To solve this problem, we first need to define the polynomial models for each investment using the formula \( x = 1 + r \). Since the interest rate is given as 3.7%, we convert this to a decimal for our calculations:
\[ r = 0.037 \implies x = 1 + r = 1 + 0.037 = 1.037 \]
Investment 1:
For Investment 1, we deposit $4,000 at the beginning of the first year. The future value is based on compounding interest over 7 years. The polynomial model can be set up as follows:
\[ C_1(x) = 4000 \cdot x^7 \]
Investment 2:
For Investment 2, we make multiple deposits at different times:
- Deposit $2,000 at the beginning of the first year.
- Deposit $1,500 at the beginning of the third year (which earns interest for 5 years).
- Deposit $500 at the beginning of the fifth year (which earns interest for 3 years).
The polynomial model for Investment 2 can be expressed as:
\[ C_2(x) = 2000 \cdot x^7 + 1500 \cdot x^5 + 500 \cdot x^3 \]
Evaluating the Models:
Now we will substitute \( x = 1.037 \) into both models to find the final amounts.
For Investment 1:
\[ C_1(1.037) = 4000 \cdot (1.037)^7 \]
Calculating \( (1.037)^7 \):
\[ (1.037)^7 \approx 1.28403 \quad \text{(using a calculator)} \]
So,
\[ C_1(1.037) \approx 4000 \cdot 1.28403 \approx 5136.12 \]
For Investment 2:
\[ C_2(1.037) = 2000 \cdot (1.037)^7 + 1500 \cdot (1.037)^5 + 500 \cdot (1.037)^3 \]
Calculating each term:
- First term: \( 2000 \cdot (1.037)^7 \approx 2000 \cdot 1.28403 \approx 2568.06 \)
- Second term: \( 1500 \cdot (1.037)^5 \)
Calculating \( (1.037)^5 \): \[ (1.037)^5 \approx 1.195618 \quad \text{(using a calculator)} \] So, \[ 1500 \cdot (1.037)^5 \approx 1500 \cdot 1.195618 \approx 1793.43 \]
- Third term: \( 500 \cdot (1.037)^3 \)
Calculating \( (1.037)^3 \): \[ (1.037)^3 \approx 1.114882 \quad \text{(using a calculator)} \] So, \[ 500 \cdot (1.037)^3 \approx 500 \cdot 1.114882 \approx 557.44 \]
Adding these together:
\[ C_2(1.037) \approx 2568.06 + 1793.43 + 557.44 \approx 4918.93 \]
Summary:
- Final amount for Investment 1: $5,136.12
- Final amount for Investment 2: $4,918.93
Interest Earned:
- For Investment 1, interest earned = \( 5136.12 - 4000 = 1136.12 \)
- For Investment 2, interest earned = \( 4918.93 - 4000 = 918.93 \)
Conclusion:
Investment 1 results in more interest earned at the end of the 7-year term.
Reasoning:
In general, investing a larger amount at one time can lead to more interest earned than dividing the investment into smaller deposits. This is because money invested earlier has more time to grow due to compounding interest. In this case, even with smaller subsequent investments, the initial larger investment in Investment 1 allowed for more compounding growth over the entire term compared to the staggered deposits in Investment 2.