Problem 112 - College Algebra

Question

A rectangular parking lot has a length that is 3 yards greater than the width. The area of the rectangular lot is 180 square yards. Find the length and the width.

Step-by-Step Solution

Answer

The width of the parking lot is 12 yards and the length is 15 yards.

Step 1 - Define the variables

Let's use \( w \) to represent the width and \( l \) for the length. According to the problem, the length is 3 yards greater than the width, hence we can write the first equation as \( l = w + 3 \). The second equation can be derived from the given area of the rectangle formula \( A = l × w \). Given the area is 180 square yards, our second equation was \( 180 = l × w \).

Step 2 - Substitute the known values

Substitute \( w + 3 \) for \( l \) in the equation defining the area to get a single equation: \( 180 = (w + 3) × w \).

Step 3 - Simplify and Solve

We expand this to get \( 180 = w^2 + 3w \). Rearranging terms we obtain the quadratic equation \( w^2 + 3w - 180 = 0 \). Solving this equation gives \( w = 12 \) and \( w = -15 \). We discard the negative solution as width cannot be negative. So, the width of the parking lot is 12 yards.

Step 4 - Determine the length

Substituting \( w = 12 \) yards into equation for \( l \), we find \( l = w + 3 = 12 + 3 = 15 \) yards.

Key Concepts

Rectangular Area Calculation

When tackling problems involving rectangular areas, it's essential to remember the fundamental formula for the area of a rectangle:

• Area \( A = ext{length} \times ext{width} \)

This formula applies to numerous real-world scenarios, such as finding the dimensions of a parking lot or a garden plot. In this exercise, the area was given as 180 square yards.
Understanding that the length is related to the width is crucial for forming equations that you can solve. In this case, the length was three yards greater than the width. Recognizing these relationships allows us to set up the equations needed to find unknown dimensions.

Variable Substitution

Variable substitution is a powerful tool in algebraic problem-solving. It allows us to simplify complex expressions and equations by replacing one variable with another expression. In this problem, we know:

• The length \( l \) is \( w + 3 \) yards.

By substituting \( w + 3 \) for \( l \) in the area equation, we get
\( 180 = (w + 3) \times w \). This substitution helps in reducing the two-variable equation into a single variable equation, making it much easier to solve. By simplifying, we convert a potentially daunting problem into something more manageable and straightforward.

Algebraic Problem Solving

Algebraic problem solving often involves manipulating equations to find unknown values. After performing the substitution, we expand and rearrange the equation:

• \( 180 = w^2 + 3w \)
• Rearranged as \( w^2 + 3w - 180 = 0 \)

This forms a quadratic equation, where the solutions represent possible values for \( w \). Solving quadratic equations can involve factoring, completing the square, or using the quadratic formula. In this instance, solving the equation gives two potential answers: \( w = 12 \) and \( w = -15 \). Since a negative width doesn't make sense physically, we discard \( w = -15 \), leaving us with the accurate solution of \( w = 12 \).
Finally, by substituting back into the original relationship, we solve for the length \( l = w + 3 \), resulting in \( l = 15 \). Understanding the logical flow of these steps is key to mastering algebra.