每个点满足所有约束。其中最佳可行整数点给出最优整数利润。
This example demonstrates the full pipeline: forming inequalities, drawing the feasible region, finding vertices, and evaluating the objective function. Master these steps and feasible-region questions become straightforward.
此例题展示了完整流程：建立不等式、绘制可行区域、求顶点、评估目标函数。掌握这些步骤后，可行区域相关题目将变得简单直接。
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