Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Writing Real-Life Problems In Exercises 47-50, solve the number problem and write a real-life problem that could be represented by this verbal model. For instance, an applied problem that could be represented by Exercise 47 is as follows. The sum of the length and width of a one-story house is 100 feet. The house has 2500 square feet of floor space. What are the length and width of the house? One number is 2 more than another number. The product of the two numbers is 440 . Find the numbers

Step-by-Step Solution

Answer

The length and width of the house are either \( (50 + 10 \times \sqrt{11}, 50 - 10 \times \sqrt{11}) \) or \( (50 - 10 \times \sqrt{11}, 50 + 10 \times \sqrt{11}) \) and the two numbers from the second problem are either \( (22, 20) \) or \( (-20, -22) \)

Step 1: Problem 1: Initialize and Write the Equations

Let \( x \) represent the length of the house and \( y \) the width of the house. From the problem, we have the two equations \(x + y = 100\) (sum of length and width) and \(xy = 2500\) (floor space).

Step 2: Problem 1: Solve the Equations

We rearrange the first equation to get \( y = 100 - x \). Substituting this into the second equation, we get \( x * (100 - x) = 2500 \). This simplifies to \( x^2 - 100x + 2500 = 0 \), a quadratic equation. Solving this quadratic equation yields \( x = 50 \pm 10 \times \sqrt{11} \). Substituting these values of x into the first equation gives \( y = 50 \mp 10 \times \sqrt{11} \). Hence the length and width of the house are \( (50 + 10 \times \sqrt{11}, 50 - 10 \times \sqrt{11}) \) and \( (50 - 10 \times \sqrt{11}, 50 + 10 \times \sqrt{11}) \).

Step 3: Problem 2: Initialize and Write the Equations

Let \( a \) and \( b \) be the two numbers such that \( a = b + 2 \) and \( ab = 440 \). These two relationships give the equations to solve.

Step 4: Problem 2: Solve the Equations

Substituting \( a = b + 2 \) into \( ab = 440 \), we get \( (b + 2)b = 440 \). Simplifying, this yields the quadratic equation \( b^2 + 2b - 440 = 0 \). Solving this quadratic equation gives \( b = 20, -22 \). Substituting these values into \( a = b + 2 \) gives \( a = 22, -20 \). Hence, the two numbers are \( (22, 20) \) and \( (-20, -22) \).

Key Concepts

Algebraic Problem Solving

Algebraic problem solving is an essential tool in mathematics, especially when dealing with real-world situations that involve unknowns. In our exercise, we have two problems where algebra helps in forming equations from given conditions.

To solve these problems, we need to understand how to translate words into mathematical expressions:

• The phrase "sum of the length and width of a house is 100 feet" translates into the equation \(x + y = 100\).
• "2500 square feet of floor space" becomes \(xy = 2500\), illustrating the concept of product.

These translations form a system of equations, which can then be manipulated to find unknown variables.

Understanding relationships and using algebra to express them in terms of equations helps unravel the unknowns. Solving involves substitution or elimination methods, sometimes leading to quadratic equations like \(x^2 - 100x + 2500 = 0\). The roots of these equations give meaningful answers that can be checked by plugging back into the original conditions.

Real-Life Applications

Applying mathematical concepts to practical scenarios is where real-life applications of quadratic equations shine. In the exercise related to finding the dimensions of a house based on its floor space, we see how equations describe a complex physical layout.

When you apply this in real life, think of:

• Architecture: Architects use these calculations to ensure spaces are optimally utilized.
• Carpentry: Craftsmen make precise cuts for materials to fit perfectly.
• Engineering: Engineers solve similar equations to determine load and stress factors.

These aren't just numbers and letters; they represent tangible situations, ensuring projects are designed accurately and efficiently.

In everyday scenarios, knowing how to set up and solve quadratic equations can be invaluable. It allows you to construct better budgets and project timelines, and even make more informed decisions based on quantifiable data.

Systems of Equations

Understanding systems of equations is crucial to solving problems with multiple unknowns. A system of equations involves two or more equations that are solved simultaneously. Each equation provides a different piece of information about the same set of variables.

In our exercise, the house dimensions are determined using:

• \(x + y = 100\), which is a linear equation representing the perimeter constraint.
• \(xy = 2500\), a quadratic equation representing the area constraint.

Solving these equations simultaneously requires methods such as substitution or elimination.

Learning how to handle systems of equations can help in fields such as economics, where demand and supply relationships need to be balanced, or computer science, where multiple equations are solved to optimize algorithms.

By understanding and mastering solving these systems, you can tackle not only academic problems but also sophisticated real-world challenges with confidence and precision.