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Newton's Method (as applied to Auto Financing)

  • #1

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. [edit] 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.[edit]
 
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Answers and Replies

  • #2
Ray Vickson
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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. [edit] 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.[edit]
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
[tex] 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).
[/tex]
So, we need to solve
[tex] 18000 = 375 \left( \frac{1+r}{r} - \frac{1}{r(1+r)^{59}} \right).[/tex]

RGV
 

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