Question

A car dealer sells a new car for \(18,000 . He also offers to

sell the same car for payments of \)375 per month for five years. What monthly interest rate is this dealer charging?

To solve this problem, you will need to use the formula for the present value A of an annuity consisting of n equal payments of size R with interest rate i per time period:

\(A = \frac{R}{i}\left( {1 - {{\left( {1 + i} \right)}^{ - n}}} \right)\)

Replacing i by x, show that

\(48x{\left( {1 + x} \right)^{60}} - {\left( {1 + x} \right)^{60}} + 1 = 0\)

Use Newton’s Methods to solve this equation.

Short answer

The monthly interest rate charged by the dealer is \(0.76286\% \).

Step-by-step solution

Newton’s formula for nth approximation of root.

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Finding the interest rate

Given in the question.

\(A = \frac{R}{i}\left( {1 - {{\left( {1 + i} \right)}^{ - n}}} \right)\)

\(A = \$ 18000\)

\(R = \$ 375\)

Since payments will be made after year and there are 12 months in years so there must be a total of 60 payments.

\(n = 60\)

i = interest rate

Substituting the values into the equation

\(18000 = \frac{{375}}{i}\left( {1 - {{\left( {1 + i} \right)}^{ - 60}}} \right)\)

According to the question replacing i by x in the equation

\(\begin{aligned}{c}18000 &= \frac{{375}}{x}\left( {1 - {{\left( {1 + x} \right)}^{ - 60}}} \right)\\18000x &= 375\left( {1 - {{\left( {1 + x} \right)}^{ - 60}}} \right)\\48x &= \left( {1 - \frac{1}{{{{\left( {1 + x} \right)}^{60}}}}} \right)\\48x &= \frac{{{{\left( {1 + x} \right)}^{60}} - 1}}{{{{\left( {1 + x} \right)}^{60}}}}\\48x{\left( {1 + x} \right)^{60}} &= {\left( {1 + x} \right)^{60}} - 1\\48x{\left( {1 + x} \right)^{60}} - {\left( {1 + x} \right)^{60}} + 1 &= 0\end{aligned}\)

Hence the equation is verified.

As we know interest cannot have a negative value so taking the initial approximation \({x_1} = 0.01\)for solving the equation by Newton’s method.

\(f\left( x \right) = 48x{\left( {1 + x} \right)^{60}} - {\left( {1 + x} \right)^{60}} + 1\)

Taking initial approximation as\({x_1} = 0.01\)

Newton’s formula:

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

\(\begin{aligned}{l}{x_2} &= {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\\{x_2} &= 0.01 - \frac{{f\left( {0.01} \right)}}{{f'\left( {0.01} \right)}}\\{x_2} &= 0.00810634\end{aligned}\)

Similarly,

\(\begin{aligned}{l}{x_3} &= 0.00810634 - \frac{{f\left( {0.00810634} \right)}}{{f'\left( {00810634} \right)}}\\{x_3} &= 0.007623071\end{aligned}\)

\(\begin{aligned}{l}{x_4} &= 0.007623071 - \frac{{f\left( {0.007623071} \right)}}{{f'\left( {0.007623071} \right)}}\\{x_4} &= 0.0076287951\end{aligned}\)

\(\begin{aligned}{l}{x_5} &= 0.0076287951 - \frac{{f\left( {0.0076287951} \right)}}{{f'\left( {0.0076287951} \right)}}\\{x_5} &= 0.0076286028\end{aligned}\)

\(\begin{aligned}{l}{x_6} &= 0.0076286028 - \frac{{f\left( {0.0076286028} \right)}}{{f'\left( {0.0076286028} \right)}}\\{x_6} &= 0.0076286028\end{aligned}\)

Since the first 6 digits are not changing so, no need for further calculations.

So, the root of the equation is \(x = 0.0076286\)

Therefore, the monthly interest charge is \(i = x = 0.0076286\)

To convert into rate multiplying by 100

Hence the monthly interest rate charge is \(0.76286\% \).