Question

Larry and Peggy are making decisions on their bank accounts. Larry wants to put the original money in an account with a higher interest rate. Peggy wants to put more money in as a principle amount because the more you start with, the more interest you will gain. Explain which method will result in more money.
@inkyvoyd

Answer 1

Oh, my knowledge in financial mathematics is very vague, I'm afraid I cannot help you :(

Answer 2

do you know anybody that does?

Answer 3

@phi

Answer 4

Is there more information provided in the question?

Answer 5

nope this is the question, this chapter was about exponential and log functions.

Answer 6

so thats how I would put it @douglaswinslowcooper

Answer 7

I think you cannot determine which method gives you higher return without knowing the exact (increased) principle, interest rate and the time of deposit.

Answer 8

Yes, @Callisto, that's why we can only show them how to calculate it.

Answer 9

Let P be the original principle, P' be the new(more) one, r be the original interest rate, r' be the new(higher) one, t be the time of deposit.

A = Pe^(rt)

Larry's Amount = Pe^(r't)
Peggy's Amount = P' e^(rt)

When their amounts are equal,
\[\frac{Pe^{r't}}{P' e^{rt}} = 1\]\[\frac{P}{P'}e^{r't-rt} = 1\]\[e^{r't-rt} = \frac{P'}{P}\]\[e^{(r'-r)t} = \frac{P'}{P}\]\[ (r'-r)t = \ln(\frac{P'}{P})\]\[ t = \frac{\ln(\frac{P'}{P})}{(r'-r)}\]

When Larry's amount > Peggy's amount,
\[\frac{Pe^{r't}}{P' e^{rt}} > 1\]\[\frac{P}{P'}e^{r't-rt} > 1\]\[e^{r't-rt} > \frac{P'}{P}\]\[e^{(r'-r)t} > \frac{P'}{P}\]\[ (r'-r)t > \ln(\frac{P'}{P})\]\[ t > \frac{\ln(\frac{P'}{P})}{(r'-r)}\]
So, when \(t > \frac{\ln(\frac{P'}{P})}{(r'-r)}\), Larry's method will result in more $$

*Note: ln (P'/P) > 0 since P' > P
ln (e^(r'-r)t) > 0 since r'-r > 0