Problem 9 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A dog kennel with four pens each of area 7 square meters is to be constructed. An exterior fence surrounding a rectangular area is to be built of fence costing $$\$ 20$$ per meter. That rectangular area is then to be partitioned by three fences that are all parallel to a single side of the original rectangle and using fence that costs $$\$ 10$$ per meter. What dimensions of pens will minimize the cost of fence used?

Step-by-Step Solution

Answer

The dimensions to minimize cost are 4 meters by 1.75 meters.

Step 1: Define Variables

Let's let the dimensions of each pen be represented by the width "x" and the length "y". Since each pen has an area of 7 square meters, we can express this as: \( x \times y = 7 \).

Step 2: Express the Dimensions of the Entire Kennel Area

The entire rectangular area needs to be fenced and thus includes the four pens. Since the pens are divided by internal fences that run parallel to one side, the total length of the external rectangle is "x" and the total width is "4y" (four pens in parallel along width).

Step 3: Write the Cost Function

The external fence, which costs \( \$20 \) per meter, surrounds the rectangle of dimensions "x" by "4y", giving a total cost of \( 20(2x + 8y) \). The internal fences, parallel to the shorter side, are three partitions each of length "x", costing \( 10 \times 3x \). The total cost is then: \( C = 20(2x + 8y) + 30x \).

Step 4: Solve for "y" in terms of "x"

Using the equation from Step 1, express "y" with "x": \( y = \frac{7}{x} \).

Step 5: Substitute "y" into the Cost Function

Substitute \( y = \frac{7}{x} \) into the cost function to find a single variable function: \[ C = 20(2x + 8(\frac{7}{x})) + 30x \].

Step 6: Simplify the Cost Equation

Simplify the cost equation: \[ C = 20(2x + \frac{56}{x}) + 30x = 40x + \frac{1120}{x} + 30x = 70x + \frac{1120}{x} \].

Step 7: Find the Minimum Cost Using Calculus

Differentiate the cost function \( C \) with respect to \( x \) and set the derivative to zero to find the critical points: \[ C'(x) = 70 - \frac{1120}{x^2} \]. Setting the derivative to zero gives: \[ 70 = \frac{1120}{x^2} \]. Solve for \( x \): \[ x^2 = \frac{1120}{70} = 16 \Rightarrow x = 4 \].

Step 8: Calculate "y" Using Critical Point

Substitute \( x = 4 \) back into the equation \( y = \frac{7}{x} \) to find "y": \[ y = \frac{7}{4} = 1.75 \].

Step 9: Verify the Solution Gives a Minimum

Perform a second derivative test to confirm a minimum: \[ C''(x) = \frac{2240}{x^3} \]. Since \( C''(x) > 0 \) for \( x = 4 \), the cost is minimized.

Key Concepts

Cost Minimization

Cost minimization is a crucial concept when dealing with real-world problems involving limited resources. In our dog kennel exercise, the goal is to construct a setup that results in the least expense for fencing. To achieve this, we need to think about minimizing how much we spend while still fulfilling all construction requirements.
Minimizing cost often involves balancing different expense sources. Here, we balance between the cost of external and internal fences. The costs per meter differ, with external fences being pricier compared to internal partitions, which are parallel to one side of the rectangle. Therefore, adjusting the dimensions so that both external and internal fencing are optimal is key to reducing overall costs.
Optimizing these two factors often leads to writing a cost function. This function lists the total fencing cost based on specific dimensions, and finding the right dimensions means finding the cost function's minimum value.

Derivative Test

The derivative test helps in finding the minimum or maximum value of a function, which is crucial in optimization problems. After formulating the cost function in our exercise, differentiating it with respect to the variable "x" reveals where the optimize points occur.
Once we have the derivative of the cost function, setting this derivative to zero helps find critical points. However, these critical points could be either a minimum or a maximum, which prompts the use of the second derivative test.

• Calculate the first derivative of the cost function.
• Set the first derivative to zero to find critical points.
• Calculate the second derivative.
• Check the sign of the second derivative at critical points - if positive, it confirms a local minimum.

In our example, the critical point yielded by the first derivative test confirmed that the cost is minimized when further affirmed by the second derivative test.

Rectangular Area Partition

Understanding how to partition a rectangular area is essential for efficiently using space, which in our example involves creating multiple dog pens within the external fencing.
These pens occupy a part of this larger rectangular section and are delimited by three internal partitions parallel to one side. The configuration of these internal partitions relative to the external sides conveys 'division' appropriately and effectively, minimizing the material needed up to the optimized point.

• The area of each pen is pre-determined (7 square meters in this case).
• These partitions allow the rectangle to serve its purpose as separate, usable spaces for the dogs.
• Understanding dimensions and total area is crucial for solving partition-related problems.

With proper dimensions for the pens, the overall setup becomes efficient in both space management and cost.