Problem 65 - Calculus A New Horizon

Question

A rectangular plot of land is to be fenced off so that the area enclosed will be \(400 \mathrm{ft}^{2} .\) Let \(L\) be the length of fencing needed and \(x\) the length of one side of the rectangle. Show that \(L=2 x+800 / x\) for \(x>0,\) and sketch the graph of \(L\) versus \(x\) for \(x>0\)

Step-by-Step Solution

Answer

The formula for the fencing is \( L = 2x + \frac{800}{x} \). L decreases and approaches 2x as x increases for x > 0.

Step 1: Understand the Problem

We know the area of the rectangle is given by \( A = x \cdot y \), where \( x \) and \( y \) are the lengths of the sides. We need to find an expression for \( L \), the perimeter, in terms of \( x \).

Step 2: Define Known Values

We are given that the area \( A = 400 \). Thus, \( x \cdot y = 400 \). Solve for \( y \): \( y = \frac{400}{x} \).

Step 3: Express Fencing Length (L) in Terms of x

The perimeter of a rectangle is given by \( L = 2x + 2y \). Substitute \( y = \frac{400}{x} \) into this equation: \( L = 2x + 2\left(\frac{400}{x}\right) = 2x + \frac{800}{x} \).

Step 4: Graph L versus x

The function \( L = 2x + \frac{800}{x} \) is defined for \( x > 0 \). It is a hyperbolic function that decreases as \( x \) increases, reaching a lower bound. To sketch it, note that for large \( x \), \( \frac{800}{x} \) approaches zero, making \( L \approx 2x \). Conversely, for small \( x \), \( \frac{800}{x} \) dominates, making \( L \) very large.

Key Concepts

Rectangular Plot

When dealing with problems like this, it's important to first visualize and understand the concept of a rectangular plot of land. Imagine a flat, four-sided figure with opposite sides being equal in length, and right angles at each corner.
In our exercise, this rectangle is specifically required to have an area of 400 square feet. This condition forms one of the essential constraints when dealing with optimization problems.

To approach this, remember the formula used to calculate the area of a rectangle. It is simply the multiplication of its two adjacent sides, mathematically expressed as:

• Area, \( A = x \cdot y \)

where \( x \) and \( y \) are the lengths of the rectangle's sides. By setting \( y = \frac{400}{x} \), we utilize the given area constraint, paving the way for further calculations involving the perimeter.

Fencing Perimeter Calculation

In the context of this problem, the perimeter refers to the total length of fencing required to enclose the rectangular plot. Calculating fences is a crucial application of understanding perimeter in real-world scenarios, particularly in agriculture and construction.
To compute the perimeter, use the well-established formula for a rectangle:

• Perimeter, \( L = 2x + 2y \)

Given that \( y = \frac{400}{x} \), substitute \( y \) into the perimeter formula:
\( L = 2x + 2\left(\frac{400}{x}\right) \)

This simplifies to:

• \( L = 2x + \frac{800}{x} \)

This equation expresses the fencing length as a function of \( x \), helping us better understand how changes in one dimension affect the total perimeter. As \( x \) increases, the term \( \frac{800}{x} \) decreases, showing how the total fencing length varies with \( x \).

Function Graphing

Graphing functions is a powerful mathematical tool that allows us to visually analyze the behavior of relationships between variables. For the function \( L = 2x + \frac{800}{x} \), graphing helps illustrate changes in the fencing length as \( x \) varies.
To graph this function, consider its behavior for various values of \( x \):

• When \( x \) is very small, \( \frac{800}{x} \) becomes very large, making \( L \) large as well.
• As \( x \) increases, \( \frac{800}{x} \) decreases, reducing its impact on \( L \) and approximating \( L \) to \( 2x \).

This results in a hyperbolic graph for \( L \) versus \( x \) where \( x > 0 \). Such a curve typically decreases, showing that as \( x \) grows larger, the fence needed reduces, following the horizontal asymptote defined by \( L = 2x \). Understanding this graphical pattern is key to mastering optimization problems where constraints influence variable relationships.