The question is not complete but it will be assumed to find the better of the two options,
A. $745.74 monthly for 36 months with 10% down and no residual value, or
B. $433.67 monthly for 36 months with $1100 down and $14200 residual value.
It is true that car dealers make it difficult to compare prices, even with a calculator and/or compound interest reference book. The challenge here is we need to compare the effective monthly interest, i, charged in each case, which is not usually listed in books, nor given by an explicit formula.
We will proceed anyway, and take up the challenge.
Case A: 745.74 for 36 months with 10% down and no residual value.
Future value of loan = (27340*0.9)(1+i)^36
Future value of payments = 745.74((1+i)^36-1)/i
Equate the two and solve for i:
(27340*0.9)(1+i)^36 = 745.74((1+i)^36-1)/i
We can solve by trial and error, which takes a certain time, or we can use Newton's method, with an initial value of i=0.004 (by graphing), where
let f(i)=(27340*0.9)(1+i)^36 - 745.74((1+i)^36-1)/i
then f'(i)=(745.74*((x+1)^36-1))/x^2-(26846.64*(x+1)^35)/x+885816.0*(x+1)^35
i1=i0-f(i0)/f'(i0)
after two iterations, we obtain i=0.0047888
Case B:
$433.67 monthly for 36 months with $1100 down and $14200 residual value.
Here, the future value, equated to zero,
g(x)=(27340-1100)*(1+i)^36 - 433.67((1+i)^36-1)/i = 0
Solving similarly by Newton's method, we have
i=0.004825936
This means that Option A is slightly more advantageous, with a lower interest rate (and also less hassle for the maintenance of the car during three years).
