Calculating Amount Financed with Dealer Fee and Down Payment

[cici190]

Alright so here is my problem. I have a customer who is wanting to finance something at my work. There is a dealer fee of 7.75% of the amount financed and the customer wants to put a down payment of 4000.That 4000 needs to include the dealer fee. The issue is that once the down payment is reduced the amount of the dealer fee goes up because the amount financed goes up. The total amount of what they are purchasing is $7794.43. Am I making any sense? Please help me!

Casey

 

[MarkFL]

Let's let $T$ be the total cost, $D$ be the down payment, $A$ be the amount financed and $F$ be the dealer fee, which is to be included in the down payment.

Now, let's split the down payment into two portions, the dealer fee and the remainder $R$ which will reduce the amount financed:

\(\displaystyle D=F+R\)

Now, the dealer fee is a portion $0\le p\le1$ of $A$, so we have:

\(\displaystyle D=pA+R\)

We also find:

\(\displaystyle A=T-R\implies A=T-(D-pA)\)

Thus, we find:

\(\displaystyle A=\frac{T-D}{1-p}\)

Plugging in the data you gave, we find the amount financed is:

\(\displaystyle A=\frac{7794.43-4000}{1-0.0775}\approx4113.20\)

 

[cici190]

Let's let $T$ be the total cost, $D$ be the down payment, $A$ be the amount financed and $F$ be the dealer fee, which is to be included in the down payment.

Now, let's split the down payment into two portions, the dealer fee and the remainder $R$ which will reduce the amount financed:

\(\displaystyle D=F+R\)

Now, the dealer fee is a portion $0\le p\le1$ of $A$, so we have:

\(\displaystyle D=pA+R\)

We also find:

\(\displaystyle A=T-R\implies A=T-(D-pA)\)

Thus, we find:

\(\displaystyle A=\frac{T-D}{1-p}\)

Plugging in the data you gave, we find the amount financed is:

\(\displaystyle A=\frac{7794.43-4000}{1-0.0775}\approx4113.20\)

Thank you so much. This is exactly what I needed!