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P2 2025 Jan Q3

A Level / Edexcel / P2

IAL 2025 Jan Paper · Question 3

题目

Problem

Figure 1 shows an open-topped container used for holding water. The container is in the shape of a cuboid and is made of sheet metal. The base of the container is a rectangle 3x3x metres by xx metres. The height of the container is yy metres as shown in Figure 1. Given that the capacity of the container is 120 m3120\text{ m}^3

(a) show that the area A m2A\text{ m}^2 of the sheet metal used to make the container is given by

A=Px2+QxA=Px^2+\frac{Q}{x}

where PP and QQ are positive constants to be found.

(4)

(b) Use calculus to find the value of xx for which AA has a stationary value, giving your answer to 3 significant figures.

(4)

(c) Find d2Adx2\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} and hence show that the value of xx found in part (b) gives the minimum value of AA.

(2)
题目中文翻译

图1所示为一个用来装水的无盖容器。容器形状为长方体，由金属薄板制成。底面为长 3x3x 米、宽 xx 米的长方形；高度为 yy 米，如图1所示。

已知容器容量为 120 m3120\text{ m}^3，

(a) 证明制作容器所需金属薄板的面积 A m2A\text{ m}^2 为

A=Px2+Qx, A=Px^2+\frac{Q}{x},

其中 PP 和 QQ 是待求的正常数。（4分）

(b) 用微积分求使 AA 取得驻值的 xx，答案保留3位有效数字。（4分）

(c) 求 d2Adx2\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2}，并由此证明 (b) 中的 xx 使 AA 取得最小值。（2分）

解答

(a)

解法一

思路

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先用体积求出 yy。这是 open-topped container，所以 sheet metal 包括底面和四个侧面，不包括顶面。

答题过程

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The volume is

120= (3x)(x)y120= 3x2yy= 40x2.\begin{align*} 120=&\,(3x)(x)y\\ 120=&\,3x^2y\\ y=&\,\frac{40}{x^2}. \end{align*}

Since the container is open-topped, the sheet metal consists of the base and four side faces.

The base area is

3x2.\begin{align*} 3x^2. \end{align*}

The two longer side faces have total area

2(3xy)=6xy,\begin{align*} 2(3xy)=6xy, \end{align*}

and the two shorter side faces have total area

2(xy)=2xy.\begin{align*} 2(xy)=2xy. \end{align*}

So

A= 3x2+6xy+2xy= 3x2+8xy= 3x2+8x(40x2)= 3x2+320x.\begin{align*} A=&\,3x^2+6xy+2xy\\ =&\,3x^2+8xy\\ =&\,3x^2+8x\bigg(\frac{40}{x^2}\bigg)\\ =&\,3x^2+\frac{320}{x}. \end{align*}

Therefore,

P=3,Q=320.\begin{align*} P=3,\qquad Q=320. \end{align*}

(b)

解法一

思路

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stationary value 要令 dAdx=0\frac{\mathrm{d}A}{\mathrm{d}x}=0。注意 320x\frac{320}{x} 要写成 320x−1320x^{-1} 来求导。

答题过程

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From part (a),

A= 3x2+320x−1.\begin{align*} A=&\,3x^2+320x^{-1}. \end{align*}

Differentiate:

dAdx= 6x−320x−2= 6x−320x2.\begin{align*} \frac{\mathrm{d}A}{\mathrm{d}x} =&\,6x-320x^{-2}\\ =&\,6x-\frac{320}{x^2}. \end{align*}

For a stationary value,

6x−320x2= 06x= 320x26x3= 320x3= 1603.\begin{align*} 6x-\frac{320}{x^2}=&\,0\\ 6x=&\,\frac{320}{x^2}\\ 6x^3=&\,320\\ x^3=&\,\frac{160}{3}. \end{align*}

Hence

x= 16033≈ 3.76(3 s.f.).\begin{align*} x=&\,\sqrt[3]{\frac{160}{3}}\\ \approx&\,3.76\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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第二导数大于 0 说明曲线在该点向上弯，因此是 minimum。这里 xx 是长度，所以 x>0x>0。

答题过程

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From

dAdx=6x−320x−2,\begin{align*} \frac{\mathrm{d}A}{\mathrm{d}x}=6x-320x^{-2}, \end{align*}

we get

d2Adx2= 6+640x−3= 6+640x3.\begin{align*} \frac{\mathrm{d}^2A}{\mathrm{d}x^2} =&\,6+640x^{-3}\\ =&\,6+\frac{640}{x^3}. \end{align*}

At the stationary point, use the exact result x3=160/3x^3=160/3:

d2Adx2= 6+640160/3= 6+12= 18>0.\begin{align*} \frac{\mathrm{d}^2A}{\mathrm{d}x^2} =&\,6+\frac{640}{160/3}\\ =&\,6+12\\ =&\,18>0. \end{align*}

Therefore, the stationary value is a minimum.