# Newton's Method (as applied to Auto Financing)

## Homework Statement

A car dealer sells a new car for $18,000. He also offers to sell the same car for monthly payments of$375.00 for five years. What monthly rate is this dealer charging?

## Homework Equations

A = [R(1 - (1 + i))^-60] / i

where A = the present value, R = the monthly payment, i = the interest rate, n = the number of monthly payments

Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 + 1 = 0.

## The Attempt at a Solution

I thought that, "Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 -1 = 0.", seemed to suggest that they wanted me to solve for i (or x), but in hindsight I don't think that's what I need to do. I don't know where to start now.  Or I suppose rather that I'm just supposed to set it equal to 0 somehow. In which case I guess I don't know what they're doing to (1 + x)^60.

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

A car dealer sells a new car for $18,000. He also offers to sell the same car for monthly payments of$375.00 for five years. What monthly rate is this dealer charging?

## Homework Equations

A = [R(1 - (1 + i))^-60] / i

where A = the present value, R = the monthly payment, i = the interest rate, n = the number of monthly payments

Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 + 1 = 0.

## The Attempt at a Solution

I thought that, "Replacing i by x, show that 48x(1 + x)^60 - (1 + x)^60 -1 = 0.", seemed to suggest that they wanted me to solve for i (or x), but in hindsight I don't think that's what I need to do. I don't know where to start now.  Or I suppose rather that I'm just supposed to set it equal to 0 somehow. In which case I guess I don't know what they're doing to (1 + x)^60.
You are using an incorrect formula. If you have 60 monthly payments of R, the payment in month 2 is discounted by 1/(1+r), that in month 3 is discounted by 1/(1+r)^2, ... and the final payment in month 60 is discounted by 1/(1+r)^59. Here, r is the monthly interest rate. So, equating the net present value to the car's price P we have
$$P = R \left( 1 + \frac{1}{1+r} + \frac{1}{(1+r)^2} + \cdots + \frac{1}{(1+r)^{59}} \right)\\ = R \left( \frac{1+r}{r} - \frac{1}{r(1+r)^{59}} \right).$$
So, we need to solve
$$18000 = 375 \left( \frac{1+r}{r} - \frac{1}{r(1+r)^{59}} \right).$$

RGV