Problem 68 - College Algebra and Calculus: An Applied Approach

Question

\(P\) dollars, invested at interest rate \(r\) compounded annually, increases to an amount \(A=P(1+r)^{2}\) in 2 years. For an investment of \(\$ 2000\) to increase to an amount greater than \(\$ 2350\) in 2 years, the interest rate must be greater than what percent?

Step-by-Step Solution

Answer

The interest rate must be greater than \((\sqrt{\frac{2350}{2000}} -1) \cdot 100\) percent.

Step 1: Substitute the given values to the formula

The first step is to substitute the values of P and A into the formula. So we get: \n\[2350 = 2000(1+r)^{2}\]

Step 2: Simplify the equation

The next step is to simplify the equation to isolate r, which is our unknown. We start by dividing both sides of the equation by 2000 to isolate the (1+r)^2 term:\n\[ \frac{2350}{2000} = (1+r)^{2}\] \nThen, take the square root on each side of the equation to isolate (1+r):\n\[\sqrt{\frac{2350}{2000}} = 1+r \]

Step 3: Solve for r

Next, subtract 1 from both sides of the equation to solve for r:\n\[ r=\sqrt{\frac{2350}{2000}} -1 \].

Step 4: Convert the interest rate to a percentage

Finally, the interest rate is often represented as a percentage, so we multiply r by 100 to get the rate as a percentage:\n\[\text{Rate} = r \cdot 100\]\nTherefore, an interest rate over the value of r will result in an amount greater than $2350 in 2 years when the initial investment is $2000.

Key Concepts

Interest Rate Calculation

Interest rate calculation is fundamental in understanding how investments grow over time. When you invest a principal amount, known as \( P \), at an interest rate \( r \), the goal is often to see how much your investment will grow or what rate will help your investment achieve a certain future value. For compound interest, which is what we’re dealing with in this scenario, the interest not only applies to the initial principal each period but also to the accumulated amount from previous periods.

The core formula used in calculating compound interest over several years is \( A = P(1+r)^n \), where \( A \) is the future amount after \( n \) years, and \( r \) is the interest rate expressed as a decimal. To calculate \( r \) when you know \( P \), \( A \), and \( n \), you rearrange the formula to solve for \( r \). It’s crucial to understand each component:

• \( A \) - the future amount you want after the investment period.
• \( P \) - the principal or original investment.
• \( n \) - the number of years the money is invested or how often it's compounded.

Investment Growth

Investment growth refers to how your money increases over time due to compounding. In the example exercise, the calculation shows how a \( \\(2000 \) investment grows into more than \( \\)2350 \) in 2 years with a certain interest rate. The formula \( A = P(1+r)^n \) is a representation of exponential growth, where the growth rate is constant, and the investment increases by a larger amount each period.

Here’s why:

• In the first year, the interest earned is added to the original investment.
• In the subsequent year, you earn interest on this new total, not just the original amount.

This means that over time, your investment doesn't just grow linearly but accelerates, feeding off its own growth. The process dramatically enhances the final amount over long periods, providing greater returns than simple interest where interest is calculated solely off the initial principal. Understanding this growth pattern helps in planning and predicting your financial future.

Algebraic Equations

Algebraic equations allow us to solve for unknown variables, like the interest rate \( r \) in our problem. To find \( r \), we use algebra to manipulate the compound interest formula. By substituting the values we have – \( P = 2000 \) and \( A = 2350 \) – into the equation \( 2350 = 2000(1+r)^2 \), we can rearrange and isolate \( r \).

Let’s break down the solution:

• First, divide each side of the equation by 2000 to get \( \frac{2350}{2000} = (1+r)^2 \). This simplifies our equation.
• Next, take the square root of both sides to simplify further, which gives \( \sqrt{\frac{2350}{2000}} = 1 + r \).
• Finally, subtract 1 from both sides to solve for \( r \), resulting in \( r = \sqrt{\frac{2350}{2000}} - 1 \).

Every step is a basic algebraic operation applied to isolate \( r \). The understanding of these operations is crucial for tackling more complex financial calculations and modeling potentially profitable investments.