Problem 54 - A Graphical Approach to Precalculus with Limits

Question

A total of \(\$ 5000\) is invested at \(2 \%, 3 \%,\) and \(4 \%\). The amount invested at \(4 \%\) equals the total amount invested at \(2 \%\) and \(3 \%\). The total interest for one year is \$145. If possible, find the amount invested at each interest rate. Interpret your answer.

Step-by-Step Solution

Answer

Invest 2000 at 2%, 500 at 3%, and 2500 at 4%.

Step 1: Define Variables

Let \( x \) be the amount invested at \(2\%\), \( y \) be the amount invested at \(3\%\), and \( z \) be the amount invested at \(4\%\). We have the equation: \( x + y + z = 5000 \) representing the total investment.

Step 2: Express Condition of Investment at 4%

According to the problem, the amount invested at \(4\%\) equals the total amount invested at \(2\%\) and \(3\%\). This means \( z = x + y \).

Step 3: Set Up Equation for Total Interest

The total interest for one year is given by the expression \(0.02x + 0.03y + 0.04z = 145\).

Step 4: Substitute for z

From Step 2, we have \( z = x + y \). Substitute \( z \) in the interest equation: \( 0.02x + 0.03y + 0.04(x + y) = 145 \).

Step 5: Simplify the Interest Equation

Simplify \( 0.02x + 0.03y + 0.04x + 0.04y = 145 \) to get \( 0.06x + 0.07y = 145 \).

Step 6: Solve for x and y Using Both Equations

We have two equations now: 1) \( x + y + z = 5000 \) 2) \( 0.06x + 0.07y = 145 \)Substituting \( z = x + y \) into the first equation gives \( x + y + x + y = 5000 \), which simplifies to \( 2x + 2y = 5000 \). Hence, \( x + y = 2500 \).

Step 7: Substitute for x and Solve for y

Using \( x + y = 2500 \), we substitute into the second equation: \( 0.06x + 0.07y = 145 \). Let \( x = 2500 - y \). Substituting gives \( 0.06(2500 - y) + 0.07y = 145 \). Simplifying, \( 150 - 0.06y + 0.07y = 145 \). Thus, \( 0.01y = 5 \). Solving gives \( y = 500 \).

Step 8: Find x from y

Since \( y = 500 \), substitute back into \( x + y = 2500 \): \( x + 500 = 2500 \), so \( x = 2000 \).

Step 9: Find z from x and y

Using \( z = x + y \), we have \( z = 2000 + 500 = 2500 \).

Step 10: Verify Solution

The total amounts are \( x = 2000 \), \( y = 500 \), \( z = 2500 \) which sum to \( 5000 \) and calculate the total interest as \( 0.02 \times 2000 + 0.03 \times 500 + 0.04 \times 2500 = 145 \). The solution is correct.

Key Concepts

Simple Interest

Simple interest is a way to calculate the earnings or costs related to an investment or loan over time. It is straightforward because it only considers the principal amount of the investment. Unlike compound interest, which can complicate the calculations by computing interest on top of interest, simple interest remains consistent. Here’s how it works:

• The formula for simple interest is given by: \[ I = P \times r \times t \]where \( I \) is the interest earned, \( P \) is the principal amount, \( r \) is the interest rate, and \( t \) is the time period, usually in years.
• In our problem, three different interest rates are applied to various investments. Each earns interest based on its unique rate over the same period, one year.
• For instance, if you invest \(2000 at an interest rate of 2%, you would calculate \( I = 2000 \times 0.02 \times 1 = 40 \), meaning \)40 in interest would be earned over that year.

Understanding simple interest is crucial because it helps quantify how much money one would gain or owe after a set period. In investment scenarios, it allows investors to predict income from their investments without dealing with more complex computations.

System of Equations

A system of equations involves finding the values of variables that satisfy all equations involved. In investment scenarios, they help us solve problems where multiple unknowns are interlinked, like in our original exercise.

• Equations are typically formulated based on the conditions provided in the problem.
• For the investment word problem, we have two primary equations:
  • \( x + y + z = 5000 \): This equation represents the total amount invested across the three different percentages.
  • \( 0.06x + 0.07y = 145 \): Represents the total interest earned from all investments.
• By solving this system, we determine exactly how much money is invested at each interest rate. This ensures all conditions of the problem are met and allow us to validate our result by checking against all criteria and original amounts.

Establishing and solving a system of equations is fundamental in many areas beyond just investments. It’s an essential skill in mathematical problem-solving that applies to numerous real-world issues.

Algebraic Substitution

Algebraic substitution is a method used to solve equations, particularly systems with two or more equations. This technique involves replacing one variable with an expression derived from another equation. Here’s how it works in the given problem context:

• We start with the expression for one variable in terms of others. For example, from our condition that \( z = x + y \), we can substitute \( z \) wherever it's required in equations.
• Using substitution helps reduce the number of variables in complex equations. In our context, substituting \( z = x + y \) into the equation for total interest simplifies our computations.
• In practice, substitution might look like converting our interest equation \( 0.06x + 0.07y = 145 \), into a less complex equation by solving for \( x \) and \( y \) directly after eliminating \( z \).
• Thus, substitution allows us to focus on one equation at a time, streamlining the process of solving for unknowns efficiently.

Algebraic substitution is beneficial because it simplifies often challenging calculations into more manageable steps. By mastering this technique, one can tackle a variety of algebraic problems with ease.