91Ó°ÊÓ

Let x be the length of the wire used to form the circle, and let (60-x) be the length of the wire used to form the square. The circumference of the circle is equal to x, and the perimeter of the square is equal to (60-x). To express the combined area A of the circle and the square in terms of x, we first need to find the radius r of the circle and the side length s of the square. - For the circle, the circumference C = 2πr, so r = x / (2π). - For the square, the perimeter P = 4s, so s = (60-x) / 4. Now, we can express the areas of the circle and the square in terms of x: - The area of the circle A_c = πr^2 = π(x/(2π))^2 = x^2 / (4π). - The area of the square A_s = s^2 = ((60-x) / 4)^2. The combined area A is the sum of the areas of the circle and the square: A(x) = A_c + A_s = (x^2 / (4π)) + ((60-x) / 4)^2.

To find the critical points of the combined area function A(x), we need to find where A'(x) = 0 or where A'(x) is undefined. First, we find the derivative of A(x) with respect to x. We can apply the chain rule and power rule to differentiate the terms: A'(x) = (1 / (2π)) * 2x - (1/4) * 2(60-x) Now, we set A'(x) = 0 and solve for x: 0 = (1 / π)x - (1/2)(60 - x) x = (60 / (1 + (2/π))) Since x is a length of wire, it must be in the range [0, 60]. Therefore, the critical points of the combined area function are at x = 0, x = 60, and x = (60 / (1 + (2/π))).

To determine whether the combined area is minimized or maximized at the critical points, we can analyze the behavior of A'(x), the derivative of the combined area function, at each critical point: - At x = 0 (all wire used for the square), there is no wire left to form a circle, so this does not give a valid circle and square configuration, and we exclude x=0 as a possibility. - At x = 60 (all wire used for the circle), there is no wire left to form a square, so this does not give a valid circle and square configuration, and we exclude x=60 as a possibility. - At x = (60 / (1 + (2/π))), A'(x) = 0, which means the combined area function has a critical point at this value of x. Now we use either the first or second derivative test to determine whether the combined area function has a minimum or maximum at x = (60 / (1 + (2/π))). Using the second derivative test: A''(x) = (1 / π) - (1/2) A''(60 / (1 + (2/π))) > 0. Since A''(x) is greater than 0, the combined area function A(x) has a minimum at x = (60 / (1 + (2/π))).

To minimize the combined area of the circle and the square, the wire should be cut at x = (60 / (1 + (2/Ï€))). To find the minimum area, we can plug this value into the combined area function A(x): A_min = A(60 / (1 + (2/Ï€))) Since the combined area function has only one valid critical point, and it is a minimum, there is no maximum combined area in this problem. In other words, we cannot maximize the combined area of the circle and the square by changing the cut position. To summarize: (a) To minimize the combined area of the circle and the square, the wire should be cut at x = (60 / (1 + (2/Ï€))), and the minimum combined area is A(60 / (1 + (2/Ï€))). (b) There is no maximum combined area in this problem, as the only valid critical point gives a minimum combined area.