A piece of wire of length 57 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to (a) minimize and (b) maximize the combined area of the circle and the square?

Step 1 ：Let's denote the length of the wire used to form the circle as \(x\), and the length of the wire used to form the square as \(57-x\).

Step 2 ：The radius of the circle is \(r = \frac{x}{2\pi}\), and the side length of the square is \(s = \frac{57-x}{4}\).

Step 3 ：The area of the circle is \(A_c = \pi r^2 = \pi \left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi}\), and the area of the square is \(A_s = s^2 = \left(\frac{57-x}{4}\right)^2\).

Step 4 ：The total area of the circle and the square is \(A = A_c + A_s = \frac{x^2}{4\pi} + \left(\frac{57-x}{4}\right)^2\).

Step 5 ：To find the minimum and maximum of \(A\), we need to find its critical points. The critical points occur where the derivative of \(A\) with respect to \(x\) is zero, or where the derivative is undefined.

Step 6 ：The derivative of \(A\) with respect to \(x\) is \(A' = \frac{x}{2\pi} - \frac{57-x}{2}\).

Step 7 ：Setting \(A' = 0\), we get \(\frac{x}{2\pi} - \frac{57-x}{2} = 0\). Solving this equation for \(x\), we get \(x = \frac{57}{2 + \frac{1}{\pi}}\).

Step 8 ：Since \(0 \leq x \leq 57\), the critical points are \(x = 0\), \(x = \frac{57}{2 + \frac{1}{\pi}}\), and \(x = 57\).

Step 9 ：Substituting these values into \(A\), we get \(A(0) = \left(\frac{57}{4}\right)^2\), \(A\left(\frac{57}{2 + \frac{1}{\pi}}\right)\), and \(A(57) = \left(\frac{57}{2\pi}\right)^2\).

Step 10 ：By comparing these values, we can find the minimum and maximum of \(A\).

Step 11 ：\(A\left(\frac{57}{2 + \frac{1}{\pi}}\right)\) gives the minimum combined area, and \(A(0) = \left(\frac{57}{4}\right)^2\) gives the maximum combined area.

Step 12 ：So, to minimize the combined area, the wire should be cut at \(x = \frac{57}{2 + \frac{1}{\pi}}\), and to maximize the combined area, the wire should be cut at \(x = 0\).