Question

Continuously Compounded Interest Quick Check
5 of 55 of 5 Items




Question
Which investment will earn the most amount of money?(1 point)
Responses

An investment of $2,800 that earns 6% interest compounded monthly over 2 years.
An investment of $2,800 that earns 6% interest compounded monthly over 2 years.

An investment of $1,200 that earns 2% interest compounded monthly over 8 years.
An investment of $1,200 that earns 2% interest compounded monthly over 8 years.

An investment of $4,200 that earns 12% interest compounded monthly over 4 years.
An investment of $4,200 that earns 12% interest compounded monthly over 4 years.

An investment of $2,400 that earns 4% interest compounded monthly over 12 years.

Answer 1

To determine which investment will earn the most money, we can use the formula for compound interest:

\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]

where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per unit \( t \).
- \( t \) is the time the money is invested for, in years.

We will calculate the total amount \( A \) for each investment.

1. **Investment 1**: $2,800 at 6% for 2 years compounded monthly.
- \( P = 2800 \)
- \( r = 0.06 \)
- \( n = 12 \)
- \( t = 2 \)
\[ A = 2800 \left(1 + \frac{0.06}{12}\right)^{12 \times 2} \]
\[ A = 2800 \left(1 + 0.005\right)^{24} \]
\[ A = 2800 \left(1.005\right)^{24} \]
\[ A \approx 2800 \times 1.12749 \approx 3,159.97 \]

2. **Investment 2**: $1,200 at 2% for 8 years compounded monthly.
- \( P = 1200 \)
- \( r = 0.02 \)
- \( n = 12 \)
- \( t = 8 \)
\[ A = 1200 \left(1 + \frac{0.02}{12}\right)^{12 \times 8} \]
\[ A = 1200 \left(1 + 0.0016667\right)^{96} \]
\[ A = 1200 \left(1.0016667\right)^{96} \]
\[ A \approx 1200 \times 1.173 \approx 1,407.60 \]

3. **Investment 3**: $4,200 at 12% for 4 years compounded monthly.
- \( P = 4200 \)
- \( r = 0.12 \)
- \( n = 12 \)
- \( t = 4 \)
\[ A = 4200 \left(1 + \frac{0.12}{12}\right)^{12 \times 4} \]
\[ A = 4200 \left(1 + 0.01\right)^{48} \]
\[ A = 4200 \left(1.01\right)^{48} \]
\[ A \approx 4200 \times 1.644 \approx 6,905.55 \]

4. **Investment 4**: $2,400 at 4% for 12 years compounded monthly.
- \( P = 2400 \)
- \( r = 0.04 \)
- \( n = 12 \)
- \( t = 12 \)
\[ A = 2400 \left(1 + \frac{0.04}{12}\right)^{12 \times 12} \]
\[ A = 2400 \left(1 + 0.0033333\right)^{144} \]
\[ A = 2400 \left(1.0033333\right)^{144} \]
\[ A \approx 2400 \times 1.490 \approx 3,576.00 \]

Now, let's summarize the computed amounts:
1. Investment 1: $3,159.97
2. Investment 2: $1,407.60
3. Investment 3: $6,905.55
4. Investment 4: $3,576.00

From the calculations, the investment that earns the most amount of money is **Investment 3: $4,200 at 12% interest compounded monthly over 4 years**.