Problem 319 - Calculus Volume 1

Question

For the following exercises, consider the construction of a pen to enclose an area. You have 400 ft of fencing to construct a rectangular pen for cattle. What are the dimensions of the pen that maximize the area?

Step-by-Step Solution

Answer

The pen should be 100 feet by 100 feet to maximize the area.

Step 1: Understand the Problem

You need to use 400 feet of fencing to construct a rectangular pen. The task is to maximize the area enclosed by the rectangular pen. The pen has two equal lengths and two equal widths.

Step 2: Set Up the Equations

The perimeter of the rectangle is given by the equation \( P = 2l + 2w \), where \( l \) is the length and \( w \) is the width. We know the perimeter is 400 feet, so we have the equation \( 2l + 2w = 400 \). Simplifying, we get \( l + w = 200 \).

Step 3: Express the Area Formula

The formula for the area \( A \) of a rectangle is given by \( A = lw \). We need to express \( w \) in terms of \( l \) using the equation from Step 2: \( w = 200 - l \). Substituting, we get \( A = l(200 - l) \).

Step 4: Maximize the Area

The area formula \( A = 200l - l^2 \) is a quadratic equation in terms of \( l \). To find the maximum area, we need to find the vertex of this parabola. The vertex of the quadratic equation \( ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). In our equation, \( a = -1 \) and \( b = 200 \), hence \( l = -\frac{200}{2(-1)} = 100 \).

Step 5: Find the Maximum Dimensions

Substitute \( l = 100 \) back into the equation \( w = 200 - l \) to find \( w \, \). Doing this gives \( w = 200 - 100 = 100 \). Thus, both the length and width are 100 feet.

Key Concepts

Rectangular Area Maximization

When constructing a rectangular pen, one primary goal is often to maximize the enclosed area. This involves using a fixed perimeter, in this case, 400 feet of fencing. The perimeter formula of a rectangle is defined as the sum of all sides, so for our pen, we have two equal lengths and two equal widths. The area of a rectangle is calculated by multiplying its length by its width.

• Apply the perimeter equation: Start with the equation for perimeter: \( P = 2l + 2w \). Given \( P = 400 \), we can set it up as \( 2l + 2w = 400 \).
• Simplify the perimeter equation to express one variable: Solving for \( l + w \), we derive \( l + w = 200 \).
• Maximize the area with a fixed perimeter: Insert this into the area formula, \( A = lw \). Replace \( w \) with \( 200 - l \) to form \( A = l(200 - l) \). This step prepares the equation for maximization.

By carefully setting equations based on given constraints, one can find the dimensions that use all available material to cover the most space.

Perimeter Equations

Perimeter equations are fundamental in optimizing problems like this where a shape's boundaries facilitate maximal utilization of space. For a rectangle, the perimeter finds the total distance around the boundary.

- **Equation Setup:** Perimeter of a rectangle is \( P = 2l + 2w \). With known perimeter, simplify to \( l + w = 200 \) after substituting given value.- **Reformulate for Optimization:** Use this setup to reformulate in terms of one variable to ease maximization, deriving \( w = 200 - l \).Using known formulas effectively helps in simplifying problems, reducing complexity by expressing boundary conditions. This strategic simplification is crucial in setting groundwork for optimization.

Quadratic Functions

Quadratic functions often appear in maximization problems due to their parabolic nature, representing curves that have a clear minimum or maximum point. When we express the area function \( A = l(200 - l) \), it becomes \( A = 200l - l^2 \). This is a quadratic equation where the standard form is \( ax^2 + bx + c \).

- **Vertex Form:** To find the maximum value, identify the vertex of the quadratic function, the peak of the parabola, using \( x = -\frac{b}{2a} \).- **Vertex Calculation:** For our function, \( a = -1 \), \( b = 200 \), we calculate \( l = -\frac{200}{2(-1)} = 100 \). This gives us the length at maximum area.Once the point of maximum (vertex) is identified, resolve for other dimensions, confirming maximum dimensions for width \( w = 100 \), resulting in a balanced rectangle where length and width are equal. Quadratic functions provide clear solutions to complex problems by locating optimal points efficiently.