Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

An item costs \(\$ 1000\) this year. This is a \(10 \%\) increase over the price last year. What was the price last year? (Caution: it was not \(\$ 900\). It would be wise to give last year s price a name like \(x\), or \(P\), or some other labeling of your choice.)

Step-by-Step Solution

Answer

The price of the item last year was approximately $909.09.

Step 1: Define the variable and the equation

Let's name last year's price as \(x\). Since there is a 10% increase over the previous year's price, the equation can be written as \(x + 0.1x = 1000\)

Step 2: Simplify the equation

The equation can be simplified to \(1.1x = 1000\)

Step 3: Solve for the variable \(x\)

The price of the item last year \(x\) can be found by reworking the equation. This would make the equation \(x = 1000/1.1\).

Step 4: Perform the calculation

Calculate the value to find \(x\) which would give us the price for last year.

Key Concepts

Percent Change Calculation

Understanding percent change is essential in various fields, from finance to everyday shopping. It measures how much a quantity has grown or shrunk as a percentage of its original amount. To calculate percent change, we apply the formula:

\[ \text{Percent Change} = \left( \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \right) \times 100\]
In the exercise scenario, we are given the new price after a 10% increase. But the twist is to think backwards, identifying the original price before the increase. Crucially, remember that a common mistake is to simply subtract the percent change from the new value. This mistake fails to account for the proportional relationship between the new value and the original value. Percent change calculations require understanding this relation to avoid such pitfalls.

Price Before Increase

In the given problem, determining the original price before a percentage increase is our goal. This involves unwinding the percent change. When a new value results from an increase on an original amount, the relationship can be expressed algebraically. The new value (in this case, $1000) is equal to the original price plus the amount of increase (10% of the original price). To express this relationship, we write:

\[ \text{New Value} = \text{Original Price} + (\text{Percent Increase} \times \text{Original Price})\]
This relationship is key to solving our problem, as it allows us to set up an equation to find the 'price before increase'. It’s important to remember that the original value serves as the base for the percentage calculation, not the other way around which is a common misunderstanding.

Setting Up Algebraic Equations

Transitioning from understanding the problem to setting up the algebraic equations is where many students falter. To bridge this gap, one must grasp the underlying relationships and translate them into an equation. In our exercise, we have a known final value (the price after increase) and an unknown original value (the price before increase). Assign a variable to represent this unknown, such as 'x'.

Next, incorporate the percent increase into the equation. Since a 10% increase is at play, represent this arithmetically as '1 + 0.1' (since 10% is 0.1 in decimal form). Multiplying this sum by the original price 'x' gives us the final price, leading to the setup:

\[1.1x = \text{New Value}\]
The correct setup of an algebraic equation is vital, as it paves the way for solving the variable in question.

Solving for x

The last phase in our journey is resolving the algebraic equation for the variable 'x', which stands for the original price before the increase. Following our established equation \(1.1x = 1000\), we need to isolate 'x' to discover its value. To do this, we divide both sides of the equation by 1.1, resulting in:

\[x = \frac{1000}{1.1}\]
Performing the division yields our answer. Solving for 'x' is essential not only for answering the problem at hand but also for building a foundation in algebraic manipulation. Remember to perform the same operation on both sides of the equation to maintain balance and arrive at the correct solution.