Problem 70 - 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 \(w\). $$ \begin{aligned} &A(l, w)=300\\\ &P(l, w)=70 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( l = 20 \) and \( w = 15 \).

Step 1: Write Down the Known Equations

We have two equations based on the information provided:1. The area of the rectangle: \( A(l, w) = l \times w = 300 \).2. The perimeter of the rectangle: \( P(l, w) = 2l + 2w = 70 \).

Step 2: Rearrange Perimeter Equation

Start by rearranging the perimeter equation to express one variable in terms of the other. From\[ 2l + 2w = 70 \]divide the entire equation by 2 to simplify, yielding:\[ l + w = 35 \]Solving for \( l \) gives:\[ l = 35 - w \] This expresses \( l \) in terms of \( w \).

Step 3: Substitute Expression into Area Equation

Substitute the expression for \( l \) from Step 2 into the area equation:\[ l \times w = 300 \]Replace \( l \) by \( (35 - w) \), so:\[ (35 - w) \times w = 300 \]Expanding gives the quadratic equation:\[ 35w - w^2 = 300 \].

Step 4: Rearrange to Form a Standard Quadratic Equation

Rearrange the quadratic equation from Step 3 to the standard form:\[ -w^2 + 35w - 300 = 0 \]Multiply through by -1 to make it easier to solve:\[ w^2 - 35w + 300 = 0 \].

Step 5: Solve Quadratic Equation for w

We will use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -35 \), and \( c = 300 \):Calculate the discriminant:\[ b^2 - 4ac = (-35)^2 - 4\times 1 \times 300 = 1225 - 1200 = 25 \]Apply the quadratic formula:\[ w = \frac{35 \pm \sqrt{25}}{2}\]\[ w = \frac{35 \pm 5}{2}\]This gives solutions:\[ w = \frac{40}{2} = 20 \]and\[ w = \frac{30}{2} = 15 \].

Step 6: Determine Corresponding Length l

For each value of \( w \), calculate \( l \) using \( l = 35 - w \).For \( w = 20 \):\[ l = 35 - 20 = 15 \]For \( w = 15 \):\[ l = 35 - 15 = 20 \].Since \( l > w \), the valid solution is \( l = 20 \), \( w = 15 \).

Key Concepts

Rectangles in Algebra

Rectangles are one of the basic shapes in geometry, but they hold great significance when we explore them in algebraic contexts. While a rectangle is simply a four-sided shape with opposite sides equal, representing them using algebra allows for solving problems involving unknown values.

In algebra, we often express the properties of rectangles through equations. These include calculating the area and perimeter of rectangles using algebraic formulas. The beauty of algebra allows us to create a system of equations that can be manipulated to find unknown dimensions such as length and width.

Because rectangles are everywhere--from the screens you are reading this on to the bricks in a wall--understanding how to solve equations related to rectangles is a valuable skill. This forms the foundation for more complex geometric problem-solving. When faced with a problem, we can set up equations reflecting the relationships given and solve for the variables we need to find.

Quadratic Equations

Quadratic equations are prevalent in algebra and are essential for solving many real-world problems. These equations are typically written in the form \( ax^2 + bx + c = 0 \). Solving for \( x \) involves finding the values that satisfy the equation.

One powerful tool for solving quadratic equations is the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula finds the solutions or 'roots' of the quadratic equation using the coefficients \( a \), \( b \), and \( c \).

In the context of our exercise, arranging the area equation into a quadratic form and applying the quadratic formula helped us find possible values for the widths \( w \) of our rectangle. The discriminant \( b^2 - 4ac \) tells us about the nature of the roots. A positive discriminant indicates two real and distinct roots, which we encountered in our solution.

Perimeter and Area

Understanding the perimeter and area of rectangles is crucial. These are fundamental properties that tell us about the size and boundary length of a rectangle.

The perimeter, calculated as \( P = 2l + 2w \), tells us the total length around the rectangle. This is useful for understanding the amount of material needed to frame or border the shape. The area, given by \( A = l \times w \), indicates the size of the surface enclosed within the rectangle. Both values depend on the length \( l \) and width \( w \).

Using these formulas, we established the system of equations necessary to solve our problem. By knowing the perimeter and the area, we could express relationships between \( l \) and \( w \), allowing us to find their specific values through substitution and algebraic manipulation. Solving these equations not only tests our algebraic skills but also showcases the practical applications of algebra in solving geometric problems.