91Ó°ÊÓ

When working on optimization problems involving geometric shapes, understanding constraints is crucial. For a rectangular garden, the perimeter is a key constraint. The perimeter of a rectangle is given by the formula \(2L + 2W = \text{total perimeter}\), where \(L\) is the length and \(W\) is the width. In this exercise, the total perimeter is 300 meters. This means that the sum of all four sides of the rectangle must equal 300 meters.
You can simplify the constraint equation by dividing it by 2, resulting in \(L + W = 150\). This equation helps in establishing a direct relationship between \(L\) and \(W\). By defining one variable in terms of the other—\(W = 150 - L\)—we can substitute it into further equations, simplifying our calculations significantly.

The goal of many geometric optimization problems is to maximize or minimize a particular measure; in this case, it's the area of the rectangular garden. The area \(A\) of a rectangle is given by the formula \(A = L \times W\). To find the maximum area, you'll need to express the area formula in terms of a single variable. Using the equation from the perimeter constraint, \(W = 150 - L\), we substitute into the area formula:
\[ A = L(150 - L) \]
This can be expanded and simplified to:
\[ A = 150L - L^2 \]
This quadratic equation represents the area of the garden in terms of its length \(L\). The next step involves finding the maximum value of this quadratic function. In simpler terms, we need to find the length \(L\) that allows the garden to cover the greatest area possible.

To maximize the area function \(A = 150L - L^2\), we apply principles from calculus, specifically derivative application. Derivatives help us find critical points—values where the function changes behavior (e.g., from increasing to decreasing). To find these points, we take the derivative of the area formula with respect to \(L\) and set it to zero:
\[ \frac{dA}{dL} = 150 - 2L \]
Setting the derivative equal to zero allows us to solve for \(L\):
\[ 150 - 2L = 0 \]
Solving for \(L\), we get:
\[ L = 75 \]
This critical point indicates that the length \(L\) of 75 meters yields the maximum area. To find the corresponding width, substitute \(L = 75\) back into the constraint equation \(W = 150 - L\):
\[ W = 150 - 75 = 75 \]
Thus, the dimensions of the rectangular garden that provide the greatest area under a perimeter constraint of 300 meters are 75 meters by 75 meters. This process elegantly demonstrates the power of calculus in solving real-world optimization problems.