Chapter 4.4, Problem 51E

(a)

To determine

To Estimate:The initial number of students.

Answer to Problem 51E

Initial number of students are 18.

Explanation of Solution

Given information: Given function $P(t)=\frac{300}{1+2^{4-t}}$

Calculation:

  $\begin{array}{l}P(t)=\frac{300}{1+2^{4-t}}\\ \text{put}t=0\\ P(0)=\frac{300}{1+2^{4-0}}\\ P(0)=\frac{300}{17}=18\text{approx}\end{array}$

Thus, the initial number of students is 18.

(b)

To determine

To Estimate: The spreading of rumour after 4 days.

Answer to Problem 51E

Rumour is spread to 52 students per day.

Explanation of Solution

Given information: Given function $P(t)=\frac{300}{1+2^{4-t}}$

Calculation:

  $\begin{array}{l}P(t)=\frac{300}{1+2^{4-t}}\\ \text{Differentiate}\text{with}\text{respect}\text{to}t\\ P\text{'}(t)=300\times \frac{d{(1+2^{4-t})}^{-1}}{d(1+2^{4-t})}\times \frac{d(1+2^{4-t})}{d(4-t)}\times \frac{d(4-t)}{dt}\\ P\text{'}(t)=300\times \frac{-1}{{(1+2^{4-t})}^2}\times 2^{4-t}\ln 2\times -1\\ P\text{'}(t)=\frac{300(2^{4-t}\ln 2)}{{(1+2^{4-t})}^2}\\ \text{at}t=4\\ P\text{'}(4)=\frac{300(2^{4-4}\ln 2)}{{(1+2^{4-4})}^2}\\ P\text{'}(4)=75\ln 2\\ P\text{'}(4)=52\text{approx}\end{array}$

Hence, the rumour is spread to 52 students per day.

(c)

To determine

To maximise: The given function.

Answer to Problem 51E

The maximum rate is 52 students per day

Explanation of Solution

Given information: Given function $P(t)=\frac{300}{1+2^{4-t}}$

Graph:

  

Interpretation:

By analysing the above graph the maximum rate occurs at about 4 days, and the rate is 52 students per day which is calculated below:

  $\begin{array}{l}P(t)=\frac{300}{1+2^{4-t}}\\ \text{Differentiate}\text{with}\text{respect}\text{to}t\\ P\text{'}(t)=300\times \frac{d{(1+2^{4-t})}^{-1}}{d(1+2^{4-t})}\times \frac{d(1+2^{4-t})}{d(4-t)}\times \frac{d(4-t)}{dt}\\ P\text{'}(t)=300\times \frac{-1}{{(1+2^{4-t})}^2}\times 2^{4-t}\ln 2\times -1\\ P\text{'}(t)=\frac{300(2^{4-t}\ln 2)}{{(1+2^{4-t})}^2}\\ \text{at}t=4\\ P\text{'}(4)=\frac{300(2^{4-4}\ln 2)}{{(1+2^{4-4})}^2}\\ P\text{'}(4)=75\ln 2\\ P\text{'}(4)=52\text{approx}\end{array}$