CalcGospel 國際數學圖譜
返回

IAL 2024 June D1 Q5

A Level / Edexcel / D1

IAL 2024 June Paper · Question 5

题目

Problem

The head of a Mathematics department needs to order three types of paper. The three types of paper are plain, lined and graph.

All three types of paper are sold in reams. (A ream is 500 sheets of paper.)

Based on the last academic year the head of department formed the following constraints.

  • At least half the paper must be lined
  • No more than 15% of the paper must be graph paper
  • The ratio of plain paper to graph paper must be 5:25:2

The cost of each ream of plain, lined and graph paper is £5, £12 and £15 respectively.

The head of department has at most £834 to spend on paper.

The head of department wants to maximise the total number of reams of paper ordered.

Let xx, yy and zz represent the number of reams of plain paper, lined paper and graph paper ordered respectively.

(a) Formulate this information as a linear programming problem in xx and yy only, stating the objective and listing the constraints as simplified inequalities with integer coefficients.

(7)

The head of department decides to order exactly 42 reams of lined paper and still wishes to maximise the total number of reams of paper ordered.

(b) Determine

(i) the total number of reams of paper to be ordered,

(ii) the number of reams of graph paper to be ordered.

(3)
题目中文翻译

数学系主任需要订购三种纸张，分别是白纸、横线纸和方格纸。

三种纸张均以“令”为单位出售。（一令纸包含 500 张。）

根据上一学年的情况，系主任列出了以下约束：

  • 至少一半的纸张必须是横线纸；
  • 方格纸不得超过纸张总量的 15%15\%；
  • 白纸与方格纸的数量比必须为 5:25:2。

每令白纸、横线纸和方格纸的价格分别为 £5、£12 和 £15。

系主任最多可花费 £834 购买纸张，并希望订购的纸张总令数最大。

设订购的白纸、横线纸和方格纸令数分别为 xx、yy 和 zz。

(a) 将这些信息表示为只含 xx 和 yy 的线性规划问题，写出目标函数，并把各项约束列为具有整数系数的最简不等式。

系主任决定恰好订购 42 令横线纸，同时仍希望订购的纸张总令数最大。

(b) 求：

(i) 订购纸张的总令数；

(ii) 订购方格纸的令数。

解答

(a)

解法一

思路

展开

先用 x:y:zx:y:z 的实际含义把三项比例限制、预算限制和目标函数写成含 x,y,zx,y,z 的式子。由白纸与方格纸之比为 5:25:2 得到 z=25xz=\frac25x，再把它代入其余各式，消去 zz。

答题过程

展开

The initial objective is to maximise

P=x+y+z.P=x+y+z.

Since at least half of all the paper must be lined,

12(x+y+z)≤y,\frac12(x+y+z)\leq y,

so

x−y+z≤0.x-y+z\leq0.

Since no more than 15%15\% of the paper may be graph paper,

320(x+y+z)≥z,\frac{3}{20}(x+y+z)\geq z,

so

3x+3y−17z≥0.3x+3y-17z\geq0.

The ratio condition and the budget constraint give

2x=5z2x=5z

and

5x+12y+15z≤834.5x+12y+15z\leq834.

From 2x=5z2x=5z,

z=25x.z=\frac25x.

Substituting this into the objective gives

P=75x+y=1.4x+y.P=\frac75x+y=1.4x+y.

Substituting it into the three constraints and simplifying gives

7x−5y≤0,19x−15y≤0,11x+12y≤834.\begin{gathered} 7x-5y\leq0,\\ 19x-15y\leq0,\\ 11x+12y\leq834. \end{gathered}

Therefore the required linear programming problem is

Maximise P=1.4x+y\boxed{\text{Maximise }P=1.4x+y}

subject to

11x+12y≤834,7x−5y≤0,19x−15y≤0,x≥0,y≥0.\boxed{ \begin{gathered} 11x+12y\leq834,\\ 7x-5y\leq0,\\ 19x-15y\leq0,\\ x\geq0,\quad y\geq0. \end{gathered} }

(b)

解法一

思路

展开

把 y=42y=42 代入上一问的三个约束，分别求出 xx 的上界。由于目标函数 1.4x+y1.4x+y 随 xx 增大而增大，应取所有上界中最小的一个。最后利用 z=25xz=\frac25x 求方格纸数量。

答题过程

展开

Substituting y=42y=42 into the constraints gives

11x+12(42)≤834⟹ x≤30,7x−5(42)≤0⟹ x≤30,19x−15(42)≤0⟹ x≤63019.\begin{align*} 11x+12(42)\leq834\Longrightarrow&\,x\leq30,\\ 7x-5(42)\leq0\Longrightarrow&\,x\leq30,\\ 19x-15(42)\leq0\Longrightarrow&\,x\leq\frac{630}{19}. \end{align*}

Hence the largest possible value of xx is 30.

(i) The total number of reams is

P=1.4(30)+42=84.P=1.4(30)+42=\boxed{84}.

(ii) Since z=25xz=\frac25x,

z=25(30)=12.z=\frac25(30)=\boxed{12}.

Therefore 12 reams of graph paper should be ordered.