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IAL 2023 June D1 Q8

A Level / Edexcel / D1

IAL 2023 June Paper · Question 8

题目

Problem

A headteacher is deciding how to allocate prizes to the students who are leaving at the end of the school year.

There are three categories of prize: academic, sport, and leadership.

  • Each academic prize costs £14, each sport prize costs £8, and each leadership prize costs £12. The total amount available to spend on all prizes is £976.
  • For every 5 academic prizes there must be at least 2 leadership prizes.
  • At least half the prizes must be academic.
  • 20% of the prizes must be for sport.

The headteacher wishes to maximise the total number of prizes.

Let xx, yy and zz represent the number of academic, sport and leadership prizes respectively.

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

(8)

Given that the headteacher awards 16 sport prizes,

(b) calculate the corresponding number of leadership prizes that the headteacher awards. You must show your working.

(2)
题目中文翻译

一位校长正在决定如何向学年末毕业的学生颁发奖品。

奖品有三个类别：学术、体育和领导力。

  • 每个学术奖品费用为 £14，每个体育奖品为 £8，每个领导力奖品为 £12。可用于所有奖品的总金额为 £976。
  • 每 5 个学术奖品必须至少有 2 个领导力奖品。
  • 至少一半的奖品必须是学术奖品。
  • 20% 的奖品必须是体育奖品。

校长希望最大化奖品总数。

设 xx、yy 和 zz 分别表示学术、体育和领导力奖品的数量。

(a) 将此表述为仅关于 xx 和 yy 的线性规划问题，说明目标并列出简化为整数系数不等式的约束条件。

已知校长颁发 16 个体育奖品，

(b) 计算校长颁发的相应领导力奖品数量。必须展示计算过程。

解答

(a)

解法一

思路

展开

先把题意逐项写成关于 x,y,zx,y,z 的目标与约束。由“体育奖品占总数的 20%20\%”可得到等式，并用它把 zz 表示成 x,yx,y，再代入目标函数和其余约束，化成只含 x,yx,y 且具有整数系数的形式。

答题过程

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Initially, the objective is to maximise

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

The constraints before eliminating zz are

14x+8y+12z≤976,14x+8y+12z\leq 976, 5z≥2x,5z\geq 2x, x≥12(x+y+z),x\geq \frac12(x+y+z),

and

y=15(x+y+z).y=\frac15(x+y+z).

The last equation gives

x−4y+z=0,x-4y+z=0,

so

z=4y−x.z=4y-x.

Therefore the objective becomes

P=x+y+(4y−x)=5y.P=x+y+(4y-x)=5y.

Substituting z=4y−xz=4y-x into the remaining conditions gives the linear programming problem

Maximise P=5y\boxed{\text{Maximise }P=5y}

subject to

x+28y≤ 488,7x−20y≤ 0,2x−5y≥ 0,x−4y≤ 0,x≥ 0,y≥ 0.\boxed{ \begin{aligned} x+28y\leq&\,488,\\ 7x-20y\leq&\,0,\\ 2x-5y\geq&\,0,\\ x-4y\leq&\,0,\\ x\geq&\,0,\\ y\geq&\,0. \end{aligned}}

(b)

解法一

思路

展开

把 y=16y=16 代入 (a) 的四个非平凡约束，分别得到 xx 的上下界。预算约束与“至少一半奖品为学术奖品”的约束共同迫使 x=40x=40，再代入 z=4y−xz=4y-x 求领导力奖品数量。

答题过程

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Substituting y=16y=16 into the constraints gives

x+28(16)≤488⟹ x≤40,7x−20(16)≤0⟹ x≤3207,2x−5(16)≥0⟹ x≥40,x−4(16)≤0⟹ x≤64.\begin{aligned} x+28(16)\leq 488\Longrightarrow&\,x\leq 40,\\ 7x-20(16)\leq 0\Longrightarrow&\,x\leq \frac{320}{7},\\ 2x-5(16)\geq 0\Longrightarrow&\,x\geq 40,\\ x-4(16)\leq 0\Longrightarrow&\,x\leq 64. \end{aligned}

Hence x=40x=40. Since z=4y−xz=4y-x,

z=4(16)−40=24.z=4(16)-40=24.

Therefore the headteacher awards

24 leadership prizes.\boxed{24\text{ leadership prizes}}.