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Re: Credit card balance transfer fee problem

• To: mathgroup at smc.vnet.net
• Subject: [mg103394] Re: [mg103184] Credit card balance transfer fee problem
• From: Kelly Jones <kelly.terry.jones at gmail.com>
• Date: Sun, 20 Sep 2009 06:20:14 -0400 (EDT)
• References: <200909101118.HAA17845@smc.vnet.net>

On 9/10/09, Kelly Jones <kelly.terry.jones at gmail.com> wrote:
>
> My credit card company loans me $10000 for a cash advance fee of 3%
> ($300), and an interest rate of 2% per year. I have to pay off the
> loan in 1 year, but my monthly minimum payment is only 3% of my
> outstanding balance. In other words, I can pay 3% of my balance for
> the first 11 months, and then pay off the remaining balance in the
> 12th month.

Thanks again to everyone who's helping.

I finally figured out that it's best to solve both equations
simultaneously:

sol = DSolve[{
  have'[t] == have[t]*Log[1+invest]-monpay*owed[t],
  owed'[t] == owed[t]*Log[1+interest]-monpay*owed[t],
  have[0] == borrowed, owed[0] == borrowed*(1+fee)
}, {have[t], owed[t]}, t]

h[t_, borrowed_, fee_, interest_, monpay_, invest_] = have[t] /. sol[[1]]
o[t_, borrowed_, fee_, interest_, monpay_, invest_] = owed[t] /. sol[[1]]

Trying to solve what value of invest makes h() and o() equal at t=1:

Solve[h[1,borrowed,fee,interest,monpay,invest] ==
      o[1,borrowed,fee,interest,monpay,invest], invest]

yields a nonsensical answer.

With arbitrary t, Mathematica can't solve it at all

Solve[h[t,borrowed,fee,interest,monpay,invest] ==
      o[t,borrowed,fee,interest,monpay,invest], invest, t]

Thoughts?

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• References:
  • Credit card balance transfer fee problem
    • From: Kelly Jones <kelly.terry.jones@gmail.com>