To determine which investment yields the greater return over 2 years, we'll use the compound interest formulas for both situations given:
1. Compounded monthly:
The formula for compound interest when interest is compounded periodically is:
[tex]\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\][/tex]
where:
- [tex]\(A\)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\(P\)[/tex] is the principal amount (the initial amount of money, which is [tex]$14,000).
- \(r\) is the annual interest rate (in decimal form, so 12% is 0.12).
- \(n\) is the number of times that interest is compounded per unit time (12 for monthly compounding).
- \(t\) is the time the money is invested for in years (2 years).
Plugging in the known values:
\[
A_{\text{monthly}} = 14000 \left(1 + \frac{0.12}{12}\right)^{12 \times 2}
\]
Upon calculating this, we find that:
\[
A_{\text{monthly}} \approx 17776.29
\]
2. Compounded continuously:
The formula for compound interest when interest is compounded continuously is:
\[
A = P e^{rt}
\]
where:
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial amount of money, which is $[/tex]14,000).
- [tex]\(r\)[/tex] is the annual interest rate (in decimal form, so 11.92% is 0.1192).
- [tex]\(t\)[/tex] is the time the money is invested for in years (2 years).
- [tex]\(e\)[/tex] is the base of the natural logarithm, approximately equal to 2.71828.
Plugging in the known values:
[tex]\[
A_{\text{continuous}} = 14000 \cdot e^{0.1192 \times 2}
\][/tex]
Upon calculating this, we find that:
[tex]\[
A_{\text{continuous}} \approx 17769.03
\][/tex]
Comparing the two amounts:
- [tex]\(\$ 17,776.29\)[/tex] for the investment compounded monthly at 12%.
- [tex]\(\$ 17,769.03\)[/tex] for the investment compounded continuously at 11.92%.
The investment that yields the greater return over 2 years is the 12% compounded monthly.
Thus, the correct answer is:
[tex]\[
\boxed{\text{12% compounded monthly}}
\][/tex]
