8.4 Question 6, 8.4.19 Points: 0 of 4 Save Find the monthly payment \( R \) needed to have a sinking fund accumulate the future value \( A \). The yearly interest rate \( r \) and the time \( t \) in years are given below. Interest is compounded monthly. \[ \mathrm{A}=\$ 4,000 ; \mathrm{r}=6.5 \% ; \mathrm{t}=4 \] \[ \mathrm{R}=\$ \] \( \square \) (Round up to the nearest cent.)

We are given the sinking fund formula that relates the future value \( A \) to the monthly payment \( R \): \[ A = R \cdot \frac{(1+i)^n - 1}{i} \] where - \( i \) is the monthly interest rate, - \( n \) is the total number of payments. ### Step 1. Identify Given Values - Future Value: \( A = \$4000 \) - Annual interest rate: \( r = 6.5\% = 0.065 \) - Time period in years: \( t = 4 \) - Since interest is compounded monthly, the number of months (payments) is: \[ n = 12t = 12 \times 4 = 48 \] - The monthly interest rate is: \[ i = \frac{r}{12} = \frac{0.065}{12} \approx 0.00541667 \] ### Step 2. Write the Formula and Solve for \( R \) The formula for the sinking fund is: \[ A = R \cdot \frac{(1+i)^n - 1}{i} \] Solve for \( R \): \[ R = A \cdot \frac{i}{(1+i)^n - 1} \] Substitute the known values: \[ R = 4000 \cdot \frac{0.00541667}{(1+0.00541667)^{48} - 1} \] ### Step 3. Compute the Denominator First, compute \( (1+i)^{n} \): \[ (1+0.00541667)^{48} \approx (1.00541667)^{48} \] To approximate, use the natural logarithm: \[ \ln(1.00541667) \approx 0.005398 \] \[ 48 \cdot 0.005398 \approx 0.2591 \] \[ (1.00541667)^{48} \approx e^{0.2591} \approx 1.296 \] Now, subtract 1: \[ (1+i)^n - 1 \approx 1.296 - 1 = 0.296 \] ### Step 4. Compute \( R \) Substitute back into the formula: \[ R \approx 4000 \cdot \frac{0.00541667}{0.296} \] First, compute the numerator: \[ 4000 \times 0.00541667 \approx 21.66668 \] Then, divide by 0.296: \[ R \approx \frac{21.66668}{0.296} \approx 73.275 \] ### Step 5. Round to the Nearest Cent Rounded to the nearest cent: \[ R \approx \$73.28 \] --- ### Final Answer \[ \boxed{\$73.28} \]