Answer:
Part A)
About $3767.34.
Part B)
About $3692.47.
Step-by-step explanation:
Part A)
Recall that compound interest is given by the formula:
[tex]\displaystyle A = P\left(1+\frac{r}{n}\right)^{nt}[/tex]
Where A is the final amount, P is the initial amount, r is the interest rate, n is the number of times compounded per year, and t is the number of years.
To obtain $4000 after two years, let A = 4000 and t = 2.
Because the account pays 3% interest compounded monthly, r = 0.03 and n = 12.
Substitute and solve for P:
[tex]\displaystyle \begin{aligned} (4000) & = P\left(1+\frac{(0.03)}{(12)}\right)^{(12)(2)} \\ \\ P & = \frac{4000}{\left(1+\dfrac{(0.03)}{(12)}\right)^{(12)(2)}} \\ \\ & \approx \$3767.34\end{aligned}[/tex]
In concluion, about $3767.34 should be deposited.
Part B)
Recall the formula for continuous compound:
[tex]\displaystyle A = Pe^{rt}[/tex]
Where e is Euler's number.
Hence, let A = 4000, r = 0.04 and t = 2. Substitute and solve for P:
[tex]\displaystyle \begin{aligned}(4000) & = Pe^{(0.04)(2)} \\ \\ P & = \frac{4000}{e^{(0.02)(4)}} \\ \\ & \approx \$3692.47 \end{aligned}[/tex]
In conclusion, about $3692.47 should be deposited.
