you have just received an inheritance of $28,000 and would like to invest it into an account. the bank…

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1 + r)^n$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $n$ is the number of years. For the 4 - year investment: $P=$28000$, $r = 0.058$, $n = 4$. $A_1=28000\times(1 + 0.058)^4$. $A_1=28000\times(1.058)^4$. $(1.058)^4=1.058\times1.058\times1.058\times1.058\approx1.24194$. $A_1=28000\times1.24194=$34774.32$. Let the annual withdrawal be $x$. Using the present - value of an ordinary annuity formula $P = x\times\frac{1-(1 + r)^{-n}}{r}$, where $P$ is the present value (the amount in the account at the start of the withdrawals), $r$ is the interest rate per period, and $n$ is the number of periods. Since the account is zeroed out at the end, we can also work backward. The future value $A_1$ is used to find the equal - annual withdrawal. But if we assume we want to find the total amount available for withdrawal over the 4 - year period, it is $A_1=$34774.32$.

For the 3 - year investment: $P = 28000$, $r=0.07083$, $n = 3$. $A_2=28000\times(1 + 0.07083)^3$. $(1 + 0.07083)^3=1.07083\times1.07083\times1.07083\approx1.22696$. $A_2=28000\times1.22696=$34354.88$.

We can also use the annuity approach. But comparing the future values of the two investments after their respective time - periods, we can see that the 4 - year investment at 5.8% compounded annually will yield a higher return. We need to find the equal - annual withdrawal amount. For the 4 - year investment: Let the annual withdrawal be $x$. Using the present - value of an ordinary annuity formula $PV = x\times\frac{1-(1 + r)^{-n}}{r}$, where $PV$ is the present value of the annuity (the amount in the account at the start of withdrawals, which is the future value of the investment). Here, $PV = 28000\times(1.058)^4$, $r = 0.058$, $n = 4$. $28000\times(1.058)^4=x\times\frac{1-(1 + 0.058)^{-4}}{0.058}$. $34774.32=x\times\frac{1 - 1.058^{-4}}{0.058}$. $1.058^{-4}=\frac{1}{(1.058)^4}\approx0.8052$. $\frac{1 - 0.8052}{0.058}=\frac{0.1948}{0.058}\approx3.36$. $x=\frac{34774.32}{3.36}\approx$10349.5$. The total amount withdrawn over 4 years is approximately $4\times10349.5=$41398$.

For the 3 - year investment: $PV = 28000\times(1.07083)^3$, $r = 0.07083$, $n = 3$. $34354.88=x\times\frac{1-(1 + 0.07083)^{-3}}{0.07083}$. $(1.07083)^{-3}=\frac{1}{(1.07083)^3}\approx0.8151$. $\frac{1 - 0.8151}{0.07083}=\frac{0.1849}{0.07083}\approx2.61$. $x=\frac{34354.88}{2.61}\approx$13162.8$. The total amount withdrawn over 3 years is approximately $3\times13162.8=$39488.4$.

The 4 - year account will yield a higher return. To find the total amount available for withdrawal (assuming zero balance at the end), we use the future - value of the investment. For the 4 - year account: $A = 28000\times(1 + 0.058)^4=28000\times1.24194 = 34774.32$. If we assume equal annual withdrawals over 4 years, using the present - value of an ordinary annuity formula $A=x\times\frac{1-(1 + 0.058)^{-4}}{0.058}$, we find $x\approx8043.59$ and the total amount withdrawn over 4 years is $4x\approx32174.36$.

For the 3 - year account: $A = 28000\times(1+ 0.07083)^3=28000\times1.22696 = 34354.88$. Using the present - value of an ordinary annuity formula $A=x\times\frac{1-(1 + 0.07083)^{-3}}{0.07083}$, we find $x\approx10685.63$ and the total amount withdrawn over 3 years is $3x\approx32056.89$.

Answer:

4 year account; $32,174.36$