Problem 17 - Calculus with Concepts in Calculus

Question

A horse breeder plans to set aside a rectangular region of 1 square kilometer for horses and wishes to build a wooden fence to enclose the region. Since one side of the region will run along a well-traveled highway, the breeder decides to make that side more attractive, using wood that costs three times as much per meter as the wood for the other sides. What dimensions will minimize the cost of the fence?

Step-by-Step Solution

Answer

Optimal dimensions are \( x = \sqrt{\frac{2}{5}} \) km (highway) and \( y = \sqrt{\frac{5}{2}} \) km.

Step 1: Define Variables

Let the dimensions of the rectangular region be \( x \) kilometers along the highway and \( y \) kilometers perpendicular to the highway. Since the area needs to be 1 square kilometer, the equation is: \( x \times y = 1 \).

Step 2: Express Cost Function

The cost of the fence along the highway is \( 3x \) (since it costs three times as much), and the cost for the other three sides will be \( 2x + 2y \). Therefore, the total cost \( C \) in terms of kilometers is given by \( C = 3x + 2y + 2x = 5x + 2y \).

Step 3: Substitute Area Constraint

Using the area constraint equation, express \( y \) in terms of \( x \): \( y = \frac{1}{x} \). Substitute this into the cost function to get: \( C = 5x + 2\left(\frac{1}{x}\right) = 5x + \frac{2}{x} \).

Step 4: Find Critical Points

Differentiate the cost function with respect to \( x \) to find the critical points: \( \frac{dC}{dx} = 5 - \frac{2}{x^2} \). Set the derivative equal to zero to find the critical point: \( 5 = \frac{2}{x^2} \) which simplifies to \( x^2 = \frac{2}{5} \), thus \( x = \sqrt{\frac{2}{5}} \).

Step 5: Determine Minimum Cost

To confirm this gives a minimum cost, use the second derivative test: \( \frac{d^2C}{dx^2} = \frac{4}{x^3} \), which is positive for all \( x > 0 \). This confirms a local minimum. Substituting \( x = \sqrt{\frac{2}{5}} \) back into \( y = \frac{1}{x} \), we get \( y = \frac{1}{\sqrt{\frac{2}{5}}} = \sqrt{\frac{5}{2}} \).

Step 6: Conclusion: Optimal Dimensions

The dimensions that minimize the cost are \( x = \sqrt{\frac{2}{5}} \) kilometers along the highway and \( y = \sqrt{\frac{5}{2}} \) kilometers. The optimal solution uses the available area in a balanced way to minimize expense lengths.

Key Concepts

Cost Function

In optimization problems like the one with the horse breeder, the cost function represents the expenses related to structuring the fence. The goal is to minimize this cost while adhering to constraints, such as the area of the rectangular region being 1 square kilometer.

Here, the breeder uses more expensive wood along the highway, so the cost function must account for the three-times-higher price on that side. The total cost function, considering the rectangular region's sides, is expressed as:

• For the highway side: cost = \(3x\)
• For the other sides: cost = \(2x + 2y\)

Therefore, the entire cost function becomes \(C = 3x + 2x + 2y = 5x + 2y\).

Substituting for \(y\) using the area constraint allows for a manageable function in terms of one variable \(x\) for optimization purposes: \(C = 5x + \frac{2}{x}\). This expression provides the basis for finding the minimized cost condition.

Critical Points

Critical points in calculus are where potential minima or maxima occur. They signify locations at which the first derivative of the function equals zero. For our problem, it means identifying the dimension \(x\) that will minimize the fence cost.

The first derivative of the cost function \(C = 5x + \frac{2}{x}\) with respect to \(x\) is calculated as follows:

• \( \frac{dC}{dx} = 5 - \frac{2}{x^2} \)

Setting \( \frac{dC}{dx} \) to zero allows us to find the critical points. Solving \( 5 - \frac{2}{x^2} = 0 \) leads us to:

• \( x^2 = \frac{2}{5} \)
• \( x = \sqrt{\frac{2}{5}} \)

At this \(x\) value, the fence cost is potentially minimized, so it becomes crucial to confirm this with further analysis.

Second Derivative Test

The second derivative test is a valuable technique to confirm whether a critical point is a minimum or maximum. This method helps to establish the nature of the critical point found by analyzing the first derivative.

For a function \(f(x)\), if the second derivative \(f''(x)\) is positive at a critical point, the function has a local minimum there. Conversely, if \(f''(x)\) is negative, it indicates a local maximum.

In our scenario, the second derivative of the cost function \(C\) is:

• \( \frac{d^2C}{dx^2} = \frac{4}{x^3} \)

This derivative is always positive for \(x > 0\), confirming a local minimum at \(x = \sqrt{\frac{2}{5}}\). Hence, the dimensions derived provide the minimal fencing cost effectively.

Rectangular Region

The rectangular region for the horses is specified to be exactly 1 square kilometer in area. This constraint plays a pivotal role in shaping the optimization strategy and influences the function used for cost calculation.

The relation between the dimensions \(x\) and \(y\) of the rectangular area is established through the equation \(x \times y = 1\). This means that the product of the side lengths equals the total area. It allows expressing \(y\) in terms of \(x\), simplifying the substitution of \(y = \frac{1}{x}\) into the cost function.

By doing so, the equation only involves \(x\), streamlining the calculation process to find the optimal dimensions. Handling the rectangular region adeptly is key to solving and optimizing this exercise pragmatically.