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Chapter 12: Problem 47

Solve each problem using a nonlinear system. The area of a rectangular rug is \(84 \mathrm{ft}^{2}\) and its perimeter is \(38 \mathrm{ft}\). Find the length and width of the rug.

Short Answer

Expert verified
The length and width of the rug are 12 feet and 7 feet (in either order).

Step by step solution

01

- Define Variables

Let the length of the rug be denoted by \( l \) and the width by \( w \).
02

- Set Up Equations

Based on the problem, we know two things:1. The area of the rectangle is given as \( lw = 84 \).2. The perimeter of the rectangle is given as \( 2l + 2w = 38 \).
03

- Simplify Perimeter Equation

Rewrite the perimeter equation to make it simpler:\( l + w = 19 \).
04

- Express Length in Terms of Width

From \( l + w = 19 \), solve for \( l \):\( l = 19 - w \).
05

- Substitute Length into Area Equation

Substitute \( l = 19 - w \) into the area equation \( lw = 84 \):\( (19 - w)w = 84 \).
06

- Solve Quadratic Equation

Expand and solve the quadratic equation:\( 19w - w^2 = 84 \)Rearrange to standard form:\( w^2 - 19w + 84 = 0 \).Using the quadratic formula, \( w = \frac{19 \pm \sqrt{19^2 - 4 \cdot 1 \cdot 84}}{2 \cdot 1} \) which simplifies to \( w = 12 \) or \( w = 7 \).
07

- Find Corresponding Length

For \( w = 12 \), the length \( l = 19 - 12 = 7 \).For \( w = 7 \), the length \( l = 19 - 7 = 12 \).
08

- Verify Solutions

Both (length, width) pairs \( (7, 12) \) and \( (12, 7) \) correctly satisfy the original conditions (area and perimeter).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic equations
To solve quadratic equations, you often need to use a specific method called the quadratic formula. The standard form of a quadratic equation is written as:
\[ax^2 + bx + c = 0\]
Here, \(a\), \(b\) and \(c\) are constants. The quadratic formula is
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula helps solve for \(x\) to find the variable values that make the equation true.
The \(\pm\) symbol means that you will get two solutions. These are often the points where a parabola intersects the x-axis. When solving real-world problems, it's important to determine which solution fits the context of the problem. For instance, in the exercise above, solving the quadratic equation
\[w^2 - 19w + 84 = 0\]
gives us the values for the width of the rectangle. Using \(19\), \(84\) and the formula, you find:
\[w = 12 \quad or \quad w = 7\]
It's essential to substitute these values back into the original context to verify their correctness and relevance.
area of a rectangle
The area of a rectangle is calculated by multiplying its length by its width. The formula is:
\[\text{Area} = l \times w\]
Given in the exercise that the area is \(84 \mathrm{ft}^2 \), we use this equation to set up one of our system equations:
\[l \times w = 84\]
Once you know the values of length and width, you can easily compute the area. This property helps in a variety of practical applications.
Example: If you have a length of 7 feet and a width of 12 feet, then:
\[\text{Area} = 7 \times 12 = 84 \mathrm{ft}^2\]
This confirms the calculation fits the problem's conditions.
perimeter of a rectangle
The perimeter of a rectangle is obtained by adding up all its sides. The formula is:
\[ P = 2l + 2w \] Given in the exercise, the perimeter is \(38 \mathrm{ft}\). We use the perimeter equation to set up our second system equation:
\[2l + 2w = 38\] This can be simplified by dividing everything by 2:
\[ l + w = 19 \] This simplified equation is easy to manage and helps in solving for one variable in terms of the other.
Example: If the width is 7 feet, as found earlier, substituting it into the perimeter equation provides the length:
\[l + 7 = 19\]and simplifies to:
\[l = 12 \]This way, both width and length fit in perfect harmony with the overall perimeter constraint.

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