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The "fencing problem" is a classic mathematical exercise that involves finding the optimal way to enclose an area with a fixed amount of fencing. This problem often arises in practical scenarios, such as building a pen for farm animals or, as in this case, creating a space in the backyard for a dog. The task seeks to determine dimensions that yield the largest possible area, given a limited amount of fencing material.
In the problem concerning Tina and Imai's backyard, we explore two scenarios. First, we calculate the maximum area enclosed with fencing on all four sides. Second, we consider using the house as one side of the boundary, taking advantage to reduce the fencing needed along that side.

When solving problems like these, one major consideration is the perimeter constraint. This tells us how much total fencing material is available. In mathematical terms, the perimeter is the sum of all the sides of the rectangle that the fence will cover.
For scenario (a), using fencing around all four sides, the perimeter constraint can be expressed as:

• \(2x + 2y = 200\)
• By simplifying, we find \(x + y = 100\)

This tells us that the sum of the length and width should equal 100 feet.
For scenario (b), with the house acting as a side, the perimeter constraint changes to:

• \(x + 2y = 200\)
• Here, the single length of the rectangle plus twice the width should equal 200 feet.

These equations guide us in forming expressions to find the maximum area within these bounds.

Calculus optimization techniques help solve the problem of maximizing or minimizing a function. In this context, the function we want to optimize is the area of the rectangle. To do this, we take the following steps:

• Express the area as a function of one variable.
• Use the perimeter constraint to rewrite the area function in terms of one variable. For example, in scenario (a), with \(x + y = 100\), we can express \(y\) as \(100 - x\) and substitute into the area equation: \(A = x(100 - x) = 100x - x^2\).
• The next step is to take the derivative of this area function. The derivative tells us how the area changes with respect to changes in dimensions.For scenario (a):\(A' = 100 - 2x\)

This derivative helps find critical points that can then be analyzed to find maximum or minimum values.

Critical points in calculus are where the derivative of a function is zero or undefined. These points are crucial in determining where a function reaches its minimum or maximum.
Finding critical points involves setting the derivative of the function to zero and solving for the variable. In this problem, the critical point for scenario (a) is found by solving \(100 - 2x = 0\), which gives \(x = 50\). This indicates that the length, \(x\), should be 50 feet for maximum area with four-sided fencing.
For scenario (b), the derivative is \(A' = 200 - 4y\). Setting this equal to zero gives \(y = 50\), suggesting the width should be 50 feet when the house forms one boundary.
By calculating the corresponding side lengths using perimeter constraints, both scenarios reveal dimensions that maximize the enclosed area, demonstrating how critical points guide optimal solutions in calculus optimization problems.