Let's solve this problem step-by-step:
### Account A:
1. Initial investment (P): [tex]$50,000
2. Annual interest rate (r): 8% or 0.08
3. Term of investment (t): 8 years
4. Compounding frequency (n): 1 time per year (since interest is compounded annually)
To calculate the final amount in Account A, we will use the compound interest formula:
\[ A = P \left( 1 + \frac{r}{n} \right)^{nt} \]
Plugging in the values:
\[ A = 50000 \left( 1 + \frac{0.08}{1} \right)^{1 \times 8} \]
\[ A = 50000 (1 + 0.08)^8 \]
\[ A = 50000 (1.08)^8 \]
Performing the calculation:
\[ (1.08)^8 \approx 1.85093 \]
\[ A \approx 50000 \times 1.85093 \]
\[ A \approx 92546.51 \]
So, the final amount in Account A after 8 years is approximately \$[/tex]92,546.51.
### Account B:
1. Initial investment (P): [tex]$50,000
2. Annual interest rate (r): 7% or 0.07
3. Term of investment (t): 10 years
To calculate the final amount in Account B, we will use the formula for continuous compounding:
\[ A = Pe^{rt} \]
Plugging in the values:
\[ A = 50000 \times e^{0.07 \times 10} \]
\[ A = 50000 \times e^{0.7} \]
We know that \( e^{0.7} \approx 2.01424 \):
\[ A \approx 50000 \times 2.01424 \]
\[ A \approx 100687.64 \]
So, the final amount in Account B after 10 years is approximately \$[/tex]100,687.64.
### Conclusion:
Comparing the final amounts:
- Account A: \[tex]$92,546.51
- Account B: \$[/tex]100,687.64
The account that earns the greatest amount of interest is Account B. Therefore, the best decision for the customer is to invest in Account B.
