Problem 59 - Algebra A Combined Function

Question

Solve. The heaviest reported door in the world is the 708.6-ton radiation shield door in the National Institute for Fusion Science at Toki, Japan. If the height of the door is 1.1 feet longer than its width, and its front area (neglecting depth) is 1439.9 square feet, find its width and height. [Interesting note: The door is 6.6 feet thick.] (Source: Guiness World Records)

Step-by-Step Solution

Answer

Width: 37.39 feet, Height: 38.49 feet.

Step 1: Define the Variables

Let's define the variables needed to solve the problem. Let\( x \) be the width of the door in feet. Then, the height of the door will be \( x + 1.1 \) feet because the height is 1.1 feet longer than the width.

Step 2: Setup the Equation

We know that the front area of the door (height multiplied by width) is 1439.9 square feet. Therefore, we can write the equation\[ x (x + 1.1) = 1439.9. \]

Step 3: Expand the Equation

Expand the equation from Step 2 to clear the parentheses:\[ x^2 + 1.1x = 1439.9. \]

Step 4: Rearrange the Quadratic Equation

Reorganize the equation to resemble the form \( ax^2 + bx + c = 0 \).\[ x^2 + 1.1x - 1439.9 = 0. \]

Step 5: Solve the Quadratic Equation Using the Quadratic Formula

Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1.1 \), and \( c = -1439.9 \). Substitute these values into the formula:\[ x = \frac{-1.1 \pm \sqrt{(1.1)^2 - 4(1)(-1439.9)}}{2(1)}. \]

Step 6: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \).\[(1.1)^2 - 4 \times 1 \times (-1439.9) = 1.21 + 5759.6 = 5760.81. \]

Step 7: Find the Square Root of the Discriminant

Calculate \( \sqrt{5760.81} \). The square root is approximately 75.8744.

Step 8: Solve for x

Substitute the square root back into the quadratic formula:\[ x = \frac{-1.1 \pm 75.8744}{2}. \] Solve to find two potential solutions for \( x \):1. \( x = \frac{-1.1 + 75.8744}{2} \approx 37.3872 \)2. \( x = \frac{-1.1 - 75.8744}{2} \approx -38.4872 \).Since negative values for width do not make sense, the width \( x \) is 37.3872 feet.

Step 9: Calculate the Height

Now that we have the width, calculate the height by adding 1.1 feet to the width.\[ \text{Height} = 37.3872 + 1.1 = 38.4872 \text{ feet}. \]

Key Concepts

Understanding Quadratic Equations

Quadratic equations are a fundamental component of algebra, often encountered in problems involving areas and dimensions. These equations are in the form of \( ax^2 + bx + c = 0 \). In this context, we used it to find the unknown dimensions of the door. A quadratic equation can have zero, one, or two real solutions, depending on the discriminant \( b^2 - 4ac \). Solving a quadratic equation involves rewriting the roots, which are the values of \( x \) that make the equation true.

• The quadratic formula allows us to solve any quadratic equation: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• In our example with the door, the equation sets the width \( x \) and incorporates the relation that height is 1.1 feet more than width, ultimately forming the equation \( x(x + 1.1) = 1439.9 \).

Understanding these equations helps solve real-world problems, such as calculating dimensions for construction.

Calculating Area Successfully

Area calculation is essential when determining dimensions of planar surfaces, like doors or walls. In the exercise, the front area of the door is crucial to forming the quadratic equation. The area formula used is straightforward:
- *Area = Width \( \times \) Height*.
By knowing one dimension (width) and the specific relationship of height being 1.1 feet greater than the width, we set up the area equation as \( x(x + 1.1) = 1439.9 \). Calculating areas precisely allows you to effectively plan and utilize materials in real-world tasks, where precision can influence cost and construction feasibility. It serves a practical function in architecture and design.

Defining Variables for Clarity

Defining variables is a crucial step in solving any algebraic problem. It simplifies complex scenarios into manageable parts. In our exercise, we define \( x \) as the width of the door. This variable stands in for an unknown value we need to find. Describing variables offers clarity, especially when multiple dimensions or factors are involved.

• Start by identifying what you need, like finding a width from a known area.
• Create relationships between variables (e.g., height is width + 1.1).

Defining variables effectively turns word problems into mathematical statements that are easier to manage and solve.

Applying Algebra in Real-World Problems

Algebra, including quadratic equations, is often applied to solve real-world challenges. The exercise offers an opportunity to understand how math can help with real-life applications such as designing and building structures. Measuring and calculating these dimensions affect engineering, architecture, and various industries.

• Real-world applications require accurate dimensioning for stability and economy.
• Solving quadratic equations helps ensure designs are possible within given constraints, like width and areas in a space.

By practicing these concepts, students can see the relevance of algebra beyond the classroom, appreciating how it provides solutions to tangible problems.