Problem 106 - College Algebra with Modeling and Visualization

Question

A rectangular pen for a pet is under construction using 100 feet of fence. (a) Find the dimensions that give an area of 576 square feet. (b) Find the dimensions that give maximum area.

Step-by-Step Solution

Answer

(a) Dimensions are 32 ft by 18 ft or 18 ft by 32 ft. (b) Maximum area dimensions are 25 ft by 25 ft.

Step 1: Determine the Variables

Let the length of the pen be denoted by \( l \) and the width by \( w \). The perimeter of the rectangular pen is 100 feet, hence the equation for the perimeter is \( 2l + 2w = 100 \).

Step 2: Express One Variable in Terms of Another

From the perimeter equation \( 2l + 2w = 100 \), we can express one variable in terms of the other. Solving for \( l \), we get:\[ l = 50 - w \]

Step 3: Write Area Equation

The area \( A \) of the rectangle is given by \( A = l \times w \). Substituting \( l = 50 - w \) into the area equation, we get:\[ A = (50 - w)w = 50w - w^2 \]

Step 4a: Solve for Specific Area (a)

To find the dimensions that yield an area of 576 square feet, set the area equation to 576 and solve:\[ 50w - w^2 = 576 \]Rearranging gives:\[ w^2 - 50w + 576 = 0 \]This is a quadratic equation, which we solve using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -50 \), and \( c = 576 \).\[ w = \frac{50 \pm \sqrt{50^2 - 4 \times 1 \times 576}}{2} \]Simplifying:\[ w = \frac{50 \pm \sqrt{2500 - 2304}}{2} = \frac{50 \pm \sqrt{196}}{2} = \frac{50 \pm 14}{2} \] This gives \( w = 32 \) or \( w = 18 \). Hence, if \( w = 32 \), \( l = 18 \) and if \( w = 18 \), \( l = 32 \).

Step 4b: Solve for Maximum Area (b)

To find the dimensions that give the maximum area, complete the square for the area function \( A = 50w - w^2 \):\[ A = - (w^2 - 50w) \]Half of 50 is 25, and 25 squared is 625. Add and subtract this inside the equation:\[ A = -(w^2 - 50w + 625 - 625) = -( (w - 25)^2 - 625 ) = 625 - (w - 25)^2 \]The maximum area occurs when \( (w - 25)^2 = 0 \), which gives \( w = 25 \). Substituting back to find \( l \), we have \( l = 50 - 25 = 25 \).

Key Concepts

Rectangular Geometry

Understanding rectangular geometry involves focusing on the properties and relationships between length, width, perimeter, and area. In a rectangle, the opposite sides are equal in length. This means the perimeter, which is the total length around the rectangle, can be calculated with the formula:

• Perimeter: \( P = 2l + 2w \)

Here, \( l \) represents the length, and \( w \) represents the width of the rectangle. Knowing this formula helps us solve problems related to fencing or borders.
When we talk about the area of a rectangle, we're discussing the space within its boundaries, calculated using:

• Area: \( A = l imes w \)

For any optimization of space, like maximizing or setting a specific area, understanding these fundamental properties is crucial.

Optimization Problems

Optimization problems often ask us to find the best solution given a set of restrictions, such as maximizing area while balancing perimeter constraints. In mathematical terms, we aim to maximize or minimize a function subject to certain conditions.
For a rectangular pen with a given amount of fencing for the perimeter, we need to decide how to set the pen's dimensions to either achieve a set area or maximize it.

• Given a fixed perimeter, the optimization problem involves using a quadratic equation derived from the area function.
• The area function’s maximum value is found by transforming it into a perfect square to reveal its vertex or peak point.

Identifying that a squared term \((w - 25)^2\) results in zero gives the maximum value, making the side lengths equal for the maximum area situation. This is a symmetric solution found directly from the problem setup, highlighting the importance of understanding quadratic functions.

Perimeter and Area Calculations

Calculating the perimeter and area are foundational tasks in solving geometric problems. When using 100 feet of fence to build a pen, decisions must be made regarding how much will be used for each dimension.
The fixed perimeter of 100 feet helps us establish a relationship between length and width \((2l + 2w = 100)\), allowing us to express one variable, such as length, in terms of another, like width. Plugging these expressions into the area formula helps us derive relevant equations:

• Perimeter constraint leads us to: \( w = 50 - l \)
• Area is calculated as: \( A = (50 - w) imes w \)
• Quadratic equations often emerge, enabling us to calculate for specific areas or maxima using the quadratic formula or completing the square.

These calculations link directly to determining optimal dimensions for the desired area outcomes, showcasing the interplay between algebra and geometry in practical scenarios.