Problem 13 - A Graphical Approach to College Algebra

Question

Solve each problem. Maximizing Area A farmer has 1000 feet of fence to enclose a rectangular area. What dimensions for the rectangle result in the maximum area enclosed by the fence?

Step-by-Step Solution

Answer

The dimensions that maximize the area are 250 feet by 250 feet.

Step 1: Define the Variables

Let the length of the rectangle be \( l \) and the width be \( w \). Since the farmer has 1000 feet of fencing, the perimeter of the rectangle can be expressed as \( 2l + 2w = 1000 \).

Step 2: Express One Variable in Terms of the Other

To eliminate one variable, solve the perimeter equation for \( w \):\[ w = 500 - l \]

Step 3: Write the Area Function

The area \( A \) of the rectangle is given by \( A = lw \). Substitute \( w = 500 - l \) into this area formula to express \( A \) in terms of \( l \):\[ A = l(500 - l) = 500l - l^2 \]

Step 4: Find the Maximum Area Using the Quadratic Formula

The area equation \( A = 500l - l^2 \) is a quadratic function in the form \( A(l) = -l^2 + 500l \). This is a downward-opening parabola, so the maximum value occurs at the vertex. The vertex of a parabola \( ax^2 + bx + c \) is given by \( l = -\frac{b}{2a} \). Here, \( a = -1 \) and \( b = 500 \), so:\[ l = -\frac{500}{2(-1)} = 250 \]

Step 5: Calculate the Corresponding Dimension

Substitute \( l = 250 \) back into the equation for \( w \):\[ w = 500 - 250 = 250 \]

Step 6: Confirm the Maximum Area

Check if these dimensions maximize the area:\[ A = 250 \times 250 = 62500 \text{ square feet} \]Since this setup uses all 1000 feet of fencing, the solution is valid.

Key Concepts

Rectangular Perimeter

A rectangle's perimeter is the total distance around it, which is the sum of all its sides. For rectangles, this is given by the formula:

• Perimeter = 2(length) + 2(width).

In the problem, the farmer has a fixed amount of fencing, 1000 feet, to use for enclosing the area. Therefore, the perimeter equation is set to 1000:

• \[ 2l + 2w = 1000 \]

This equation helps establish a relationship between the length and width, allowing us to express one variable in terms of the other. By finding this relationship, we can further delve into optimizing the rectangle's area.

Quadratic Function

A quadratic function is a type of polynomial where the highest degree of the variable is 2. Its general form is:

• \[ ax^2 + bx + c \]

In this exercise, the area as a function of length is given by:

• \[ A(l) = -l^2 + 500l \]

Here, our task is to maximize the area, which means finding the peak of this quadratic function. Because the coefficient of \( l^2 \) is negative (\( a = -1 \)), we know the parabola opens downwards. Therefore, the maximum area corresponds to the vertex, or top, of the parabola. Recognizing this setup allows us to manipulate the equation to find the optimal length that maximizes the area.

Vertex of a Parabola

The vertex of a parabola provides key information about its maximum or minimum value. For a quadratic equation in the form \( ax^2 + bx + c \), the vertex formula helps determine this critical point:

• \[ x = -\frac{b}{2a} \]

In our fencing problem, the equation \( A = -l^2 + 500l \) features:

• \( a = -1 \)
• \( b = 500 \)

Plugging these into the vertex formula gives us the length at which area is maximized:

• \[ l = -\frac{500}{2(-1)} = 250 \]

This calculation means that at 250 feet in length, the rectangle reaches its maximum possible area with the given perimeter constraints.

Fencing Problem

A classic fencing problem involves maximizing the area enclosed by a given length of fencing. Such problems typically provide a fixed perimeter and ask for the optimal dimensions that create the largest possible rectangle. In this scenario, we worked through it step-by-step:

• Used the provided perimeter to establish a relationship between length and width.
• Formulated an area function in terms of one variable using the perimeter equation.
• Calculated using the vertex of the quadratic function to find the optimal length.
• Finally, ensured the dimensions utilize the entire fence, validating the solution.

This practical approach can be applied to various contexts, ensuring optimal use of resources to enclose maximum space.