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IAL 2025 Jan D1 Q8

A Level / Edexcel / D1

IAL 2025 Jan Paper · Question 8

题目

Problem

Figure 4 shows the constraints of a linear programming problem in xx and yy.

The unshaded area, including its boundaries, forms the feasible region RR.

An objective line has been drawn and labelled on the graph.

(a) State the four inequalities that define the feasible region.

(2)

The maximum value of the objective function is 6289\dfrac{628}{9}.

The minimum value of the objective function is 2607\dfrac{260}{7}.

(b) Determine the objective function, showing your working clearly.

(5)

The practical solution to the linear programming problem requires integer values.

(c) State the minimum integer solution for this problem.

(1)
题目中文翻译

图 4 显示了 xxyy 的线性规划问题的约束条件。

未阴影区域(包括其边界)构成可行域 RR

图上已画出并标注了一条目标线。

(a) 写出定义可行域的四个不等式。

目标函数的最大值为 6289\dfrac{628}{9}

目标函数的最小值为 2607\dfrac{260}{7}

(b) 确定目标函数,清楚展示运算过程。

该线性规划问题的实际解需要整数值。

(c) 说明此问题的最小整数解。

解答

(a)

解法一

思路

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对每条边界线在可行域内选一个测试点,判断可行域位于直线的哪一侧。图中 RR 位于 y=3x+4y=3x+4 的下方、x+2y=24x+2y=24 的上方,并位于另外两条直线的下方。

答题过程

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The four inequalities defining the feasible region are

y3x+4,x+2y24,2x+5y100,4x+y64.\boxed{ \begin{gathered} y\leqslant 3x+4,\qquad x+2y\geqslant 24,\\ 2x+5y\leqslant 100,\qquad 4x+y\leqslant 64. \end{gathered} }

(b)

解法一

思路

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先求目标线在连续可行域中取得最大值和最小值的两个顶点。设目标函数为 P=ax+byP=ax+by,把两个顶点及题目给出的极值代入,联立求出 aabb

答题过程

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The maximum occurs at the intersection of

4x+y=644x+y=64

and

2x+5y=100.2x+5y=100.

Solving these simultaneously gives

(x,y)=(1109,1369).(x,y)=\left(\frac{110}{9},\frac{136}{9}\right).

The minimum occurs at the intersection of

x+2y=24x+2y=24

and

y=3x+4.y=3x+4.

Therefore,

x+2(3x+4)=24,x+2(3x+4)=24,

so

(x,y)=(167,767).(x,y)=\left(\frac{16}{7},\frac{76}{7}\right).

Let the objective function be

P=ax+by.P=ax+by.

Using the given maximum and minimum values,

1109a+1369b=6289,167a+767b=2607.\begin{align*} \frac{110}{9}a+\frac{136}{9}b=&\,\frac{628}{9},\\ \frac{16}{7}a+\frac{76}{7}b=&\,\frac{260}{7}. \end{align*}

Hence

55a+68b=314,4a+19b=65.\begin{align*} 55a+68b=&\,314,\\ 4a+19b=&\,65. \end{align*}

Eliminating aa gives 773b=2319773b=2319, so b=3b=3. Substitution then gives a=2a=2. Therefore, the objective function is

P=2x+3y.\boxed{P=2x+3y}.

(c)

解法一

思路

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官方评分方案把连续最小点附近的整数点 (3,11)(3,11) 指定为本小题答案。不过,按题目印出的四个约束直接检验时,(4,10)(4,10) 也在可行域中,而且目标值更小。因此下面同时保留官方给分答案与这项数学核对结果。

答题过程

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The official mark scheme gives the minimum integer solution as

(3,11).\boxed{(3,11)}.

At this point,

P=2(3)+3(11)=39.P=2(3)+3(11)=39.

However, direct substitution shows that (4,10)(4,10) satisfies all four printed constraints:

103(4)+4,4+2(10)=24,2(4)+5(10)=58100,4(4)+10=2664.\begin{gathered} 10\leqslant 3(4)+4,\qquad 4+2(10)=24,\\ 2(4)+5(10)=58\leqslant100,\\ 4(4)+10=26\leqslant64. \end{gathered}

It also gives

P=2(4)+3(10)=38<39.P=2(4)+3(10)=38<39.

Thus, for the linear programming problem exactly as printed, the mathematical minimum integer solution is

(4,10),\boxed{(4,10)},

which is inconsistent with the official answer (3,11)(3,11).