To solve the question, we would be making use of the compound interest formula. This is given as;
![A=P(1+(r)/(n))^(nt)](https://img.qammunity.org/2023/formulas/mathematics/high-school/39foo2gerf9tf1ffk32zwshrn339mz02kv.png)
A=final amount = $20,000
P=initial principal balance
r=interest rate =3%
n=number of times interest applied per time period
t=number of time periods elapsed = 6 years
Part A
If the interest is compounded annually, n =1
Therefore;
![\begin{gathered} 20000=P(1+(0.03)/(1))^(1*6) \\ 20000=P(1+0.03)^6 \\ (1.03)^6P=20000 \\ P=(20000)/((1.03)^6) \\ P=16749.69 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2kzuybe92somnlstgge5gp76pk3sssqfgl.png)
Answer: The minimum amount would be $16749.69
Part B
If the interest is compounded monthly, n =12
![\begin{gathered} 20000=P(1+(0.03)/(12))^(6*12) \\ 20000=P((12+0.03)/(12))^(72) \\ ((12.03)/(12))^(72)P=20000 \\ P=(20000)/(((12.03)/(12))^(72)) \\ P=16709.16 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w6dyr47juk35nt924121j74l312z5uzqnj.png)
Answer: The minimum amount would be $16709.16
Part C
If the interest is compounded monthly, n =365
![\begin{gathered} 20000=P(1+(0.03)/(365))^(365*6) \\ 20000=P((365+0.03)/(365))^(2190) \\ ((365.03)/(365))^(2190)P=20000 \\ P=(20000)/(((365.03)/(365))^(2190)) \\ P=16705.53 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e4r0pes3wa1w1b8pafc0zdr5tm8716s4ge.png)
Answer: The minimum amount would be $16705.53