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### Section 6.5 : Applications

1. We have $10,000 to invest for 44 months. How much money will we have if we put the money into an account that has an annual interest rate of 5.5% and interest is compounded 1. quarterly 2. monthly 3. continuously Show All Solutions Hide All Solutions a quarterly Show Solution From the problem statement we can see that, $P = 10000\hspace{0.25in}r = \frac{{5.5}}{{100}} = 0.055\hspace{0.25in}t = \frac{{44}}{{12}} = \frac{{11}}{3}$ Remember that the value of $$r$$ must be given as a decimal, i.e. the percentage divided by 100. Also remember that $$t$$ must be in years and so we’ll need to convert the months we are given to years. For this part we are compounding interest rate quarterly and that means it will compound 4 times per year and so we also then know that, $m = 4$ At this point all that we need to do is plug into the equation and run the numbers through a calculator to compute the amount of money that we’ll have. $A = 10000{\left( {1 + \frac{{0.055}}{4}} \right)^{\frac{{11}}{3}\left( 4 \right)}} = 10000{\left( {1.01375} \right)^{\frac{{44}}{3}}} = 10000\left( {1.221760422} \right) = 12217.60$ So, we’ll have$12,217.60 in the account after 44 months.

b monthly Show Solution

From the problem statement we can see that,

$P = 10000\hspace{0.25in}r = \frac{{5.5}}{{100}} = 0.055\hspace{0.25in}t = \frac{{44}}{{12}} = \frac{{11}}{3}$

Remember that the value of $$r$$ must be given as a decimal, i.e. the percentage divided by 100. Also remember that $$t$$ must be in years and so we’ll need to convert the months we are given to years.

For this part we are compounding interest rate monthly and that means it will compound 12 times per year and so we also then know that,

$m = 12$

At this point all that we need to do is plug into the equation and run the numbers through a calculator to compute the amount of money that we’ll have.

$A = 10000{\left( {1 + \frac{{0.055}}{{12}}} \right)^{\frac{{11}}{3}\left( {12} \right)}} = 10000{\left( {1.00453333} \right)^{44}} = 10000\left( {1.222876562} \right) = 12228.77$

So, we’ll have $12,228.77 in the account after 44 months. c continuously Show Solution From the problem statement we can see that, $P = 10000\hspace{0.25in}r = \frac{{5.5}}{{100}} = 0.055\hspace{0.25in}t = \frac{{44}}{{12}} = \frac{{11}}{3}$ Remember that the value of $$r$$ must be given as a decimal, i.e. the percentage divided by 100. Also remember that $$t$$ must be in years and so we’ll need to convert the months we are given to years. For this part we are compounding continuously and so we won’t have an $$m$$ and will be using the other equation and all we have all we need to do the computation so, $A = 10000{{\bf{e}}^{\left( {0.055} \right)\left( {\frac{{11}}{3}} \right)}} = 10000{{\bf{e}}^{0.2016666667}} = 10000\left( {1.223440127} \right) = 12234.40$ So, we’ll have$12,234.40 in the account after 44 months.