Problem 320 - Calculus Volume 1

Question

For the following exercises, consider the construction of a pen to enclose an area. You have 800 ft of fencing to make a pen for hogs. If you have a river on one side of your property, what is the dimension of the rectangular pen that maximizes the area?

Step-by-Step Solution

Answer

Dimensions are 400 feet by 200 feet.

Step 1: Understanding the Problem

We need to maximize the area of a rectangular pen using 800 ft of fencing, with a river on one side as a natural boundary.

Step 2: Setting Up the Equation

Let's denote the length of the pen parallel to the river as \( L \) and the width as \( W \). The total fencing used thus will be on only three sides, so the fencing constraint is \( 2W + L = 800 \).

Step 3: Express Area in Terms of One Variable

The area \( A \) of the rectangle is given by \( A = L \times W \). From the constraint equation, solve for \( L \): \( L = 800 - 2W \). Substitute this in the area equation: \( A = W(800 - 2W) = 800W - 2W^2 \).

Step 4: Finding Maximum Area

The function \( A = 800W - 2W^2 \) is a quadratic equation in the form of \( ax^2 + bx + c \), where \( a = -2, b = 800 \). The maximum value happens at \( W = -\frac{b}{2a} = \frac{-800}{2(-2)} = 200 \).

Step 5: Calculate the Corresponding Length

Substitute \( W = 200 \) back into \( L = 800 - 2W \) to find \( L = 800 - 2(200) = 400 \). Thus, the dimensions are \( 400 \) feet by \( 200 \) feet.

Step 6: Conclusion

Given 800 feet of fencing with a river on one side, the dimensions 400 feet by 200 feet maximize the enclosed area.

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial function with the highest degree being two. In simpler terms, it is expressed in the form of \( ax^2 + bx + c \), where \( a, b, \text{ and } c \) are constants and \( x \) represents the variable. These functions generally create a parabola when graphed, which is a symmetrical curve. It's important to understand the nature of this curve:

• If the coefficient \( a \) is positive, the parabola opens upwards, resembling a U shape.
• If \( a \) is negative, the parabola opens downwards, like an inverted U.

The vertex of this parabola represents either a maximum or a minimum point.
For our fencing problem, the quadratic function is used to find the dimensions that give us the maximum area. The function derived, \( A = 800W - 2W^2 \), represents the area in terms of width \( W \), and using this, we can find the point where the area is maximized.

Fencing Problems

Fencing problems often appear in optimization word problems and involve using a fixed amount of resources — in this case, fencing — to enclose an area. These problems are about maximizing or minimizing some aspect of construction.
For a rectangular pen next to a river, one side does not require fencing, reducing the number of sides that need material. An efficient strategy is to express any dimensions in terms of a single variable to use calculus or algebra to optimize — here, maximizing the area.
In our example, because the total available fencing is 800 ft and we are only using it for three sides, we used an equation to express the total fenced length in terms of width \( W \). This setup is crucial in determining the best way to use limited resources.

Constraint Equations

Constraint equations are mathematical expressions that describe the limitations or restrictions of a problem. In optimization problems, constraints often define the boundaries within which we must work.
In the case of the fencing problem, the constraint equation is \( 2W + L = 800 \). This equation stems from the fact that the fencing is only along three sides, with the length \( L \) and twice the width \( 2W \) summing to the total available fencing.
Such equations are vital because they allow us to combine the constraints with the objective — here, maximizing area. By rearranging the constraint equation to express one variable in terms of another, we can substitute back into the equation for area. This allows the use of a quadratic function to find the optimal solution.