Problem 14 - Contemporary Precalculus

Question

A rectangular field will be fenced on all four sides. Fencing for the north and south sides costs \(\$ 5\) per foot and fencing for the other two sides costs \(\$ 10\) per foot. What is the maximum area that can be enclosed for \(\$ 5000 ?\)

Step-by-Step Solution

Answer

Answer: The maximum area that can be enclosed with a budget of $5,000 is 31,250 square feet.

Step 1: Define variables

Let \(x\) be the length of the north and south sides and \(y\) be the length of the other two sides. We know that the cost of fencing for the north and south sides is \(5x\) per side, and the cost of fencing the other two sides is \(10y\) per side.

Step 2: Write the cost equation

In order to find the optimal allocation of the $5000 budget, we need to write an equation representing the total cost of fencing the rectangular field. This can be written as: \(C = 5x + 5x + 10y + 10y = 10x + 20y\).

Step 3: Write the constraint equation

The given constraint for the cost is that it must not exceed $5000. Hence, we have: \(10x + 20y \leq 5000\).

Step 4: Express one variable in terms of the other

To express the area of the field as a function of one variable, we will first solve for either \(x\) or \(y\) in the constraint equation. Solving for \(y\), we get: \(y = \frac{5000 - 10x}{20}\).

Step 5: Write the area equation

The area of the rectangular field is given by the product of its sides \(x\) and \(y\). Using the expression for \(y\) from the previous step, the area of the field can be written as: \(A = x\left(\frac{5000 - 10x}{20}\right) = \frac{5000x - 10x^2}{20}\).

Step 6: Differentiate the area equation

In order to find the maximum area, we need to differentiate the area equation with respect to \(x\) and set the derivative equal to zero. This will give us the critical points where the maximum area may occur. Differentiating, we get: \(\frac{dA}{dx} = \frac{5000 - 20x}{20}\).

Step 7: Find the critical points

Setting the derivative equal to zero, we can find the critical points: \(\frac{5000 - 20x}{20} = 0 \Rightarrow 5000 - 20x = 0 \Rightarrow x = 250\).

Step 8: Calculate the corresponding y-value

Using the value of \(x\) found in the previous step, we can find the corresponding value of \(y\). Using the equation from Step 4: \(y = \frac{5000 - 10(250)}{20} = \frac{2500}{20} = 125\).

Step 9: Find the maximum area

Now that we have the dimensions of the field that give the maximum area, we can calculate the maximum area itself using the area equation from Step 5: \(A_{max} = \frac{5000(250) - 10(250)^2}{20} = 250 \cdot 125 = 31,\!250\,\text{square feet}\). Hence, the maximum area that can be enclosed for \(5000 is 31,250\,\text{square feet}\).

Key Concepts

Rectangular Field Fencing Problem

When faced with the challenge of fencing a rectangular field, the primary goal is often to maximize the area that can be enclosed under specific constraints. In the context of precalculus, this optimization problem presents a practical application of mathematical concepts.

Consider a scenario where different sides of a field have distinct fencing costs, like in our exercise where the north and south sides are more economical (abla5 per foot) than the other two sides (abla10 per foot). Here, the fencing problem becomes trickier, as one must consider both dimensions and cost-effectiveness when planning the fence layout.

The educational takeaway is recognizing that real-world problems can be translated into mathematical models, allowing for strategic decision-making. By using algebraic expressions and equations, these problems become solvable, demonstrating the power and utility of mathematics in practical scenarios.

Cost Constraint Equation

A cost constraint equation is essential in optimization problems, serving as the guiding principle for acceptable solutions. It's akin to a budget that cannot be exceeded, much like our abla5000 limit for fencing the field.

In our example, the total cost is expressed as C = 10x + 20y, encapsulating the costs for all four sides of the field. This equation starkly reflects how the constraint binds the feasible combinations of the field's length (x) and width (y).

Breaking down the concept further, the equation's 10x and 20y components are derived from the cost-per-foot of the respective sides, providing a foundation on which we layer our budgetary limit. Through such equations, cost considerations become transparent and calculable, enabling students to understand budgetary impacts on decision-making in various contexts.

Differentiation for Maximum Area

Differentiation is a formidable tool in calculus, used to determine the rate at which one quantity changes with respect to another. For optimization problems, differentiation helps find the maximum or minimum values of functions.

In our fencing problem, we aim to maximize the field's area (A) subject to a cost constraint. By expressing A as a function of length (x), and differentiating A with respect to x, we obtain dA/dx, the derivative representing the rate of change of area with changing length.

Setting this derivative to zero gives us the critical points, which are potential candidates for maximum or minimum values. Finding these points provides clear insights into how a parcel of land can be most effectively utilized, a consideration crucial in not just mathematics, but in fields like architecture and urban planning. Through this exercise, students not only grasp the process of differentiation but also comprehend its vital role in making efficient use of resources.