Question

Michele wants to design a floor that will have a length and width that add up to 30 feet. She also wants the area of the floor to be 216 square feet. What will the dimensions of the floor need to be? (A) 10 feet by 20 feet (B) 12 feet by 18 feet (C) 14 feet by 16 feet (D) 17 feet by 17 feet

Short answer

The floor dimensions are 12 feet by 18 feet (option B).

Step-by-step solution

Define Variables

Let's define the variables: Let \( l \) be the length and \( w \) be the width of the floor. According to the problem, the sum of the length and width is 30 feet, so \( l + w = 30 \).

Set up the Area Equation

The area \( A \) of the floor is given as 216 square feet. The formula for the area of a rectangle is \( A = l \times w \). So, \( lw = 216 \).

Solve for One Variable

We have the equations: \( l + w = 30 \) and \( lw = 216 \). From the first equation, express one variable in terms of the other: \( l = 30 - w \).

Substitute and Solve Quadratic Equation

Substitute \( l = 30 - w \) into the area equation: \((30 - w)w = 216\). This simplifies to the quadratic equation \( w^2 - 30w + 216 = 0 \).

Solve the Quadratic Equation

Solve the quadratic equation \( w^2 - 30w + 216 = 0 \) using the quadratic formula: \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -30, c = 216 \). Substitute these values into the formula to find \( w \).

Calculation with Quadratic Formula

Calculate the discriminant: \( b^2 - 4ac = (-30)^2 - 4(1)(216) = 900 - 864 = 36 \). The square root of 36 is 6. Now, calculate \( w \): \( w = \frac{30 \pm 6}{2} \). This gives the solutions \( w = 18 \) and \( w = 12 \).

Determine Dimensions

If \( w = 18 \), then \( l = 30 - 18 = 12 \). If \( w = 12 \), then \( l = 30 - 12 = 18 \). Both yield the dimensions of 12 feet by 18 feet.

Key concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra that describe a particular kind of polynomial equation. They are typically written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the variable being solved for. The unique aspect of quadratic equations is that they involve the square of the unknown variable, which often results in two solutions.

Solving quadratic equations can be approached in several ways, such as factoring, completing the square, or using the quadratic formula. In the floor design problem, after substituting the known values into the equation \( w^2 - 30w + 216 = 0 \), it becomes a quadratic equation (where \( a = 1, b = -30, \) and \( c = 216 \)). Using the quadratic formula, we found two possible width values for the floor: \( w = 18 \) feet and \( w = 12 \) feet. Both solutions could be valid, leading to two sets of dimensions.

Geometry

Geometry is a branch of mathematics that deals with shapes, sizes, and the properties of space. In problems related to physical spaces, such as designing floors, it becomes crucial in calculating areas and understanding the layout of a space.

For a rectangular floor, the relationship between its length, width, and area is given by the formula: \[A = l \times w\]where \( A \) represents the area, \( l \) is the length, and \( w \) is the width.

In Michele's problem, we use this formula to express the area in terms of variables. Since the floor's length and width must add up to 30 feet, we set up equations that link these geometric properties. This allows us to use algebra to find precise measurements for construction, showcasing how geometry and algebra often work hand in hand.

Problem Solving Steps

Problem-solving in math is like crafting a recipe; break it down into clear, manageable steps. This strategy leads to a solution without feeling lost or overwhelmed. Let's take a closer look at Michele's floor design problem and the steps followed to find the solution.

• Step 1: Define Variables - Establish what each variable represents. Here, define \( l \) as the length and \( w \) as the width.
• Step 2: Set Up Equations - Use the problem's conditions to form equations. Relate the sum of dimensions and the area using two core equations: \( l + w = 30 \) and \( lw = 216 \).
• Step 3: Express One Variable - Solve one equation for a variable to substitute it into another, transforming the problem into a more straightforward equation.
• Step 4: Substitute and Simplify - Insert the expression into the other equation, reducing it to a single quadratic equation, making it easier to solve.
• Step 5: Solve the Quadratic Equation - Apply the quadratic formula or another method to find the possible values for the unknown variable(s).
• Step 6: Verify Solutions - Substitute the found values back into the original conditions to ensure they make sense and solve the problem.

Each step of this process builds on the last, ensuring a structured approach to problem-solving and increased confidence in finding correct solutions.