Problem 58 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

COMPOUND INTEREST How much should you invest now at an annual interest rate of \(6.25 \%\) so that your balance 10 years from now will be \(\$ 2,000\) if interest is compounded: a. Monthly b. Continuously

Step-by-Step Solution

Answer

Invest approximately \(1099.85 for monthly compounding and \)1070.72 for continuous compounding.

Step 1 - Understand the formula for compound interest

For monthly compounding, the formula is: \[ A = P \times \bigg(1 + \frac{r}{n}\bigg)^{nt} \] where \(A\) is the future value, \(P\) is the principal amount, \(r\) is the annual interest rate, \(n\) is the number of times interest is compounded per year, and \(t\) is the time in years.

Step 2 - Convert and substitute values for monthly compounding

Given:\(A = 2000\)\(r = 0.0625\)\(n = 12\)\(t = 10\) years. Substitute these values into the formula: \[ 2000 = P \times \bigg(1 + \frac{0.0625}{12}\bigg)^{12 \times 10} \]

Step 3 - Solve for P in monthly compounding

Solve the equation: \[ 2000 = P \times \bigg(1 + \frac{0.0625}{12}\bigg)^{120} \]\[ P = \frac{2000}{\bigg(1 + \frac{0.0625}{12}\bigg)^{120}} \]\[ P \approx \frac{2000}{(1.005208)^{120}} \]\[ P \approx \frac{2000}{1.819396} \]\[ P \approx 1099.85 \]

Step 4 - Understand the formula for continuous compounding

For continuous compounding, the formula is: \[ A = P \times e^{rt} \] where \(e\) is the base of the natural logarithm.

Step 5 - Substitute values for continuous compounding

Given:\(A = 2000\)\(r = 0.0625\)\(t = 10\) years. Substitute these values into the formula: \[ 2000 = P \times e^{(0.0625 \times 10)} \]

Step 6 - Solve for P in continuous compounding

Solve the equation: \[ 2000 = P \times e^{0.625} \]\[ P = \frac{2000}{e^{0.625}} \]\[ P \approx \frac{2000}{1.868246} \]\[ P \approx 1070.72 \]

Key Concepts

annual interest rate

The annual interest rate is a critical component in calculating compound interest. It represents the percentage increase in the principal amount over a period of one year.
Let's break down the importance of the annual interest rate in simpler terms: It shows how much extra money you will earn from your investment yearly.
In the given exercise, the annual interest rate is 6.25%.
This means for every \(100 you invest, you expect to gain \)6.25 after one year.
When dealing with compound interest, the annual rate needs to be adjusted depending on how often interest is compounded.
For monthly compounding, the annual rate is divided by 12, translating it into a monthly rate.
For continuous compounding, the rate is integrated into an exponential function using the base of natural logarithms. Understanding this rate is crucial for accurate calculations of future investment value.

monthly compounding

Monthly compounding is when the interest on an investment is calculated and added to the principal balance once every month.
This happens 12 times a year and results in earning 'interest on interest' as the months progress.
Using the provided exercise as an example, the formula for monthly compounding is: \( A = P \times \bigg(1 + \frac{r}{n}\bigg)^{nt} \).
Here, A is the amount of money you want in the future, P is the present amount (or principal), r is the annual interest rate in decimal form, n is the number of times interest is compounded per year, and t is the total time in years.
By substituting the values given: A = 2000, r = 0.0625, n = 12, and t = 10 years, you can solve for P, the amount you should invest now.
After breaking the calculation down: \( P \approx \frac{2000}{(1.005208)^{120}} \), which simplifies to \( P \approx 1099.85 \).
So, to have \(2000 in 10 years with monthly compounding at a rate of 6.25%, you should invest approximately \)1099.85 now.

continuous compounding

Continuous compounding assumes that interest is being added to the principal balance an infinite number of times per year.
This means that the investment grows at every possible moment.
The formula for continuous compounding is: \(A = P \times e^{rt} \), where e is the base of the natural logarithm (approximately 2.71828).
Continuing with the example from the exercise: A = 2000, r = 0.0625, and t = 10 years.
Substitute these values into the formula: \( A = 2000 = P \times e^{0.625} \), then solve for P: \( P \approx \frac{2000}{1.868246} \).
This calculation results in \ P \approx 1070.72 \.
Therefore, to have \(2000 in 10 years with continuous compounding at an annual interest rate of 6.25%, you should invest about \)1070.72 right now.