Problem 41 - College Algebra with Corequisite Support

Question

For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question. Find the dimensions of an American football field. The length is \(200 \mathrm{ft}\) more than the width, and the perimeter is \(1,040 \mathrm{ft}\). Find the length and width. Use the perimeter formula \(P=2 L+2 W\).

Step-by-Step Solution

Answer

Width: 160 ft, Length: 360 ft.

Step 1: Identify Variables

Let's define the variables. Let \( L \) be the length of the football field, and \( W \) be the width. We are given that the length \( L \) is 200 feet more than the width \( W \). This can be expressed as the equation \( L = W + 200 \).

Step 2: Substitute Length Equation

Substitute the expression for \( L \) from the first equation into the perimeter formula. The perimeter formula is \( P = 2L + 2W \), and we have \( L = W + 200 \). So the formula becomes: \[ P = 2(W + 200) + 2W.\]

Step 3: Simplify the Perimeter Expression

Expand and simplify the expression: \[ P = 2(W + 200) + 2W = 2W + 400 + 2W = 4W + 400.\]

Step 4: Solve for Width

The perimeter \( P \) is given as 1040 feet. Set up the equation and solve for \( W \): \[ 1040 = 4W + 400.\] Subtract 400 from both sides to get: \[ 640 = 4W.\]Divide both sides by 4:\[ W = 160.\]

Step 5: Find Length Using Width

Substitute the value of \( W \) back into the equation for \( L \): \[ L = W + 200 = 160 + 200 = 360.\]

Step 6: Verify the Solution

Substitute \( L = 360 \) and \( W = 160 \) back into the perimeter formula to verify the solution: \[ P = 2(360) + 2(160) = 720 + 320 = 1040.\] This matches the given perimeter, confirming the dimensions are correct.

Key Concepts

dimensions of a rectangle

The dimensions of a rectangle are its length and width. Understanding these measurements is crucial when dealing with problems about rectangles, like finding the perimeter. In rectangles, we often differentiate between

• Length (L): The measure of the longer side.
• Width (W): The measure of the shorter side.

These measurements help in defining the shape and size of the rectangle.
When solving problems, it's important to know how the length and width relate to each other.
For example, in the problem of a football field, we knew that the length is 200 feet more than the width. This additional information helped us set up an equation to find the exact measurements. Recognizing the relationships between dimensions is often key to solving such exercises.

solving linear equations

Solving linear equations is a fundamental mathematical skill needed to tackle various real-life problems. A linear equation is an equation of the first degree, meaning it has no variable exponents higher than one.
It's typically in the form of:

• **Standard Form:** \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants.
• **Objective:** Find the value of the variable that satisfies the equation.

To solve a linear equation, you usually need to perform operations to isolate the variable on one side of the equation.
For example, in the football field problem, we had the equation \( 1040 = 4W + 400 \).
By subtracting 400 from both sides and then dividing both sides by 4, we isolated \( W \), leading to \( W = 160 \). These steps demonstrated how solving linear equations can lead to practical solutions, like determining dimensions.

substitution method

The substitution method is a technique for solving systems of equations or when you have related equations. This method involves solving one equation for one variable and then substituting that expression into another equation.
Here's how it works:

• Solve one equation for one variable: Start with one of your equations and express one variable in terms of the other.
• Substitute the expression: Take this expression and replace the variable in the other equation with it.

In the example setting of the football field problem, we knew \( L = W + 200 \).
By substituting \( L \) in the perimeter formula \( P = 2L + 2W \) as \( P = 2(W + 200) + 2W \), we were able to solve for the width using the perimeter equation. Once \( W \) was found, substituting it back to find \( L \) became straightforward.
This method is efficient and reduces the complexity of having two unknowns. It’s especially useful when dealing with straightforward dependencies between variables.