Problem 8 - Thomas Calculus

Question

The shortest fence \(A 216 \mathrm{m}^{2}\) rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed?

Step-by-Step Solution

Answer

The dimensions are 18 m by 12 m with a total fence length of 72 m.

Step 1: Define Variables

Let the length of the rectangle be \( l \) and the width be \( w \). Since the area of the rectangle is 216 \( \text{m}^2 \), we have the equation \( lw = 216 \).

Step 2: Express Fence Length

The total fence length includes both the outer perimeter and the additional fence dividing the garden parallel to the width. The perimeter with the dividing fence is given by \( 2l + 3w \).

Step 3: Solve for One Variable

Since \( lw = 216 \), we can express \( l \) in terms of \( w \) as \( l = \frac{216}{w} \).

Step 4: Substitute and Simplify

Substitute \( l = \frac{216}{w} \) into the fence length expression to get it in terms of \( w \): \( 2\left(\frac{216}{w}\right) + 3w \). Simplifying this gives \( \frac{432}{w} + 3w \).

Step 5: Differentiate Fence Function

To minimize the total fence length, differentiate \( f(w) = \frac{432}{w} + 3w \) with respect to \( w \). The derivative is \( f'(w) = -\frac{432}{w^2} + 3 \).

Step 6: Find Critical Points

Set \( f'(w) = 0 \) to find critical points. Solve \( -\frac{432}{w^2} + 3 = 0 \), giving \( \frac{432}{w^2} = 3 \). Thus, \( w^2 = 144 \), so \( w = 12 \) m.

Step 7: Determine Dimensions of Rectangle

Substitute \( w = 12 \) m into \( l = \frac{216}{w} \) to find \( l = \frac{216}{12} = 18 \) m.

Step 8: Calculate Minimum Fence Length

Using \( w = 12 \) m and \( l = 18 \) m, the minimum fence is \( 2l + 3w = 2(18) + 3(12) = 36 + 36 = 72 \) m.

Key Concepts

Rectangular Area Problems

Rectangular area problems in calculus often involve finding the dimensions of a rectangle that optimize a particular mathematical criterion, such as minimizing perimeter or maximizing area.
In our exercise, we are given a fixed area for a rectangle, which is 216 square meters. This rectangle needs to be divided equally by a fence, and we are tasked with minimizing the length of fencing needed.
To tackle this kind of problem, we start by defining variables: let the length be \( l \) and the width be \( w \). We know the area can be represented as \( lw = 216 \). By manipulating these variables, we explore solutions that satisfy the area condition.
By expressing one variable in terms of the other, such as \( l = \frac{216}{w} \), we can rewrite the optimization function, in this case, the total fence length formula.
This transformation is essential as it allows us to apply calculus methods to find an optimal solution by focusing on a single variable, leading to more straightforward differentiation and analysis.

Perimeter Minimization

Perimeter minimization involves finding the smallest possible perimeter of a shape under certain constraints, such as a fixed area.
In the pea patch example, we seek to minimize the total fence length, including the outer perimeter and the interior dividing fence.
The fence length is influenced by both the outer edges and the need to divide the rectangle, giving us the formula \( 2l + 3w \). We alter this to a single-variable function by substituting the expression for the length \( l = \frac{216}{w} \) into the equation.
This substitution yields the function \( f(w) = \frac{432}{w} + 3w \), which represents the total fence length now expressed entirely in terms of the width \( w \).
Thus, our task becomes analyzing this function to find its minimum value. Perimeter minimization is about finding the balance between the two components \( 2l + 3w \) while ensuring the total area is maintained.

Critical Points Calculation

Critical points are found where the derivative of a function equals zero or is undefined. In optimization problems, these points help identify where the minimum or maximum values occur.
In our scenario, we took the derivative of the fence length function \( f(w) = \frac{432}{w} + 3w \) to obtain \( f'(w) = -\frac{432}{w^2} + 3 \).
Setting the derivative equal to zero, i.e., \( f'(w) = 0 \), helps us find the width \( w \) that results in the minimum fence length. Solving \( -\frac{432}{w^2} + 3 = 0 \) leads us to \( w^2 = 144 \) and ultimately \( w = 12 \).
With this width, substitution back into our equation for \( l \) gives us \( l = 18 \). These dimensions ensure the shortest amount of fencing needed, confirming this occurrence at a critical point.
Finding critical points is crucial for solving optimization problems in calculus, as it pinpoints exactly where changes in conditions yield optimal results.