To determine the minimum value of k , the number of compounding periods per year, such that Bob's savings reach the target of \$500,000 by the time he retires at age 65, we will use the compound interest formula and a graphical approach with a TI-84 calculator.

Understanding the Compound Interest Formula

The formula for compound interest is given by:

FV = PV \left( 1 + \frac{r}{100k} \right)^{kn} \,\,\,\,\,\left\lbrace\text{SL 1.4}\right\rbrace

where:

FV is the future value of the investment, which is \$500,000 .
PV is the present value or initial investment, which is \$46,000 .
r is the annual interest rate, which is 6% .
k is the number of compounding periods per year.
n is the number of years the money is invested, which is 40.

Setting Up the Equation

We need to find k such that:

500\,000 = 46\,000 \left( 1 + \frac{6}{100k} \right)^{40k}

Graphing the Function

Enter the Function:
- On your TI-84 calculator, press the Y= button to access the function editor.
- Enter the function for the future value of Bob's savings:
  Y_1 = 46\,000 \left( 1 + \frac{6}{100X} \right)^{40X}
- Enter the target value as a horizontal line:
  Y_2 = 500\,000
Adjust the Viewing Window:
- Press the WINDOW button.
- Set the window settings to view a reasonable range for k and the future value:
  - X_{\text{min}} = 0
  - X_{\text{max}} = 10
  - Y_{\text{min}} = 0
  - Y_{\text{max}} = 600\,000
Graph the Functions:
- Press the GRAPH button to display the graphs of Y_1 and Y_2 .
Find the Intersection:
- Press 2nd and then TRACE to access the CALC menu.
- Select 5:intersect.
- Use the arrow keys to move the cursor close to the point where the two graphs intersect.
- Press ENTER three times to calculate the intersection point.

Interpreting the Result

The X -value of the intersection point represents the minimum value of k needed for Bob's savings to reach \$500,000 . You should find k \approx 5.09 .
Since k must be an integer and we need the savings to exceed \$500,000 , round k UP to the nearest whole number.

By following these steps, you will find that the minimum value of k is 6 .

Problem Bank - Financial Mathematics

Problem Bank - Financial Mathematics