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Imagine you have a garden that you want to fence. Instead of a typical four-sided fence, we’ll use an existing building to form one side. This means we just need three sides of the actual fence. The setup forms a rectangle, with one long side along the building and two shorter sides extending from it. This type of enclosure is practical and involves fewer materials, as the building itself serves as part of the boundary. In our exercise, our goal is to create a rectangular pen with a total area of 100 square feet. How much fencing will we need? This is the optimization problem we need to solve.

In this context, the width, denoted as \( W \), is the dimension running perpendicular to the building. The length, \( L \), runs parallel to the building. To find the amount of fence needed, we focus on expressing everything in terms of these dimensions.

The area of a rectangle is calculated by multiplying its length and width. Since we aim to achieve a rectangular pen with an area of exactly 100 square feet, we set up the equation \( W \cdot L = 100 \). This equation links our choices of width and length to the desired area. It's essential because it shows the relationship between \( W \) and \( L \). Once you know one, you can find the other.

From this equation, we can solve for one variable in terms of the other. For example, solving for \( L \) allows us to express the length as \( L = \frac{100}{W} \). This substitution simplifies our future calculations since it reduces the complexity to just one main variable, \( W \). This makes it more efficient when trying to minimize resources used, specifically, the fencing.

Now, let's focus on how to determine how much fence is needed. The formula for the total length of the fence required is \( F = 2W + L \). This equation captures the essence of what needs to be constructed: two widths and one length. The reason behind this formula is straightforward—since the length of the building is already enclosed, you only need two widths and a single length of fencing.

Substituting \( L = \frac{100}{W} \) into our fence formula gives us \( F = 2W + \frac{100}{W} \). This new equation allows us to consider the fence needed in terms of \( W \) alone. It also opens the door to further calculations, such as graphing and finding the minimum amount of fencing necessary.

To visualize how changing the width, \( W \), affects the required fence, we plot \( F = 2W + \frac{100}{W} \). This graph shows the relationship between \( W \) and \( F \), which helps identify the optimal dimension that minimizes the fencing. By plotting, we usually observe a curve that can help spot where the minimum cost occurs.

The graph's shape tells us about the function's behavior, showing how increases in \( W \) initially reduce fencing costs before eventually increasing again as \( W \) becomes too large. Calculus tools, like finding the derivative \( F'(W) = 2 - \frac{100}{W^2} \), help pinpoint the exact point where the fence is minimized. Setting this derivative equal to zero and solving gives us the optimal width that results in the least amount of fencing required, ensuring we use resources efficiently. By differentiating and solving, we find that the minimum occurs at \( W = 5\sqrt{2} \). This correspondence between graph and calculus confirms our solution. The elegance of this function and its graphical representation lies in its ability to express an optimisation problem and its solution succinctly.