Final answer:
To maximize the enclosed area for three rectangles with 2300 feet of fencing, the dimensions should be x = 287.5 feet (length) and y = 362.5 feet (width).
Step-by-step explanation:
The problem given is an optimization problem in which we aim to find the dimensions x and y that maximize the enclosed area for three adjacent rectangles using a fixed amount of fencing. To maximize the area with 2300 feet of fencing for three equally sized rectangles, we can consider the total perimeter required for the three rectangles. Here, x represents the length of each rectangle, and y represents the width. Since there are three rectangles with two widths in common, the total perimeter P used will be P = 2y + 4x.
According to the question, the total amount of fencing is 2300 feet, so we have the equation 2y + 4x = 2300. To solve this, we want to express y in terms of x: y = (2300 - 4x) / 2. The total enclosed area A for the three rectangles is A = 3xy. Substituting the expression for y in terms of x, we get A = 3x((2300 - 4x) / 2). Simplifying, we have A = 3x(1150 - 2x).
To maximize the area, we need to find the derivative of A with respect to x and set it to zero to find the critical points. After finding the derivative A' = 3(1150 - 4x), we set A' = 0 to find that x = 287.5 feet. Substituting x back into the equation for y, we get y = 362.5 feet. Therefore, the dimensions that maximize the area are x = 287.5 feet and y = 362.5 feet.