Chapter 3.5, Problem 44E

a .

To determine

To estimate: sales in 2020 using the given graph.

Answer to Problem 44E

40000 units

Explanation of Solution

Given:

The sales began to drop according to the model

  $S=\frac{500,000}{1+0.1e^{kt}}$.........(1)

Where S represents the number of units sold and t represents the year, with $t=0$ corresponding to 2010.

The graph representing this model is:

  

It is given that $t=0$ corresponding to 2010, so $t=10$ corresponds to the year 2020.

Now, it the given graph, draw the line $t=10$ and at the point where it intersects the graph, draw a line parallel to x axis passing through this point.

The point where this line intersects the y-axis gives the number of units sold in the year 2020.

By looking at the graph it can be estimated that the sales of the year 2020 are nearly 40,000.

b.

To determine

To find: the value of k.

Answer to Problem 44E

  $k=0.4743$

Explanation of Solution

Given:

The sales began to drop according to the model

  $S=\frac{500,000}{1+0.1e^{kt}}$.........(1)

Where S represents the number of units sold and t represents the year, with $t=0$ corresponding to 2010.

Also in the year 2014, sales were 300,000.

Formula used:

  $\ln e^x=x$

The year 2014 corresponds to $t=4$ .

The given population model gives the population in thousands Substitute $S=300,000$ and $t=4$ in equation (1), to find value of k.

  $\begin{array}{c}300,000=\frac{500,000}{1+0.1e^{4k}}\\ 1+0.1e^{4k}=\frac{500000}{300000}\\ 1+0.1e^{4k}=\frac{5}{3}\end{array}$

  $\begin{array}{c}0.1e^{4k}=\frac{5}{3}-1\\ 0.1e^{4k}=\frac{2}{3}\\ e^{4k}=\frac{2}{3\times 0.1}\\ e^{4k}=\frac{20}{3}\end{array}$

Take natural logarithms on both sides,

  $\begin{array}{c}\ln e^{4k}=\ln \frac{20}{3}\\ 4k=\ln \frac{20}{3}\\ k=\frac{1}{4}\ln \frac{20}{3}\\ k\approx 0.4743\end{array}$

Substituting $k=0.4743$ in equation (1), the sales are given by,

  $S=\frac{500,000}{1+0.1e^{0.4743t}}$

c.

To determine

To estimate: sales in 2020 using the model.

Answer to Problem 44E

40,071units

Explanation of Solution

Given:

The sales began to drop according to the model

  $S=\frac{500,000}{1+0.1e^{0.4743t}}$.........(2)

Where S represents the number of units sold and t represents the year, with $t=0$ corresponding to 2010.

It is given that $t=0$ corresponding to 2010, so $t=10$ corresponds to the year 2020.

So, substitute $t=10$ in equation (2),

  $\begin{array}{c}S=\frac{500,000}{1+0.1e^{0.4743\left(10\right)}}\\ \approx 40,071.16\end{array}$

Thus, sales is 2020 are nearly 40,071 units.

The sales in 2020 as obtained from part (a) are 40,000.

So, comparing the two, it can be said that both part (a) and part (c) provides nearly the same result.