Problem 69 - College Algebra with Modeling and Visualization

Question

The area of a rectangle with length l and width \(w\) is computed by \(A(l, w)=l w,\) and its perimeter is calculated by \(P(l, w)=2 l+2 w .\) Assume that \(l>w\) and use the method of substitution to solve the system of equations for \(l\) and \(w\). $$ \begin{array}{l} A(l, w)=35 \\ P(l, w)=24 \end{array} $$

Step-by-Step Solution

Answer

Length is 10 and width is 3.5.

Step 1: Representation of Equations

We are given two equations: the area equation \( A(l, w) = lw = 35 \) and the perimeter equation \( P(l, w) = 2l + 2w = 24 \). We need to solve these for \( l \) and \( w \).

Step 2: Express one variable in terms of the other

From the area equation \( A(l, w) = lw = 35 \), express \( l \) in terms of \( w \): \( l = \frac{35}{w} \).

Step 3: Substitute into Perimeter Equation

Substitute \( l = \frac{35}{w} \) into the perimeter equation: \[ 2(\frac{35}{w}) + 2w = 24 \].

Step 4: Simplify and Solve for \( w \)

Simplify the equation from the substitution step: \[ \frac{70}{w} + 2w = 24 \]. Multiply the entire equation by \( w \) to eliminate the fraction: \[ 70 + 2w^2 = 24w \]. Rearrange terms: \[ 2w^2 - 24w + 70 = 0 \].

Step 5: Solve the Quadratic Equation

Solve \( 2w^2 - 24w + 70 = 0 \) using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2, b = -24, c = 70 \). Calculate the discriminant: \( (-24)^2 - 4 \times 2 \times 70 = 100 \). Thus, the solutions for \( w \) are \( w = \frac{24 \pm 10}{4} \) giving \( w = 8.5 \) and \( w = 3.5 \).

Step 6: Choose the correct \( w \) value

Since \( l > w \), we select \( w = 3.5 \) because the larger solution would not satisfy \( l > w \) given our perimeter constraint.

Step 7: Calculate \( l \) using \( w = 3.5 \)

Substitute back into the expression for \( l \): \( l = \frac{35}{3.5} = 10 \).

Step 8: Conclusion and Verification

The dimensions of the rectangle are \( l = 10 \) and \( w = 3.5 \). Verify by checking: \( lw = 35 \) and \( 2l + 2w = 24 \). Both conditions hold true.

Key Concepts

Substitution Method

The substitution method is a powerful tool used to solve systems of equations. The basic idea is to express one of the variables in terms of the other using one of the equations.
This allows you to substitute this expression into the second equation, thereby working with only one variable. This makes the system easier to solve.
In the given exercise, we start with two equations:

• Area equation: \(A(l, w) = lw = 35\)
• Perimeter equation: \(P(l, w) = 2l + 2w = 24\)

We express the length \(l\) in terms of the width \(w\) from the area equation, giving us \(l = \frac{35}{w}\).
This substitution transforms the complex system into a simpler equation for \(w\) that can be solved independently.

Rectangles

A rectangle is a four-sided polygon with opposite sides equal in length and all interior angles at 90 degrees. This shape is characterized by its length and width.
In geometry, understanding the properties of rectangles is crucial because they are among the simplest and most versatile shapes. When dealing with rectangles, you typically want to calculate:

• Area, which indicates the amount of space within the rectangle.
• Perimeter, which is the total distance around the edges of the rectangle.

Applications of rectangular calculations extend to many real-world problems, such as flooring areas or fencing surrounding rectangular gardens. Understanding how to formulate and solve equations involving rectangles is essential for anyone studying algebra or geometry.

Area and Perimeter

To grasp the concepts of area and perimeter well, it's important to distinguish their roles in problem-solving. The area of a rectangle is the measure of the space enclosed by its sides. It is calculated using the formula \(A = l \times w\), where \(l\) is the length and \(w\) is the width. This tells us how many square units fit within the rectangle.
Conversely, the perimeter is the total length around the rectangle. The formula for perimeter is \(P = 2l + 2w\). This shows us how much material is needed if we were outlining or fencing the rectangle.
In exercises like the one above, these formulas help to form equations which, when solved, reveal the unknown dimensions of rectangles given specific conditions.

Quadratic Equation

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). They are essential in algebra for finding unknown values that satisfy the equation given the coefficients \(a\), \(b\), and \(c\). In our exercise, solving for the width involves dealing with a quadratic equation derived from substituting variables. We rearrange the equation \(2w^2 - 24w + 70 = 0\) to standard form.
The solution involves using the quadratic formula: \[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Given \(a = 2\), \(b = -24\), and \(c = 70\), we compute the discriminant \((b^2 - 4ac)\) to find the values of \(w\). The two solutions provide the potential widths, from which the valid one is chosen based on problem constraints, such as \(l > w\). This method confirms understanding and solution of quadratic equations.