题目

Problem

A bakery makes three types of doughnut. These are ring, jam and custard. The bakery has the following constraints on the number of doughnuts it must make each day.

The total number of doughnuts made must be at least 200
They must make at least three times as many ring doughnuts as jam doughnuts
At most 70% of the doughnuts the bakery makes must be ring doughnuts
At least a fifth of the doughnuts the bakery makes must be jam doughnuts

It costs 8 pence to make each ring doughnut, 10 pence to make each jam doughnut and 14 pence to make each custard doughnut. The bakery wants to minimise the total daily costs of making the required doughnuts.

Let $x$ represent the number of ring doughnuts, let $y$ represent the number of jam doughnuts and let $z$ represent the number of custard doughnuts the bakery makes each day.

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

(6)

On a given day, instead of making at least 200 doughnuts, the bakery requires that exactly 200 doughnuts are made. Furthermore, the bakery decides to make the minimum number of jam doughnuts which satisfy all the remaining constraints.

Given that the bakery still wants to minimise the total cost of making the required doughnuts, use algebra to

(b) (i) calculate the number of each type of doughnut the bakery will make on that day,

(ii) calculate the corresponding total cost of making all the doughnuts.

(5)

题目中文翻译

一家面包店制作三种类型的甜甜圈：环形、果酱和奶油。面包店对每天必须制作的甜甜圈数量有以下约束。

制作的甜甜圈总数必须至少 200 个
环形甜甜圈必须至少是果酱甜甜圈的三倍
面包店制作的甜甜圈中最多 70% 必须是环形甜甜圈
面包店制作的甜甜圈中至少五分之一必须是果酱甜甜圈

制作每个环形甜甜圈成本为 8 便士，每个果酱甜甜圈为 10 便士，每个奶油甜甜圈为 14 便士。面包店希望最小化制作所需甜甜圈的每日总成本。

设 $x$ 表示环形甜甜圈数量， $y$ 表示果酱甜甜圈数量， $z$ 表示面包店每天制作的奶油甜甜圈数量。

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

在某一天，面包店要求恰好制作 200 个甜甜圈，而不是至少 200 个。此外，面包店决定制作满足所有剩余约束的最少数量的果酱甜甜圈。

已知面包店仍然希望最小化制作所需甜甜圈的总成本，使用代数方法

(b) (i) 计算面包店当天将制作的每种甜甜圈的数量，

(ii) 计算制作所有甜甜圈的相应总成本。

解答

(a)

解法一

思路

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先直接写出总数、环形甜甜圈与果酱甜甜圈数量关系。两个百分比约束都以总数 $x+y+z$ 表示，再移项并清除分母，化成整数系数不等式。目标函数必须明确写成最小化每日成本。

答题过程

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The objective is to

\boxed{\text{minimise }P=8x+10y+14z},

where $P$ is measured in pence.

The total number of doughnuts gives

x+y+z\geq 200.

There must be at least three times as many ring doughnuts as jam doughnuts, so

3y\leq x.

Since at most $70\%$ of the doughnuts may be ring doughnuts,

\begin{aligned} x\leq&\,\frac{7}{10}(x+y+z) \\ 10x\leq&\,7x+7y+7z \\ 3x\leq&\,7y+7z. \end{aligned}

Since at least one fifth of the doughnuts must be jam doughnuts,

\begin{aligned} y\geq&\,\frac15(x+y+z) \\ 5y\geq&\,x+y+z \\ x+z\leq&\,4y. \end{aligned}

Therefore the complete set of constraints is

\boxed{ \begin{aligned} x+y+z&\geq 200, \\ 3y&\leq x, \\ 3x&\leq 7y+7z, \\ x+z&\leq 4y, \\ x,y,z&\geq 0, \end{aligned}}

where $x$ , $y$ and $z$ are integers.

(b)(i)

解法一

思路

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总数固定为 200 时，果酱甜甜圈至少占五分之一，所以最少为 40 个。环形甜甜圈最多占七成，即最多 140 个。固定 $y=40$ 后，以 $z=160-x$ 代入成本函数，可见 $x$ 越大成本越低，因此取允许的最大值 $x=140$ ，再求出 $z$ 。

答题过程

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Since exactly $200$ doughnuts are made,

x+y+z=200.

The jam-doughnut constraint gives

y\geq\frac15(200)=40.

The minimum permitted number of jam doughnuts is therefore $y=40$ . Also, the ring-doughnut constraint gives

x\leq\frac{7}{10}(200)=140.

With $y=40$ , we have $z=160-x$ . Hence

\begin{aligned} P=&\,8x+10(40)+14(160-x) \\ =&\,2640-6x. \end{aligned}

To minimise $P$ , $x$ must be as large as possible, so $x=140$ . Thus

z=200-140-40=20.

The remaining constraint is satisfied because $140\geq3(40)$ . Therefore the bakery makes

\boxed{ \begin{aligned} 140&\text{ ring doughnuts},\\ 40&\text{ jam doughnuts},\\ 20&\text{ custard doughnuts}. \end{aligned}}

(b)(ii)

解法一

思路

展开

把 (b)(i) 得到的三种数量代入以便士为单位的目标函数，再把 1800 便士换算成英镑。

答题过程

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The total cost is

\begin{aligned} P=&\,8(140)+10(40)+14(20) \\ =&\,1800\text{ pence} \\ =&\,\pounds18. \end{aligned}

Hence the corresponding total cost is

\boxed{\pounds18}.

IAL 2020 June D1 Q8

题目

解答

(a)

解法一

思路

答题过程

(b)(i)

解法一

思路

答题过程

(b)(ii)

解法一

思路

答题过程