Final answer:
The dimensions that maximize the area of a rectangle with a perimeter of 24 feet would be a square with sides of 6 feet, resulting in an area of 36 square feet. The correct answer is C.
Step-by-step explanation:
The question posed is about finding the dimensions that would maximize the area of a rectangle with a given perimeter. This is a classic optimization problem in mathematics, typically covered in high school algebra or calculus. To solve this, we first need to recognize that the perimeter (P) of a rectangle is given by P = 2l + 2w, where l is the length and w is the width. For a fixed perimeter of 24 feet, this equation becomes 24 = 2l + 2w, which can be rearranged to l = 12 - w.
Next, we set up the area maximization scenario. The area (A) of a rectangle is given by A = l × w. Substituting l with 12 - w, we get A = (12 - w)w. To maximize A, we take the derivative of A with respect to w and set it equal to zero. This yields w = 6, implying that l is also 6 because l + w should be 12 (half the perimeter). Therefore, a square of 6 feet by 6 feet would maximize the area, which would be 36 square feet.
In a similar geometrical question or mathematical modeling scenario, such as the one involving a farmer fencing off land or determining the scale factor for a drawing, the principle remains the same—use given parameters to find dimensions, whether maximizing area or fitting within constraints. For instance, if a farmer knows three sides of a plot, they can calculate the fourth side to enclose the maximum area. In the example mentioned, if Marta has a square with a side length of 4 inches, and then another square with dimensions that are twice the original, the area of the larger square will be four times that of the smaller square because area scales with the square of the side length (8 inches squared is 64 square inches, compared to the original 16 square inches).