- Express the perimeter constraint \(2l + 2w = 100\) as \(l = 50 - w\).
- Substitute this expression into the area formula \(A = l \times w\) to get \(A(w) = (50 - w)w\).
- Expand the expression to obtain \(A(w) = 50w - w^2\).
- Match the result to option (b), which is \(\boxed{b}\).
Explanation
- Problem Analysis and Given Data
We are tasked with finding the quadratic equation for the area \(A\) of a rectangular pen in terms of its width \(w\), given that the total fencing available (the perimeter) is 100 meters. Let \(l\) be the length and \(w\) be the width of the rectangle. The perimeter \(P\) of a rectangle is given by the formula \(P = 2l + 2w\). Since the farmer has 100 meters of fencing, we set \(2l + 2w = 100\).
- Expressing Length in Terms of Width
We start by isolating the length \(l\) in terms of the width \(w\). Starting from the equation \(2l + 2w = 100\), we divide the entire equation by 2:
\[l + w = 50\]
Now, we solve for \(l\):
\[l = 50 - w\]
- Deriving the Quadratic Equation for Area
The area \(A\) of a rectangle is the product of its length \(l\) and width \(w\). Substituting our expression for \(l\) into the area formula:
\[A = l \times w = (50 - w) \times w\]
Distributing \(w\) across the terms inside the parentheses:
\[A(w) = 50w - w^2\]
Rearranging the terms to standard quadratic form, we get \(A(w) = -w^2 + 50w\), which matches the form given in option (b).
- Final Conclusion
We have found that the area \(A\) as a function of the width \(w\) is \(A(w) = 50w - w^2\). Comparing this result with the given options:
A\(A(w) = w^2 - 100w\)
B\(A(w) = 50w - w^2\)
C\(A(w) = w^2 - 50w\)
D\(A(w) = 100w - w^2\)
The correct option is (b). The smallest possible value for \(w\) is greater than 0, and the largest is less than 50, ensuring the dimensions remain positive. The final answer is \(\boxed{b}\).
Examples
Understanding how to model the area of a rectangle as a function of its side lengths is a fundamental skill in geometry and optimization. For example, a gardener planning a vegetable patch with a fixed amount of fencing uses this exact logic to maximize the growing space. By expressing area as a quadratic function \(A(w) = -w^2 + 50w\), one can find the vertex of the parabola to determine the dimensions that provide the maximum area, ensuring the most efficient use of materials.