Chapter 3.5, Problem 35E

a.

To determine

Find the linear model using given data.

Answer to Problem 35E

The linear model is $V=-300t+1150$ .

Explanation of Solution

Given:

The computer costs $\$1150$ new has a book value of $\$550$ after $2$ years.

Calculation:

The linear model is $V=mt+b$ .

Slope $\left(m\right)=\frac{y_2-y_1}{x_2-x_1}$

  $\begin{array}{l}=\frac{550-1150}{2-0}\\ =-\frac{600}{2}\\ =-300\end{array}$

Now

  $\begin{array}{l}V=mt+b\\ \Rightarrow 1150=-300t+b\end{array}$

Substitute value of $t=0$ in the above equation.

  $\begin{array}{l}\Rightarrow 1150=-300\cdot 0+b\\ \Rightarrow 1150=b\end{array}$

Hence the linear model is $V=-300t+1150$ .

b.

To determine

Find the exponential model using given data.

Answer to Problem 35E

The exponential equation is $V=1150e^{-0.37t}$ .

Explanation of Solution

Given:

The computer costs $\$1150$ new has a book value of $\$550$ after $2$ years.

Calculation:

  $V=a{e}^{kt}$

Substitute the value of $V=1150$ and $t=0$ in the equation.

  $\begin{array}{l}\Rightarrow 1150=a{e}^{0\cdot k}\\ \Rightarrow 1150=a\end{array}$

Now

  $\Rightarrow V=1150e^{kt}---\left(i\right)$

Substitute the value of $V=550$ and $t=2$ in the equation $\left(i\right)$ .

  $\begin{array}{l}\Rightarrow 550=1150e^{k\cdot 2}\\ \Rightarrow \frac{550}{1150}=\frac{1150}{1150}{e}^{k\cdot 2}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{550}{1150}=\ln e^{2k}\\ \Rightarrow \ln \frac{11}{23}=\ln e^{2k}\\ \Rightarrow -0.74=2k\\ \Rightarrow -\frac{0.74}{2}=\frac{2}{2}k\\ \Rightarrow k=-0.37\end{array}$

Hence the exponential equation is $V=1150e^{-0.37t}$ .

c.

To determine

Graph both models using a graphing utility and check that which model depreciates faster in the first two years.

Answer to Problem 35E

The right answer is exponential model.

Explanation of Solution

Given:

The computer costs $\$1150$ new has a book value of $\$550$ after $2$ years.

Calculation:

The linear model is $V=-300t+1150$ .

The exponential equation is $V=1150e^{-0.37t}$ .

The graph of both equations in the same window is given below.

  

The graph shows that the exponential model depreciates faster in the first two years.

Hence the right answer is exponential model.

d.

To determine

Find the book values of the computer after the given years using both models.

Answer to Problem 35E

The values for linear model are $\left\{850,250\right\}$ and the values for exponential model are $\left\{794.34,379\right\}$

Explanation of Solution

Given:

The given years are one and three.

Calculation:

For linear model.

The linear model is $V=-300t+1150$ .

Substitute the value of $t=1$ in the equation.

  $V=-300\cdot 1+1150=850$

Substitute the value of $t=3$ in the equation.

  $V=-300\cdot 3+1150=250$

For exponential model $V=1150e^{-0.37t}$ .

Substitute the value of $t=1$ in the equation.

  $V=1150e^{-0.37\cdot 1}=1150e^{-0.37}=1150\cdot 0.69=794.34$

Substitute the value of $t=3$ in the equation.

  $V=1150e^{-0.37\cdot 3}=1150e^{-1.11}=1150\cdot 0.329=378.9\approx 379$

Hence the values for linear model are $\left\{850,250\right\}$ and the values for exponential model are $\left\{794.34,379\right\}$

e.

To determine

Show that the advantages and disadvantages of using each model for a buyer and a seller.

Explanation of Solution

Given:

The computer costs $\$1150$ new has a book value of $\$550$ after $2$ years.

Calculation:

The linear model is $V=-300t+1150$ .

The exponential equation is $V=1150e^{-0.37t}$ .

The graph of both equations in the same window is given below.

  

The graph shows that the exponential model depreciates faster in the first two years.

For a buyer $\to$ a buyer prefers the lowest price to buy the book. The buyer prefers the exponential model for first two years because the exponential model depreciates faster in the first two years. After that, the linear model depreciates faster. So, a buyer prefers the linear model after the first two years.

For a seller $\to$ a seller prefers the highest price to sell the book. The seller prefer thelinear model for first two years because thelinear model depreciates slower in the first two years. After that, the exponential model depreciates slower. So, a buyer prefers the exponential model after the first two years.