Final answer:
The maximum area of a right triangle made from a 36 inches long wire is obtained when the two legs are equal, which occurs in an isosceles right triangle. The calculated dimensions of these legs do not match any of the provided answer choices.
Step-by-step explanation:
The problem at hand requires us to find the dimensions of a right triangle with the maximum possible area when the triangle is formed from a 36 inches long wire. Firstly, we observe that the perimeter of the right triangle will be equal to the length of the wire, which is 36 inches. To maximize the area of a right triangle with a given perimeter, we can use the fact that the triangle will have proportions equivalent to an isosceles right triangle, where the two legs are equal, and the hypotenuse is longer by a factor of √2. Therefore, we set up equations to express the sides in terms of a single variable, say x, where x is the length of one of the legs, and √2x is the length of the hypotenuse. The perimeter P is the sum of the sides, x + x + √2x = 36 inches. Solving for x gives us x = 36 / (2 + √2) inches. Substituting this value back into the area formula for a right triangle, A = (1/2)bh where b and h are the base and height of the triangle, gives us the maximum area. By solving these equations, we determine that the legs of the right triangle with the maximum area are equal, and neither of the answer choices directly corresponds to the calculated dimensions as they imply an isosceles right triangle where each leg is 18 inches.