Problem 9 - The Calculus with Analytic Geometry

Question

A rectangular field, having an area of \(2700 \mathrm{yd}^{2}\), is to be enclosed by a fence, and an additional fence is to be used to divide the field down the middle. If the cost of the fence down the middle is \(\$ 2\) per running yard, and the fence along the sides costs \(\$ 3\) per running yard, find the dimensions of the field so that the cost of the fencing will be the least.

Step-by-Step Solution

Answer

The dimensions of the field are 45 yards by 60 yards.

Step 1: Define Variables

Let the length of the field be denoted as \(l\) (in yards) and the width be denoted as \(w\) (in yards).

Step 2: Write the Area Equation

The area of the rectangular field is given by \(l \times w = 2700\, \text{yd}^2\).

Step 3: Express One Variable in Terms of the Other

From \(l \times w = 2700\), we can write \(w = \frac{2700}{l}\).

Step 4: Write the Cost Equation

The cost to enclose the field and divide it down the middle includes the perimeter fencing plus the middle fencing. Therefore, the total cost \(C\) in dollars is:\[ C = 3(2l + 2w) + 2l = 6l + 6w + 2l = 8l + 6w \].

Step 5: Substitute for \(w\)

Substitute \(w = \frac{2700}{l}\) into the cost equation:\[ C = 8l + 6\left(\frac{2700}{l}\right) = 8l + \frac{16200}{l} \].

Step 6: Find the Derivative

To minimize the cost, take the derivative of the cost function with respect to \(l\) and set it to zero:\(\frac{dC}{dl} = 8 - \frac{16200}{l^2} = 0\).

Step 7: Solve the Derivative Equation

Solve \(8 - \frac{16200}{l^2} = 0\):\[ 8l^2 = 16200 \implies l^2 = \frac{16200}{8} = 2025 \implies l = \sqrt{2025} = 45 \].

Step 8: Determine \(w\)

Substitute \(l = 45\) back into our equation for \(w\):\[ w = \frac{2700}{45} = 60 \].

Step 9: Verify the Minimum

Check the second derivative to ensure it's a minimum: \(\frac{d^2C}{dl^2} = \frac{32400}{l^3}\). For \(l = 45\), this is positive, thus confirming a minimum.

Key Concepts

headline of the respective core concept

To solve optimization problems in calculus, we often start by finding an **area equation**. Here, we are dealing with a rectangular field, where the length is denoted by \( l \) and the width by \( w \). The given area of the field is 2700 square yards, which can be described using the equation \( l \times w = 2700 \). This equation represents the total area of the rectangle.

We need to find the dimensions \( l \) and \( w \) that satisfy this condition. By expressing one variable in terms of the other, we simplify the problem. So, we can write \( w = \frac{2700}{l} \). This substitution will later help us in formulating the cost function and solving for the dimensions that minimize the cost.

headline of the respective core concept

The **cost function** combines the total costs associated with the fences. In this problem, we have different costs for different parts of the fence. The fence along the perimeter costs \(3 \) dollars per yard, and the fence down the middle costs \(2 \) dollars per yard. To find the total cost, we first note the different components:

• The total length of the perimeter fencing: \( 2l + 2w \)
• The middle fence length: \( l \)

Next, we formulate the total cost \( C \) as follows:

\[ C = 3(2l + 2w) + 2l = 6l + 6w + 2l = 8l + 6w \]
Using our equation for \( w \), substitute \( w = \frac{2700}{l} \) into the cost function to get:

\[ C = 8l + 6\frac{2700}{l} = 8l + \frac{16200}{l} \].

This represents the total cost with respect to \( l \).

headline of the respective core concept

To minimize the cost, we use the **derivative** of the cost function. Taking the derivative with respect to \( l \), we get:

\[ \frac{dC}{dl} = 8 - \frac{16200}{l^2} \].

Setting the derivative to zero and solving for \( l \):

\[ 8 - \frac{16200}{l^2} = 0 \] \[ 8l^2 = 16200 \] \[ l^2 = \frac{16200}{8} = 2025 \] \[ l = \frac{2025} = 45 \]

Thus, the length \( l \) is 45 yards. Now, substitute this back into the equation for \( w \) to find the width:

\[ w = \frac{2700}{45} = 60 \].

Finally, check the second derivative to ensure we have a minimum:

\[ \frac{d^2C}{dl^2} = \frac{32400}{l^3} \].

For \( l = 45 \), the second derivative is positive, confirming a minimum cost. Therefore, the field's optimal dimensions are 45 yards by 60 yards.