Understanding Compound Interest: Solving Problems and Maximizing Profits

[seiferseph]

I'm having some trouble with this problem
what i did was

(15500 + 1000t)*e^0.05t > 1000
so basically, when the money you are making is greater than how much you would make by keeping the cards

i know the first t = when you sell it, but what is the second t? that should be the amount of time the money is left in the account, but it doesn't give anything. Also, I'm having trouble just solving this. I would appreciate any help, thanks!

 

[CarlB]

It seems to me that what you're trying to do is to maximize profit. Profit is defined by how much you make in one year. In the one case the profit is $1000 per year. In the other, it's 5%, <<<compounded continuously>>>. You need to compare these things. Don't forget "continuously".

Carl

 

[Tide]

IOW, at what point would the interest on the value of the card collection exceed $1,000/year.

 

[seiferseph]

I don't understand,what is wrong with what i was doing?

 

[Tide]

You have the appreciated value compounding interest continuously. According to your statement of the problem, the collection is either appreciating or you're taking the proceeds and then investing it.

Try it this way. At the end of n years, the value of the collection is $15,500 + 1000 n$. If you sold at the end of n years and deposited the money, how much would that money be worth one year later with interest compounded continuously? What would be the interest earned at the end of that year? How big does n have to be in order to produce more than $1,000 interest in that year?

 

[seiferseph]

You have the appreciated value compounding interest continuously. According to your statement of the problem, the collection is either appreciating or you're taking the proceeds and then investing it.
Try it this way. At the end of n years, the value of the collection is $15,500 + 1000 n$. If you sold at the end of n years and deposited the money, how much would that money be worth one year later with interest compounded continuously? What would be the interest earned at the end of that year? How big does n have to be in order to produce more than $1,000 interest in that year?

isn't that just

compound interset: P = P(o)*e^rt, where t = 1
and since you will be putting 15,500 + 1000 n

(15500 + 1000n)*e^0.05 > 1000 ?

because you will get 15500 + 1000n from selling, which you will put in the bank, and you want the profit made in one year to be greater than how much you would make by just keeping the cards

 

[Tide]

That looks good to me!

 

[seiferseph]

That looks good to me!

i tried solving that, but I got a negative number, and the answer is should be 4.5

 

[Tide]

Actually, you should be solving

$$(15000+1000n)(e^{0.05} - 1) = 1000$$

since it's the interest we're concerned with. You should find n is about 4.

 

[seiferseph]

Actually, you should be solving
$$(15000+1000n)(e^{0.05} - 1) = 1000$$
since it's the interest we're concerned with. You should find n is about 4.

thats what someone told me, why is it (e^0.05 - 1) though? where did the -1 come from? i did try that and got the right answer, but i don't understand where it came from

 

[CarlB]

An easier way to set up the problem is to compare the rates at which the amount is increasing. I don't think that

$$(15000+1000n)(e^{0.05} - 1) = 1000$$

is correct. Instead, the equation should be:

$$(15000+1000n)(1.05 - 1) = 1000$$

So I get 5 years exactly. The reason is that you are not comparing the amount earned per year, but are instead looking at the amount compounded continuously. As soon as that amount exceeds the rate of $1000 per year (or about $3 per day), you sell and put the money in the bank.

Have I made this clear? What I'm saying is that the point in time in which you convert over is determined by a comparison of rates, not a comparison of how the two choices will do over the following 12 months. Consequently, the continuous interest rate (.05) shouldn't be converted to an annual rate by the (e^.05 - 1). There's nothing special in the problem about a "year". The year is just the unit of measurement for time in this case.

You can convert the problem into a problem where the unit of measurement is the second. To do that, you divide the $1000 by the number of seconds in a year, and you divide the 5% by the same amount. When you're done, you'll solve the problem the same way and since what you're looking for is a higher rate, you'll get the same answer, but in seconds instead of years.

I'm going against the grain here. Maybe I'm confused. It's pretty clear that if you leave the collection alone for 5 years you will end up with a total value of $20,000. Why don't you figure out how much you'll have if instead you stop at 4.5 years and then put the resulting $19,500 in the bank at 5% interest compounded continuously? I don't have a calculator handy right now so work it out.

By the way, I find that playing "Railroad Tycoon" is the best way to learn how to use a financial calculator.

Carl

 

[seiferseph]

Oh, i get it now, Thanks so much for all the help!