题目

Problem

A clothing shop sells a particular brand of shirt, which comes in three different sizes, small, medium and large.

Each month the manager of the shop orders x small shirts, y medium shirts and z large shirts. The manager forms constraints on the number of each size of shirts he will have to order. One constraint is that for every 3 medium shirts he will order at least 5 large shirts.

(a) Write down an inequality, with integer coefficients, to model this constraint.

(2)

Two further constraints are

$x+y+z\geqslant250$ and $x\leqslant0.2(x+y+z)$

(b) Use these two constraints to write down statements, in context, that describe the number of different sizes of shirt the manager will order.

(3)

The cost of each small shirt is £6, the cost of each medium shirt is £10 and the cost of each large shirt is £15.

The manager must minimise the total cost of all the shirts he will order.

(1)

Initially, the manager decides to order exactly 150 large shirts.

(d) (i) Rewrite the constraints, as simplified inequalities with integer coefficients, in terms of x and y only.

(ii) Represent these constraints on Diagram 1 in the answer book. Hence determine, and label, the feasible region R.

(6)

(e) Use the objective line method to find the optimal vertex, V, of the feasible region. You must make your objective line clear and label V.

(2)

(f) Write down the number of each size of shirt the manager should order. Calculate the total cost of this order.

(2)

Later, the manager decides to order exactly 50 small shirts and exactly 75 medium shirts instead of 150 large shirts.

(g) Find the minimum number of large shirts the manager should order and show that this leads to a lower cost than the cost found in (f).

(2)

(Total 18 marks)

题目中文翻译

一家服装店销售某个品牌的衬衫，该衬衫有三种不同尺码：小号、中号和大号。

每月店铺经理订购 x 件小号衬衫、y 件中号衬衫和 z 件大号衬衫。经理对每种尺码的衬衫数量形成约束条件。其中一个约束是每订购 3 件中号衬衫，至少要订购 5 件大号衬衫。

(a) 写出一个具有整数系数的不等式来建模此约束。

另外两个约束是

$x+y+z\geqslant250$ 和 $x\leqslant0.2(x+y+z)$

(b) 使用这两个约束写出在实际背景下描述经理将订购的不同尺码衬衫数量的陈述。

每件小号衬衫的成本为 £6，每件中号衬衫的成本为 £10，每件大号衬衫的成本为 £15。

经理必须最小化他将订购的所有衬衫的总成本。

最初，经理决定恰好订购 150 件大号衬衫。

(d) (i) 将约束改写为仅关于 x 和 y 的简化整数系数不等式。

(ii) 在答案册的 Diagram 1 上表示这些约束。由此确定并标记可行区域 R。

(e) 使用目标线法找到可行区域的最优顶点 V。你必须清楚地画出你的目标线并标记 V。

(f) 写出经理应订购的每种尺码衬衫的数量。计算此订单的总成本。

后来，经理决定改为恰好订购 50 件小号衬衫和 75 件中号衬衫，而不是 150 件大号衬衫。

(g) 求经理应订购的大号衬衫的最少数量，并证明这比 (f) 中找到的成本更低。

解答

(a)

解法一

思路

展开

每 3 件中号衬衫至少对应 5 件大号衬衫，所以“大号数量／中号数量”至少为 $5/3$ 。交叉相乘即可得到整数系数不等式。

答题过程

展开

The constraint is

z\geqslant\frac{5}{3}y,

so, with integer coefficients,

\boxed{5y\leqslant3z}.

(b)

解法一

思路

展开

第一条约束限制三种尺码的总订购量；第二条把小号数量与全部衬衫总数比较，因此必须在陈述中明确“全部衬衫的 20%”，不能只说 0.2 倍。

答题过程

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The manager must order at least $250$ shirts in total.
At most $20\%$ of all the shirts ordered may be small shirts.

(c)

解法一

思路

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总成本等于各尺码的单位成本乘订购量后相加，题目要求将这个总成本最小化。

答题过程

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Let $C$ be the total cost. The objective is

\boxed{\text{Minimise }C=6x+10y+15z}.

(d)(i)

解法一

思路

展开

把 $z=150$ 分别代入三条约束，再移项并化成整数系数形式。三条边界分别对应总数量、大号与中号的比例、小号占总数的比例。

答题过程

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Substituting $z=150$ gives

x+y+150\geqslant250 \quad\Longrightarrow\quad \boxed{x+y\geqslant100},

5y\leqslant3(150) \quad\Longrightarrow\quad \boxed{y\leqslant90},

and

\begin{aligned} x\leqslant&\,0.2(x+y+150)\\ 5x\leqslant&\,x+y+150\\ 4x-y\leqslant&\,150. \end{aligned}

Therefore the third simplified constraint is

\boxed{4x-y\leqslant150}.

(d)(ii)

解法一

思路

展开

依次画出三条边界直线，并根据不等号方向选取公共半平面。可行域位于 $x+y=100$ 上方、 $y=90$ 下方及 $y=4x-150$ 上方，是由三条直线围成的三角形。

答题过程

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Draw the boundary lines using the following points:

Boundary	Two points on the line	Feasible side
$x+y=100$	$(0,100),(100,0)$	$x+y\geqslant100$
$y=90$	$(0,90),(60,90)$	$y\leqslant90$
$4x-y=150$	$(37.5,0),(60,90)$	$4x-y\leqslant150$

The feasible region $R$ is the triangle with vertices

(10,90),\qquad(60,90),\qquad(50,50).

(e)

解法一

思路

展开

当 $z=150$ 固定时，只需最小化 $6x+10y$ 。目标线斜率为 $-0.6$ ；将目标线平行移向较小成本方向，最后接触可行域的位置是两条斜边的交点。

答题过程

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An objective line such as

6x+10y=300

has gradient $-0.6$ . Moving a parallel objective line away from the origin from smaller objective values, it first meets the feasible region at the intersection of

x+y=100

and

4x-y=150.

Solving gives

5x=250,

so the optimal vertex is

\boxed{V=(50,50)}.

(f)

解法一

思路

展开

由最优顶点读出 $x=y=50$ ，并结合本阶段固定的 $z=150$ 。将三种订购量代入原目标函数计算总成本。

答题过程

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The manager should order $50$ small shirts, $50$ medium shirts and $150$ large shirts. The total cost is

\begin{aligned} C=&\,6(50)+10(50)+15(150)\\ =&\,300+500+2250\\ =&\,\boxed{\pounds3050}. \end{aligned}

(g)

解法一

思路

展开

把新的 $x=50,y=75$ 代回全部三条原约束。三条约束都给出 $z\geqslant125$ ，因此最少订购 125 件大号衬衫；再代入成本函数并与 (f) 比较。

答题过程

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From the ratio constraint,

5(75)\leqslant3z \quad\Longrightarrow\quad z\geqslant125.

The total-number constraint gives

50+75+z\geqslant250 \quad\Longrightarrow\quad z\geqslant125.

The small-shirt constraint also gives

\begin{aligned} 50\leqslant&\,0.2(50+75+z)\\ 250\leqslant&\,125+z\\ z\geqslant&\,125. \end{aligned}

Hence the minimum number of large shirts is $125$ . The resulting cost is

\begin{aligned} C=&\,6(50)+10(75)+15(125)\\ =&\,300+750+1875\\ =&\,\pounds2925. \end{aligned}

Since $\pounds2925<\pounds3050$ , this is cheaper than the order found in (f).

IAL 2019 June D1 Q5

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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(d)(i)

解法一

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(d)(ii)

解法一

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(e)

解法一

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(f)

解法一

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(g)

解法一

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