Final answer:
To minimize the material costs for fencing three sides of a playground with the fourth side being the house, the dimensions where length is equal to width would initially be set to 66.67 feet each, based on the available 200 feet of fencing material. However, without additional details or a calculus-based solution, this is a guideline rather than an exact answer.
Step-by-step explanation:
The homeowner has a total of 200 feet of fencing material to enclose three sides of a rectangular playground, with the house forming the fourth side. To minimize the material costs, the homeowner must find the dimensions that provide the largest area for a given perimeter, which is a classic optimization problem in mathematics. According to the fencing material constraint, if we denote the length of the side parallel to the house as L and the lengths of the two other sides as W, the total length of the fence used is 2W + L = 200 feet.
To find the dimensions that minimize cost, we use the principle that for a given perimeter, a rectangle's area is maximized when the rectangle is a square. However, since one side is formed by the house, we are looking for a situation where L is equal to W. We set up the equation 2W + W = 200 and solve for W, which also gives us the value of L because they are equal in this case. So W = L = 200/3 feet. But this is the scenario for a square, and we can't assume the side of the house equals the optimal width; thus, we usually do calculus to find the actual minimum, but the problem does not provide enough details for a calculus solution. So, based on the information given, the homeowner should consider using 66.67 feet for the width (W) and 66.67 feet for the length parallel to the house (L) as an initial guideline, subject to adjustment based on the actual length of the house side.