题目

Problem

A new mobile phone is released for sale.

The total sales $N$ of this phone, in thousands, is modelled by the equation

N=125-Ae^{-0.109t}\qquad t\ge 0

where $A$ is a constant and $t$ is the time in months after the phone was released for sale.

Given that when $t=0$ , $N=32$

(a) state the value of $A$ .

(1)

Given that when $t=T$ the total sales of the phone was $100\,000$

(b) find, according to the model, the value of $T$ . Give your answer to 2 decimal places.

(3)

(c) Find, according to the model, the rate of increase in total sales when $t=7$ , giving your answer to 3 significant figures.

Solutions relying entirely on calculator technology are not acceptable.

(2)

The total sales of the mobile phone is expected to reach $150\,000$ .

Using this information,

(d) give a reason why the given equation is not suitable for modelling the total sales of the phone.

(1)

题目中文翻译

一款新手机开始发售。

该手机的总销量 $N$ （单位：千）由方程

N=125-Ae^{-0.109t}\qquad t\ge 0

建模，其中 $A$ 是常数， $t$ 是手机发售后的月份数。

已知当 $t=0$ 时， $N=32$ 。

(a) 写出 $A$ 的值。

已知当 $t=T$ 时，手机总销量为 $100\,000$ 。

(b) 根据该模型求 $T$ 的值，答案精确到小数点后 2 位。

不接受完全依赖计算器技术的解法。

该手机的总销量预计会达到 $150\,000$ 。

(d) 利用这一信息，说明为什么给定方程不适合用来建模该手机的总销量。

解答

(a)

解法一

思路

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把初始条件 $t=0,\ N=32$ 代入模型，并利用 $e^0=1$ 求出 $A$ 。

答题过程

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When $t=0$ and $N=32$ ,

\begin{align*} 32=&\,125-Ae^0\\[4mm] 32=&\,125-A. \end{align*}

Therefore,

\boxed{A=93}.

(b)

解法一

思路

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$N$ 的单位是“千”，所以总销量 $100\,000$ 对应 $N=100$ ，不能把 $100\,000$ 直接代入模型。代入后先把指数项单独放在一边，再取自然对数求 $T$ 。

答题过程

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Since $N$ is measured in thousands, total sales of $100\,000$ correspond to $N=100$ . Using $A=93$ ,

\begin{align*} 100=&\,125-93e^{-0.109T}\\[4mm] 93e^{-0.109T}=&\,25\\[4mm] e^{-0.109T}=&\,\frac{25}{93}. \end{align*}

Taking natural logarithms,

\begin{align*} -0.109T=&\,\ln\bigg(\frac{25}{93}\bigg)\\[4mm] T=&\,\frac{\ln(93/25)}{0.109}\\[4mm] =&\,12.0525\ldots \end{align*}

Hence, according to the model,

\boxed{T=12.05\text{ months}}.

(c)

解法一

思路

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增长率是 $\frac{\mathrm{d}N}{\mathrm{d}t}$ 。对指数模型求导并代入 $t=7$ ；所得数值的单位仍是“千部／月”，最后换算成“部／月”。

答题过程

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Using $A=93$ ,

N=125-93e^{-0.109t}.

Therefore,

\frac{\mathrm{d}N}{\mathrm{d}t} =0.109(93)e^{-0.109t}.

When $t=7$ ,

\begin{align*} \frac{\mathrm{d}N}{\mathrm{d}t} =&\,0.109(93)e^{-0.109(7)}\\[4mm] =&\,4.72653\ldots \end{align*}

This is $4.72653\ldots$ thousand phones per month. Hence, to $3$ significant figures, the rate of increase in total sales is

\boxed{4730\text{ phones per month}}.

(d)

解法一

思路

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考察 $t$ 很大时模型的极限。指数项趋近于零，所以模型预测的总销量趋近于 $125$ 千部，无法达到题目预期的 $150$ 千部。

答题过程

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As $t\to\infty$ ,

e^{-0.109t}\to 0,

N=125-93e^{-0.109t}\to 125.

Thus the model has an upper limit of $125\,000$ phones, which is below the expected total sales of $150\,000$ . Therefore, the equation is not suitable for modelling the total sales of the phone.

IAL 2023 Oct Q4

题目

解答

(a)

解法一

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答题过程

(b)

解法一

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答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程