Problem 39 - Precalculus: Functions and Graphs

Question

The perimeter of a rectangle is 40 inches, and its area is 96 in \(^{2}\). Find its length and width.

Step-by-Step Solution

Answer

The length and width are 12 inches and 8 inches.

Step 1: Understand the problem

We know the perimeter of the rectangle is 40 inches, and the area is 96 square inches. We need to find the length, denoted as \(l\), and the width, denoted as \(w\).

Step 2: Write down the relevant formulas

The formulas we need are: the perimeter \(P\) of a rectangle is given by \(P = 2l + 2w\), and the area \(A\) is given by \(A = l \times w\).

Step 3: Substitute known values into formulas

Substitute the known values, \(P = 40\) and \(A = 96\), into the respective equations: \(2l + 2w = 40\) and \(lw = 96\).

Step 4: Solve the perimeter equation for one variable

Rearrange the perimeter equation to express \(l\) in terms of \(w\): divide the entire equation by 2 to get \(l + w = 20\), then rearrange to get \(l = 20 - w\).

Step 5: Substitute the expression into the area equation

Substitute \(l = 20 - w\) into the area equation: \( (20 - w)w = 96 \).

Step 6: Form and solve the quadratic equation

Expand and rearrange the equation to form the quadratic equation: \(w^2 - 20w + 96 = 0\). Now solve it using the quadratic formula: \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -20\), and \(c = 96\).

Step 7: Calculate the solutions of the quadratic equation

Calculate the discriminant: \(b^2 - 4ac = 400 - 384 = 16\). Substitute back into the quadratic formula: \(w = \frac{20 \pm 4}{2}\). This gives \(w = 12\) or \(w = 8\).

Step 8: Determine the corresponding length values

For \(w = 12\), substitute back into \(l = 20 - w\) to find \(l = 8\).For \(w = 8\), substitute back to find \(l = 12\).Thus, the possible dimensions are \((l, w) = (12, 8)\) or \((8, 12)\).

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total distance around the edges of the rectangle. To calculate it, you need to sum up twice the length and twice the width. The formula is straightforward:
\[ P = 2l + 2w \]
where \( l \) is the length and \( w \) is the width.

• This formula essentially accounts for all four sides of the rectangle.
• Note that the perimeter is always measured in linear units, such as inches or meters.

Understanding the perimeter is important, especially in problems where you need to optimize or calculate dimensions based on given measurements.

Area of a Rectangle

The area of a rectangle tells us how much space is enclosed within its boundaries. It's calculated by multiplying the length by the width.
The formula is:
\[ A = l \times w \]

• Area is always expressed in square units, like square inches or square meters.
• It gives you a sense of the rectangle's surface space.

The area aids in defining how much surface the rectangle covers and helps in applications like calculating flooring, painting walls, or even determining screen sizes for devices.

Quadratic Equation

Solving for dimensions of a rectangle often involves quadratic equations, especially when both the perimeter and area are given.
Quadratic equations are of the form:
\[ ax^2 + bx + c = 0 \]
In this problem, once we express one dimension in terms of the other using the perimeter formula, we substitute it into the area equation, resulting in a quadratic equation like:
\[ w^2 - 20w + 96 = 0 \]

• The quadratic formula is used to solve these equations: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
• It's critical to calculate the discriminant \( (b^2 - 4ac) \) correctly to avoid errors in the solution.
• This method helps in finding either the length or width, as it may yield two possible values due to the nature of \( \pm \) in the formula.

Quadratic equations form the backbone of solving complex real-world problems in geometry.

Geometry Problem Solving

Geometry problems often look complicated, but break them into smaller parts, and they become solvable.
Problem-solving in geometry requires:

• Understanding the given data and what needs to be calculated.
• Writing down the relevant formulas for the perimeter and area.
• Simplifying complex expressions, such as solving equations or using substitution methods.

For the given problem, we started with known perimeter and area values of a rectangle. We utilized mathematical operations and algebraic manipulation to express one dimension in terms of the other to make it simpler to form and solve equations. By solving the quadratic equation, we found multiple possible solutions and verified them to ensure correctness. Mastering these skills can help you excel in various geometry tasks and enhance your overall problem-solving abilities in math.