Chapter 3.2, Problem 77E

a.

To determine

By using the model to approximate the lengths of a mortgage.

Answer to Problem 77E

  $\boxed{t_1=30Years}$ and $\boxed{t_2=10Years}$ .

Explanation of Solution

Given information:

The model $t=16.625\ln \left(\frac{x}{x-750}\right),x>750$ approximates the length of a home mortgage of $\$150,000$ at $6\%$ in terms of the monthly payment. In the model, $t$ is the length of the mortgage in years and $x$ is the monthly payment in dollars.

Use the model to approximate the lengths of a $\$150,000$ mortgage at $6\%$ when the monthly payment is $\$897.72$ and when the monthly payment is $\$1659.24$ .

Calculation:

Here, the model given is:

  $t=16.625\ln \left(\frac{x}{x-750}\right),x>750$ Where, $t$ is the length of the mortgage in years and $x$ is the monthly payment in dollars.

Now, when

  $\begin{array}{l}{x}_1=\$897.72\\ t_1=16.625\ln \left(\frac{897.72}{897.72-750}\right)\end{array}$

  $\boxed{t_1=30Years}$

Hence, the length is $\boxed{t_1=30Years}$ .

Now, when

  $\begin{array}{l}{x}_2=\$1659.24\\ t_2=16.625\ln \left(\frac{1659.24}{1659.24-750}\right)\end{array}$

  $\boxed{t_2=10Years}$

Hence, the length is $\boxed{t_2=10Years}$ .

b.

To determine

To approximate the total amounts paid over the term of the mortgage.

Answer to Problem 77E

  $\boxed{A_1=\$323,179}$ and $\boxed{A_2=\$199,109}$ .

Explanation of Solution

Given information:

The model $t=16.625\ln \left(\frac{x}{x-750}\right),x>750$ approximates the length of a home mortgage of $\$150,000$ at $6\%$ in terms of the monthly payment. In the model, $t$ is the length of the mortgage in years and $x$ is the monthly payment in dollars.

Approximate the total amounts paid over the term of the mortgage with a monthly payment of $\$897.72$ and with a monthly payment of $\$1659.24$ .

Calculation:

Here, the total amounts paid over the term of mortgage with a monthly payment of:

  $\begin{array}{l}{x}_1=\$897.72\\ A_1=x_1{t}_1\\ A_1=897.72\times 12\times 30\end{array}$

  $\boxed{A_1=\$323,179}$

Hence, the amount is $\boxed{A_1=\$323,179}$ .

Now, when

  $\begin{array}{l}{x}_2=\$1659.24\\ A_2=x_2{t}_2\\ A_2=1659.24\times 12\times 10\end{array}$

  $\boxed{A_2=\$199,109}$

Hence, the amount is $\boxed{A_2=\$199,109}$ .

c.

To determine

To approximate the total interest charges.

Answer to Problem 77E

  $\boxed{I_1=\$173,179}$ and $\boxed{I_2=\$49,109}$ .

Explanation of Solution

Given information:

The model $t=16.625\ln \left(\frac{x}{x-750}\right),x>750$ approximates the length of a home mortgage of $\$150,000$ at $6\%$ in terms of the monthly payment. In the model, $t$ is the length of the mortgage in years and $x$ is the monthly payment in dollars.

Approximate the total interest charges for a monthly payment of $\$897.72$ and for a monthly payment of $\$1659.24$ .

Calculation:

Here, the interest charged for:

  $\begin{array}{l}{x}_1=\$897.72\\ I_1=A_1-150000\\ I_1=323179-150000\end{array}$

  $\boxed{I_1=\$173,179}$

Hence, the interest is $\boxed{I_1=\$173,179}$ .

Now, when

  $\begin{array}{l}{x}_1=\$1659.24\\ I_1=A_1-150000\\ I_1=199109-150000\end{array}$

  $\boxed{I_2=\$49,109}$

Hence, the interest is $\boxed{I_2=\$49,109}$ .

d.

To determine

To find the vertical asymptote for the model.

Answer to Problem 77E

  $\boxed{x=750}$

Explanation of Solution

Given information:

The model $t=16.625\ln \left(\frac{x}{x-750}\right),x>750$ approximates the length of a home mortgage of $\$150,000$ at $6\%$ in terms of the monthly payment. In the model, $t$ is the length of the mortgage in years and $x$ is the monthly payment in dollars.

What is the vertical asymptote for the model? Interpret its meaning in the context of the problem.

Calculation:

Hence, the vertical asymptote of the model is $\boxed{x=750}$ , as the monthly payment has to be greater than $\$750$ .