题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The number of squirrels in a forest is being studied.

The number of squirrels, $N$ , in the forest, $t$ years after the start of the study, is modelled by the equation

N=\frac{4000e^{0.1t}}{19+e^{0.2t}}\qquad t\ge 0

Use the equation of the model to answer parts (a), (b), (c) and (d).

(a) Find the number of squirrels in the forest at the start of the study.

(1)

(b) Find $\dfrac{dN}{dt}$ .

(2)

The number of squirrels in the forest is at a maximum when $t=T$ .

Using the answer to part (b),

(2)

(d) Hence find the maximum number of squirrels in the forest.

Show your working and give your answer to the nearest whole number.

(3)

题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

现正研究森林中的松鼠数量。

研究开始后 $t$ 年时，森林中的松鼠数量 $N$ 由下式建模：

N=\frac{4000e^{0.1t}}{19+e^{0.2t}}\qquad t\ge 0

利用该模型回答(a)、(b)、(c)、(d)。

(a) 求研究开始时森林中的松鼠数量。

(b) 求 $\dfrac{dN}{dt}$ 。

当 $t=T$ 时，森林中的松鼠数量达到最大。

利用(b)的结果，

(d) 由此求森林中松鼠数量的最大值。

写出过程，并将答案取最接近的整数。

解答

(a)

At the start of the study, $t=0$ .

N=\frac{4000e^{0.1(0)}}{19+e^{0.2(0)}} =\frac{4000}{19+1} =200

So there were

\boxed{200}

squirrels at the start of the study.

(b)

Let

N=\frac{4000e^{0.1t}}{19+e^{0.2t}}

Use the quotient rule:

\frac{\mathrm{d}N}{\mathrm{d}t} =\frac{(19+e^{0.2t})(400e^{0.1t})-(4000e^{0.1t})(0.2e^{0.2t})}{(19+e^{0.2t})^2}

Simplify the numerator:

\begin{aligned} (19+e^{0.2t})(400e^{0.1t})-(4000e^{0.1t})(0.2e^{0.2t}) &=7600e^{0.1t}+400e^{0.3t}-800e^{0.3t} \\ &=7600e^{0.1t}-400e^{0.3t} \\ &=400e^{0.1t}(19-e^{0.2t}) \end{aligned}

Therefore

\boxed{ \frac{\mathrm{d}N}{\mathrm{d}t} =\frac{400e^{0.1t}(19-e^{0.2t})}{(19+e^{0.2t})^2} }

(c)

At the maximum point, when $t=T$ ,

\frac{\mathrm{d}N}{\mathrm{d}t}=0

Since

400e^{0.1T}>0

and

(19+e^{0.2T})^2>0

we need

19-e^{0.2T}=0

Hence

e^{0.2T}=19

\boxed{A=19}

(d)

From part (c),

e^{0.2T}=19

Taking square roots,

e^{0.1T}=\sqrt{19}

Substitute these into the model:

\begin{aligned} N_{\max} &=\frac{4000e^{0.1T}}{19+e^{0.2T}} \\ &=\frac{4000\sqrt{19}}{19+19} \\ &=\frac{4000\sqrt{19}}{38} \\ &=458.8\ldots \end{aligned}

Therefore the maximum number of squirrels is

\boxed{459}

to the nearest whole number.

IAL 2025 Oct Q5

题目

解答

(a)

(b)

(c)

(d)