To determine which investment plan generates the most interest after three years, we can calculate the future value of the investment under each plan using the formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
Where:
- \( A \) = the amount of money accumulated after n years, including interest.
- \( P \) = the principal amount (the initial amount of money, which is $4,500).
- \( r \) = the annual interest rate (decimal).
- \( n \) = the number of times that interest is compounded per year.
- \( t \) = the number of years the money is invested for.
Let's calculate the future value for each plan.
-
5.0% compounded semi-annually
- \( P = 4500 \)
- \( r = 0.05 \)
- \( n = 2 \)
- \( t = 3 \)
\[ A = 4500 \left(1 + \frac{0.05}{2}\right)^{2 \cdot 3} = 4500 \left(1 + 0.025\right)^{6} = 4500 \left(1.025\right)^{6} \]
Calculating: \[ A \approx 4500 \times 1.159274 = 5226.73 \]
-
4.9% compounded quarterly
- \( P = 4500 \)
- \( r = 0.049 \)
- \( n = 4 \)
- \( t = 3 \)
\[ A = 4500 \left(1 + \frac{0.049}{4}\right)^{4 \cdot 3} = 4500 \left(1 + 0.01225\right)^{12} = 4500 \left(1.01225\right)^{12} \]
Calculating: \[ A \approx 4500 \times 1.158187 = 5226.84 \]
-
4.8% compounded monthly
- \( P = 4500 \)
- \( r = 0.048 \)
- \( n = 12 \)
- \( t = 3 \)
\[ A = 4500 \left(1 + \frac{0.048}{12}\right)^{12 \cdot 3} = 4500 \left(1 + 0.004\right)^{36} = 4500 \left(1.004\right)^{36} \]
Calculating: \[ A \approx 4500 \times 1.148882 = 5179.00 \]
-
5.1% compounded yearly
- \( P = 4500 \)
- \( r = 0.051 \)
- \( n = 1 \)
- \( t = 3 \)
\[ A = 4500 \left(1 + \frac{0.051}{1}\right)^{1 \cdot 3} = 4500 (1.051)^{3} \]
Calculating: \[ A \approx 4500 \times 1.157529 = 5208.88 \]
Now, let's summarize the results:
- 5.0% compounded semi-annually: $5,226.73
- 4.9% compounded quarterly: $5,226.84
- 4.8% compounded monthly: $5,179.00
- 5.1% compounded yearly: $5,208.88
The plan with the highest future value after three years is the 4.9% compounded quarterly plan, which generates approximately $5,226.84.