Problem 60 - Elementary Algebra

Question

The length of a rectangle is \(6 \mathrm{ft}\) more than three times its width. The area of the rectangle is \(297 \mathrm{ft}^{2}\). Find the length and width.

Step-by-Step Solution

Answer

Width: 9 ft, Length: 33 ft

Step 1: Define Variables

Let the width of the rectangle be denoted as \(w\). According to the problem, the length (\(l\)) is 6 feet more than three times the width. Therefore, \(l = 3w + 6\).

Step 2: Write the Area Formula

The area of a rectangle is given by the formula \( \text{Area} = \text{length} \times \text{width} \). We know that the area is 297 square feet, so \((3w + 6) \times w = 297 \).

Step 3: Form an Equation

Substitute the length expression into the area formula: \[ (3w + 6)w = 297 \]. Simplify this to get a quadratic equation: \[ 3w^2 + 6w = 297 \].

Step 4: Simplify and Solve the Equation

Subtract 297 from both sides to set the equation to 0: \[ 3w^2 + 6w - 297 = 0 \]. Divide the whole equation by 3 to simplify: \[ w^2 + 2w - 99 = 0 \].

Step 5: Factor the Quadratic Equation

Factor the quadratic equation \(w^2 + 2w - 99 \). The factors are \[(w + 11)(w - 9) = 0 \].

Step 6: Solve for Width

Set each factor equal to zero and solve for \(w\): \[ w + 11 = 0 \] or \[ w - 9 = 0 \]. Thus, \[ w = -11 \] (which is not a valid width) or \[ w = 9 \] (valid width).

Step 7: Find the Length

Substitute \(w = 9 \) back into the length formula: \[ l = 3(9) + 6 = 27 + 6 = 33 \].

Step 8: Conclusion

The width of the rectangle is \(9 \) feet and the length is \(33 \) feet.

Key Concepts

Quadratic Equations

A quadratic equation is a second-degree polynomial equation in a single variable. The general form is: \[ ax^2 + bx + c = 0 \] where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. In this exercise, our goal is to find the width and length of a rectangle when given its area. We first set up an equation based on the dimensions provided and manipulate it into a quadratic form for easier solving. Quadratic equations can be solved using various methods, like factoring, completing the square, or using the quadratic formula. Here, we use factoring.

Area of a Rectangle

The area of a rectangle is calculated by multiplying its length by its width. Expressed algebraically: \[ \text{Area} = \text{Length} \times \text{Width} \] In our exercise, we're given the area as 297 square feet. To find the width (\(w\)) and length (\(l\)), we start by using the relationship provided in the problem: \( l = 3w + 6 \). This allows us to set up the foundational equation: \[ (3w + 6)w = 297 \] By solving this, we find the dimensions of the rectangle.

Algebraic Expressions

Algebraic expressions involve numbers, variables, and arithmetic operations. In our exercise, expressions like \(3w + 6\) and \(297\) are used to define relationships between variables. Identifying these relationships is critical for setting up equations that represent the problem at hand. By substituting known values and simplifying, we convert the given problem scenario into a usable mathematical form. This step-wise manipulation of algebraic expressions helps us move closer to finding the solution.

Factoring

Factoring is the process of breaking down a complex expression into simpler pieces (factors) that, when multiplied together, give the original expression. For our quadratic equation \(w^2 + 2w - 99 = 0\), factoring helps us solve for \(w\). We look for two numbers that multiply to \(-99\) and add to \(2\). These numbers turn out to be \(11\) and \(-9\). Thus, we write the equation as: \[ (w + 11)(w - 9) = 0 \] Solving these factors, we get the possible widths (\(w=-11\) and \(w=9\)), but since width cannot be negative, we use \(w=9\). Factoring simplifies solving quadratic equations efficiently and is a crucial skill in algebra.