Problem 16 - A Graphical Approach to College Algebra

Question

Solve each problem. Area of a Rectangular Region A farmer wishes to enclose a rectangular region bordering a barn with fencing, as shown in the diagram. Suppose that \(x\) represents the length of each of the three parallel pieces of fencing. She has 600 feet of fencing available. (a) What is the length of the remaining piece of fencing in terms of \(x ?\) (b) Determine a function \(\mathscr{A}\) that represents the total area of the enclosed region. Give any restrictions on \(x\) (c) What dimensions for the total enclosed region would give an area of \(22,500\) square feet? Determine the answer analytically. (d) Use a graph to find the maximum area that can be enclosed. (IMAGE CAN'T COPY)

Step-by-Step Solution

Answer

(a) Remaining length: \(600 - 3x\). (b) \(\mathscr{A}(x) = 600x - 3x^2, 0 < x < 200\). (c) Dimensions: \(x = 150\) or \(x = 50\). (d) Maximum area at \(x = 100\).

Step 1: Understanding What We Have

The barn serves as one side of the rectangle, while three pieces of fencing each with length \(x\) will form the parallel sides. The remaining piece of fencing will close the rectangle.

Step 2: Express Remaining Fencing

Since the total fencing available is 600 feet, we have 3 pieces of fencing, each of length \(x\). Thus, the remaining piece for the other side is \(600 - 3x\).

Step 3: Define Area Function

The area \(\mathscr{A}\) of the rectangular region is the product of its length and width. Here, \(\mathscr{A}(x) = x \times (600 - 3x)\). Simplifying gives: \(\mathscr{A}(x) = 600x - 3x^2\).

Step 4: Identify Restrictions on \(x\)

Since \(x\) must be positive and form valid lengths, one constraint is \(600 - 3x > 0\). Solving gives \(x < 200\). Additionally, \(x > 0\). So \(0 < x < 200\) are the constraints.

Step 5: Solve for Specific Area

To find dimensions giving an area of 22,500 square feet, solve \(22,500 = 600x - 3x^2\). Rearranging gives \(3x^2 - 600x + 22,500 = 0\). Divide by 3: \(x^2 - 200x + 7500 = 0\). Using the quadratic formula: \(x = \frac{200 \pm \sqrt{200^2 - 4 \cdot 1 \cdot 7500}}{2 \cdot 1}\). Calculate to find \(x = 150\) or \(x = 50\).

Step 6: Verify Solution for Dimensions

For \(x = 150\), the other side length is \(600 - 3(150) = 150\). For \(x = 50\), the other length can be calculated similarly.

Step 7: Find Maximum Area

The function \(\mathscr{A}(x) = 600x - 3x^2\) is a quadratic. The maximum area occurs at the vertex. The vertex \(x\) value is given by \(x = -\frac{b}{2a}\) where \(a = -3\) and \(b = 600\). Thus, \(x = 100\). Substitute back to find the maximum area.

Step 8: Graph the Function

Plot \(\mathscr{A}(x) = 600x - 3x^2\). Observe that the maximum value occurs at \(x = 100\), and use the graph to confirm.

Key Concepts

Rectangular Area

A rectangular area is a space where the length and width form a rectangle, often used in practical scenarios like fencing or layout planning. For our example, the farmer wishes to create a rectangular area using fencing, with the barn acting as one side of the rectangle.
The length of this side is defined by three parallel fencing pieces, each with a length denoted as \(x\).
Understanding how these pieces of fencing work together is crucial. The **total perimeter** of the enclosed area needs to account for all the pieces required to surround and complete the rectangle. This is why determining the remaining fencing length in terms of \(x\) is critical, as it allows us to control and adjust different dimensions of the rectangle to achieve the desired effect. As with any problem dealing with area, the product of length and width gives the total area of the space.

Fencing Problem

The fencing problem involves calculating the optimal dimensions of a fence-enclosed area given a finite amount of resources: the fencing material. In the scenario given, the farmer has 600 feet of fencing material. This total length is split between three pieces each of length \(x\), and one remaining piece.
To express the length of the remaining piece of fencing, we subtract the combined length of the three pieces (\(3x\)) from the total available length (600 feet). Therefore, the expression for the remaining piece becomes \(600 - 3x\).
This problem highlights the constraint on the lengths, represented mathematically as inequalities in the provided solution. By setting up these equations or inequalities, we can solve for the optimal values of \(x\) that satisfy all requirements, such as fitting all fencing material and achieving the desired area. It's a good real-world problem-solving exercise that combines algebraic expressions with spatial reasoning.

Maximum Area

Finding the maximum area of a rectangular region with given constraints involves understanding quadratic functions and their properties. In this context, we aim to maximize \( \mathscr{A}(x) = 600x - 3x^2 \), a quadratic function concave down due to the negative \(-3x^2\) term.
The solution of the problem reveals that the maximum area occurs at the vertex of the parabola representing the area function. The vertex is found using the formula \( x = -\frac{b}{2a} \), where for the equation \( ax^2 + bx + c \), \( a = -3 \) and \( b = 600 \).
Calculating, we find \( x = 100 \). Substituting back gives the maximum area, confirmed both mathematically and graphically by plotting \( \mathscr{A}(x) \). The importance here is to see how calculus and algebra inform optimal solutions in real-world applications, making these tools invaluable for maximizing efficiency and usage in planning and construction tasks. Such exercises enhance deeper understanding of the relationships within geometric, algebraic, and practical constructs.