Problem 26 - Applied Calculus

Question

A landscape architect plans to enclose a 3000 square-foot rectangular region in a botanical garden. She will use shrubs costing \(\$ 45\) per foot along three sides and fencing costing \(\$ 20\) per foot along the fourth side. Find the minimum total cost.

Step-by-Step Solution

Answer

The minimum total cost is approximately \(\$ 8377.07\).

Step 1: Understand the Problem

We need to enclose a rectangular region of 3000 square feet using two materials for the sides: shrubs (costing \(\\( 45\) per foot) for three sides and fencing (costing \(\\) 20\) per foot) for the remaining side.

Step 2: Define Variables

Let \(x\) be the length of the side where the fencing is used, and \(y\) be the length of the side where the shrubs are used. Thus, the three sides made of shrubs are \(2y + x\).

Step 3: Set Up Equations

The area of the rectangle is given by \(xy = 3000\). This implies \(y = \frac{3000}{x}\).

Step 4: Formulate the Cost Function

The cost \(C\) consists of \(20x\) for the fencing and \(45(2y + x)\) for the shrubs. Therefore, the total cost is \(C = 20x + 45(2y + x) = 20x + 45(2\frac{3000}{x} + x)\).

Step 5: Simplify the Cost Function

Using \(y = \frac{3000}{x}\), we have \(C = 20x + 90\frac{3000}{x} + 45x = 65x + \frac{270000}{x}\).

Step 6: Find Critical Points

Differentiate the cost function to find the minimum cost. The derivative is \(C'(x) = 65 - \frac{270000}{x^2}\). Set \(C'(x) = 0\), giving \(65 = \frac{270000}{x^2}\). Solve for \(x\) to find \(x = \sqrt{\frac{270000}{65}}\).

Step 7: Calculate x Value

\(x = \sqrt{4153.846} \approx 64.47\).

Step 8: Calculate y Using x

Substitute \(x = 64.47\) into \(y = \frac{3000}{x}\) to find \(y \approx 46.55\).

Step 9: Calculate Minimum Cost

Substitute \(x = 64.47\) and \(y \approx 46.55\) back into \(C = 65x + \frac{270000}{x}\) to find the minimum cost. \(C \approx 65(64.47) + \frac{270000}{64.47} \approx 4190.55 + 4186.52 \approx 8377.07\).

Key Concepts

Cost Minimization

When tackling cost minimization problems, the goal is to find the least expensive way to achieve a desired outcome. Here, we focus on enclosing a specific area with a given constraint—using two different materials for the sides of a rectangular garden. Each material has its own cost per unit of length: shrubs are expensive at $45 per foot, whereas fencing is cheaper at $20 per foot. To minimize total expenses, we need smart allocation of these materials.

Think of cost minimization like a puzzle, where you balance different factors to reach the lowest possible expense. Consider not just the mathematical equations but also practical decisions, like which side of the garden will use the cheaper material.

• Identify materials and their costs.
• Choose variables to represent the quantities that affect cost.
• Formulate a cost function to combine all expenses.

This systematic approach is essential in optimizing costs effectively.

Rectangular Geometry

Understanding the geometry of rectangles is fundamental when dealing with real-world optimization problems. In our scenario, the landscape architect is working with a rectangular garden that needs specific sides to be enclosed with shrubs and fencing.

A rectangle’s area is determined by the product of its length and width. Given a fixed area, we are tasked with determining the lengths of these sides effectively. With the fixed area of 3000 square feet, using the equation \( xy = 3000 \) helps elucidate the relationship between the rectangle's dimensions.

Here are some essential geometric concepts to keep in mind:

• The area equation \( A = l \times w \), where \( l \) and \( w \) are the rectangle's length and width.
• How changing one dimension affects the other when the area is constant.
• Application of algebra to solve for unknown dimensions based on other given values.

This foundation in rectangular geometry helps determine the optimal dimensions needed for cost minimization.

Derivative Calculations

Derivative calculations play a crucial role in finding the optimal solution in optimization problems. They help us identify critical points that can potentially represent minimum or maximum values of a function—in this case, the cost function.

To find these points, we take the derivative of the cost function with respect to one of the variables. In our problem, the derivative \( C'(x) = 65 - \frac{270000}{x^2} \) provides a tool to locate these points by setting \( C'(x) = 0 \) and solving for \( x \).

Important steps in this process include:

• Understand what a derivative represents: the rate of change of a function regarding its variable.
• Find critical points by setting the derivative equal to zero and solving the equation.
• Determine whether these points are minima or maxima using the second derivative test or analyzing the context.

Using derivatives allows us to pinpoint the least expensive configuration, ensuring the solution is both mathematically sound and cost-efficient.