Problem 40 - Precalculus Mathematics for Calculus

Question

A rectangular bedroom is \(7 \mathrm{ft}\) longer than it is wide. Its area is \(228 \mathrm{ft}^{2} .\) What is the width of the room?

Step-by-Step Solution

Answer

The width of the room is 12 feet.

Step 1: Define Variables

Let the width of the bedroom be denoted by \( w \) (in feet). Since the length is described as being 7 feet longer than the width, let the length be \( l = w + 7 \).

Step 2: Set Up the Area Equation

The area of a rectangle is given by the formula: \( \, \text{Area} = \text{length} \times \text{width} \,\). Given that the area of the bedroom is \( 228 \), we can set up the equation: \((w)(w + 7) = 228 \).

Step 3: Formulate the Quadratic Equation

Expand the equation from the previous step: \(w^2 + 7w = 228\). Rearrange it into a standard quadratic form: \(w^2 + 7w - 228 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 7\), \(c = -228\), to solve \(w^2 + 7w - 228 = 0\).

Step 5: Calculate the Discriminant

Calculate the discriminant: \(b^2 - 4ac = 7^2 - 4 \times 1 \times (-228) = 49 + 912 = 961\).

Step 6: Apply the Quadratic Formula

Compute \(w = \frac{-7 \pm \sqrt{961}}{2 \times 1}\). Since \(\sqrt{961} = 31\), we have: \(\frac{-7 \pm 31}{2}\).

Step 7: Determine the Possible Widths

Calculate the potential solutions: \(w = \frac{-7 + 31}{2} = 12\) and \(w = \frac{-7 - 31}{2} = -19\). The negative solution is not feasible for width, so \(w = 12\).

Key Concepts

Rectangular Area Problems

Rectangular area problems are a common real-world application of quadratic equations. They involve figuring out unknown dimensions of a rectangle when the area is given. This requires understanding the formula for the area of a rectangle, which is the product of its length and width:

• Area = Length × Width

In the original exercise, the room's length is defined in terms of its width. By knowing this relationship and the total area, we can set up an equation to solve for the unknown width. This type of problem is both practical and common, as it helps in situations such as planning room layouts or determining the materials needed for construction projects.
To handle these problems, it's important to break down the scenario and set up an algebraic equation that demonstrates the relationship between the dimensions and the area. Doing this allows the transformation of a real-life situation into a math problem that can be methodically solved step by step.

Solving Equations

The problem-solving process often involves transitioning a word problem into a mathematical equation. For rectangular area problems like the one given, the approach starts with defining variables for unknown quantities. Using:

• Width as \( w \)
• Length as \( w + 7 \)

This step sets up an equation based on area: \(w(w + 7) = 228\). Expanding this, it turns into a quadratic equation: \(w^2 + 7w - 228 = 0\).
Solving quadratic equations is a central skill in algebra, usually tackled by:

• Factoring the equation, if possible.
• Completing the square.
• Using the quadratic formula: \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In this context, the quadratic formula is used, calculated by finding the discriminant \(b^2 - 4ac\), which helps determine the number and type of solutions. In this problem, solving using this formula results in two solutions, but only the positive one is viable for the dimensions of a room.

Real-world Applications

Quadratic equations aren't just academic exercises; they offer solutions to various real-world problems. In everyday situations, such as determining room dimensions, we effectively apply mathematical principles to practical scenarios.
Consider these applications:

• Architecture: Calculating space allocations.
• Gardening: Planning out plot dimensions.
• Interior design: Arranging furniture within given spaces.

Understanding how to set up and solve equations helps transform abstract math into concrete decision-making tools.
This is particularly evident in rectangular area problems where measurements must be accurate enough to guide real decisions, such as buying building materials or decorating spaces. Grasping these concepts is key to harnessing the power of math in practical and meaningful ways.