Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Investment Portfolio A total of $$\$ 50,000$$ is invested in two municipal bonds that pay \(6.75 \%\) and \(8.25 \%\) simple interest. The total annual interest is $$\$ 3900$$. How much is invested in each bond?

Step-by-Step Solution

Answer

So, \$15000 is invested at 6.75% and \$35000 is invested at 8.25%

Step 1 - Presenting given information in terms of variables

Suppose the amount invested at 6.75% is \(x\) and the amount invested at 8.25% is \(y\). From the problem, there are two equations we can get. First, the total investment: \(x + y = 50000\) Second, the total annual interest: \(0.0675x + 0.0825y = 3900\)

Step 2 - Simplifying equations

The second equation is not very handy in its current form. Multiply the equation by 100 to get rid of the decimal, this gives us: \(6.75x + 8.25y = 390000\)

Step 3 - Solving the equation

Now we can use substitution method to solve the system. We can express \(x\) from the first equation: \(x = 50000 - y\) Now we substitute \(x\) in the second equation: \(6.75(50000 - y) + 8.25y = 390000\)

Step 4 - Simplifying and solving for y

Now simplify the equation and solve for \(y\): \(337500 - 6.75y + 8.25y = 390000 \) which simplifies to \(337500 + 1.5y = 390000\) Solving for \(y\) we get that \(y = 35000\) (money invested at 8.25%)

Step 5 - Find the value for x

Now, we can substitute \(y = 35000\) into \(x + y = 50000\) to find \(x\), which gives us \(x = 15000\) (money invested at 6.75%)

Key Concepts

Simple Interest

Understanding simple interest is crucial when dealing with investments or loans. Simple interest is a quick way of calculating interest charged or earned on a principal amount. It's straightforward, as it doesn't take into account the effect of compounding.

• Principal (P): The initial amount of money invested or borrowed.
• Rate (R): The percentage of the principal charged as interest each period.
• Time (T): The time period over which the interest is calculated.

The formula for simple interest is: \[ I = P \times R \times T \]In the context of our problem, if you invest $$50,000 in bonds offering different interests, the interest earned is based solely on the principal and rate, without recurring effects of reinvestment.

System of Equations

A system of equations involves finding values for variables that satisfy multiple equations simultaneously. In our investment problem, we have two equations:

• The total investment equation: \( x + y = 50000 \)
• The annual interest equation: \( 0.0675x + 0.0825y = 3900 \)

We need to find values of \(x\) and \(y\), representing the amounts invested at different interest rates. Solving a system like this helps determine how much is invested in each bond, ensuring both conditions (total investment and interest) are met.

Substitution Method

The substitution method is a systematic way of solving a system of equations. It involves solving one equation for one variable and then substituting that expression into the other equation. Let's go through it.
We start with the equation \(x + y = 50000\).

• Express \(x\) in terms of \(y\): \(x = 50000 - y\)
• Substitute \(x\) in the interest equation: \(6.75(50000 - y) + 8.25y = 390000\)
• This substitution turns the problem into a single-variable equation, making it simpler to solve.

By solving for \(y\), you subsequently find \(x\) by plugging the result back into the first equation. This method is particularly useful in problems involving linear equations.

Algebraic Expressions

Algebraic expressions form the core of mathematical problem-solving. They consist of variables, numbers, and operations. These expressions allow us to model real-world situations, like our investment portfolio problem.
In our scenario:

• The expression \(x + y = 50000\) is straightforward and represents total investment.
• The more complex \(0.0675x + 0.0825y = 3900\) models interest earnings, requiring multiplication and addition to solve.

Manipulating these expressions through basic operations like addition, subtraction, and factoring helps solve for unknowns in a logical, step-by-step manner.