Chapter 3.5, Problem 44E

a.

To determine

Find the value of $k$ and complete the given model.

Answer to Problem 44E

The complete model is $s=\frac{500,000}{1+0.4e^{0.046t}}$ .

Explanation of Solution

Given:

The manufacturer noted that sales began to dropaccording to the model $s=\frac{500,000}{1+0.4e^{kt}}$ . In $2011$ , $300,000$ units were sold. $t=7$ represents $2007$ .

Calculation:

  $s=\frac{500,000}{1+0.4e^{kt}}$

Substitute the value of $s=300,000$ and $t=11$ in the given function.

  $\Rightarrow 300,000=\frac{500,000}{1+0.4e^{k\cdot 11}}$

Multiply both sides by $1+0.4e^{11k}$ .

  $\begin{array}{l}\Rightarrow 300000\left(1+0.4e^{11k}\right)=\frac{500000}{\left(1+0.4e^{11k}\right)}\cdot \left(1+0.4e^{11k}\right)\\ \Rightarrow 300000\left(1+0.4e^{11k}\right)=500000\end{array}$

Divide both sides by $300000$ .

  $\begin{array}{l}\Rightarrow \frac{300000}{300000}\left(1+0.4e^{11k}\right)=\frac{500000}{300000}\\ \Rightarrow 1+0.4e^{11k}=\frac{5}{3}\end{array}$

Subtract $1$ from both sides.

  $\begin{array}{l}\Rightarrow 1+0.4e^{11k}-1=\frac{5}{3}-1\\ \Rightarrow 0.4e^{11k}=\frac{2}{3}\end{array}$

Divide both sides by $0.4$ .

  $\begin{array}{l}\Rightarrow \frac{0.4}{0.4}{e}^{11k}=\frac{2}{3\cdot 0.4}\\ \Rightarrow e^{11k}=\frac{2}{1.2}\end{array}$

Taking natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln e^{11k}=\ln \frac{2}{1.2}\\ \Rightarrow 11k=0.5108\end{array}$

Divided both sides by $11$ .

  $\begin{array}{l}\Rightarrow \frac{11}{11}k=\frac{0.5108}{11}\\ \Rightarrow k=0.046\end{array}$

Hence the complete model is $s=\frac{500,000}{1+0.4e^{0.046t}}$ .

b.

To determine

Find the sales in given year.

Answer to Problem 44E

The sales in $2015$ is $278166.28$ unit.

Explanation of Solution

Given:

The manufacturer noted that sales began to drop according to the model $s=\frac{500,000}{1+0.4e^{kt}}$ . The given year is $2015$ . $t=7$ represents $2007$ .

Calculation:

  $s=\frac{500,000}{1+0.4e^{kt}}$

Substitute the value of $s=300,000$ and $t=11$ in the given function.

  $\Rightarrow 300,000=\frac{500,000}{1+0.4e^{k\cdot 11}}$

Multiply both sides by $1+0.4e^{11k}$ .

  $\begin{array}{l}\Rightarrow 300000\left(1+0.4e^{11k}\right)=\frac{500000}{\left(1+0.4e^{11k}\right)}\cdot \left(1+0.4e^{11k}\right)\\ \Rightarrow 300000\left(1+0.4e^{11k}\right)=500000\end{array}$

Divide both sides by $300000$ .

  $\begin{array}{l}\Rightarrow \frac{300000}{300000}\left(1+0.4e^{11k}\right)=\frac{500000}{300000}\\ \Rightarrow 1+0.4e^{11k}=\frac{5}{3}\end{array}$

Subtract $1$ from both sides.

  $\begin{array}{l}\Rightarrow 1+0.4e^{11k}-1=\frac{5}{3}-1\\ \Rightarrow 0.4e^{11k}=\frac{2}{3}\end{array}$

Divide both sides by $0.4$ .

  $\begin{array}{l}\Rightarrow \frac{0.4}{0.4}{e}^{11k}=\frac{2}{3\cdot 0.4}\\ \Rightarrow e^{11k}=\frac{2}{1.2}\end{array}$

Taking natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln e^{11k}=\ln \frac{2}{1.2}\\ \Rightarrow 11k=0.5108\end{array}$

Divided both sides by $11$ .

  $\begin{array}{l}\Rightarrow \frac{11}{11}k=\frac{0.5108}{11}\\ \Rightarrow k=0.046\end{array}$

The complete model is $s=\frac{500,000}{1+0.4e^{0.046t}}$ .

Substitute the value of $t=15$ in the model.

  $\begin{array}{l}s=\frac{500,000}{1+0.4e^{0.046t}}\\ \Rightarrow s=\frac{500,000}{1+0.4e^{0.046\cdot 15}}\\ =\frac{500,000}{1+0.797486}\\ =\frac{500,000}{1.797486}\\ =278166.28\end{array}$

Hence the sales in $2015$ is $278166.28$ unit.