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Melissa has decided to invest to help with her retirement savings. How much would she have to invest to have [tex]$\$[/tex] 114,700[tex]$ after 20 years, assuming an interest rate of $[/tex]3.42\%[tex]$ compounded daily?

Do not round any intermediate computations, and round your final answer to the nearest dollar. Assume there are 365 days in each year.

\[
\$[/tex] \square
\]

Financial Formulas:

[tex]\[
\begin{array}{l}
I = \text{Pr} T \\
A = P(1 + r t) \\
A = P\left(1 + \frac{r}{n}\right)^{n t} \\
A = P e^{r t} \\
A = \frac{M \left[\left(1 + \frac{r}{n}\right)^{n t} - 1\right]}{\frac{r}{n}} \\
M = \frac{P \left(\frac{r}{12}\right)}{1 - \left(1 + \frac{r}{12}\right)^{-12 t}}
\end{array}
\][/tex]

Where:

[tex]\[
\begin{array}{l}
P = \text{principal} \\
I = \text{simple interest} \\
r = \text{annual interest rate} \\
t = \text{time or term of investment or loan (in years)} \\
A = \text{future value or amount accumulated} \\
n = \text{number of times interest is compounded per year} \\
e = \text{Euler's number} \\
M = \text{installment payment or monthly payment}
\end{array}
\][/tex]


Sagot :

To determine how much Melissa needs to invest to achieve a future value of \[tex]$114,700 after 20 years with an annual interest rate of 3.42% compounded daily, we will use the compound interest formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \(A\) is the future value of the investment, which is \$[/tex]114,700.
- [tex]\(P\)[/tex] is the principal amount (the initial amount to be invested).
- [tex]\(r\)[/tex] is the annual interest rate (in decimal form), which is 0.0342.
- [tex]\(n\)[/tex] is the number of times interest is compounded per year, which is 365 for daily compounding.
- [tex]\(t\)[/tex] is the number of years the money is invested, which is 20 years.

We need to rearrange the formula to solve for [tex]\(P\)[/tex]:

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

Let's substitute the given values into the formula and solve for [tex]\(P\)[/tex]:

1. Substitute the given values into the formula:
[tex]\[ P = \frac{114700}{ \left(1 + \frac{0.0342}{365}\right)^{365 \times 20} } \][/tex]

2. Compute the base of the exponent:
[tex]\[ 1 + \frac{0.0342}{365} = 1 + 0.00009369863 \approx 1.00009369863 \][/tex]

3. Compute the exponent:
[tex]\[ 365 \times 20 = 7300 \][/tex]

4. Now compute the full expression in the denominator:
[tex]\[ \left(1.00009369863\right)^{7300} \approx 1.981314167 \][/tex]

5. Now, plug this back into our formula for [tex]\(P\)[/tex]:
[tex]\[ P = \frac{114700}{1.981314167} \approx 57878.85197906927 \][/tex]

Finally, we round the result to the nearest dollar:
[tex]\[ P \approx 57879 \][/tex]

So, Melissa would need to invest \[tex]$57,879 to have \$[/tex]114,700 after 20 years with an annual interest rate of 3.42% compounded daily.