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(D) Let the length of the wire used for the square be $x$. Then the length of the wire used for the circle is $8-x$. For the square,the perimeter is $

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$\frac{16}{\pi+4}$

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(D) Let the length of the wire used for the square be $x$. Then the length of the wire used for the circle is $8-x$. For the square,the perimeter is $4a = x$,so the side $a = \frac{x}{4}$. The area of the square is $A_1 = a^2 = \frac{x^2}{16}$. For the circle,the circumference is $2\pi r = 8-x$,so the radius $r = \frac{8-x}{2\pi}$. The area of the circle is $A_2 = \pi r^2 = \pi \left(\frac{8-x}{2\pi}\right)^2 = \frac{(8-x)^2}{4\pi}$. The total area $A(x) = \frac{x^2}{16} + \frac{(8-x)^2}{4\pi}$. To find the minimum,differentiate with respect to $x$: $A'(x) = \frac{2x}{16} + \frac{2(8-x)(-1)}{4\pi} = \frac{x}{8} - \frac{8-x}{2\pi}$. Set $A'(x) = 0$: $\frac{x}{8} = \frac{8-x}{2\pi} \implies \pi x = 32 - 4x \implies x(\pi+4) = 32 \implies x = \frac{32}{\pi+4}$. Substitute $x$ back into $A(x)$: $A = \frac{1}{16} \left(\frac{32}{\pi+4}\right)^2 + \frac{1}{4\pi} \left(8 - \frac{32}{\pi+4}\right)^2 = \frac{64}{(\pi+4)^2} + \frac{1}{4\pi} \left(\frac{8\pi+32-32}{\pi+4}\right)^2 = \frac{64}{(\pi+4)^2} + \frac{64\pi^2}{4\pi(\pi+4)^2} = \frac{64}{(\pi+4)^2} + \frac{16\pi}{(\pi+4)^2} = \frac{16(4+\pi)}{(\pi+4)^2} = \frac{16}{\pi+4}$.

$A$ wire of length $8 \text{ units}$ is cut into two parts which are bent respectively in the form of a square and a circle. The least value of the sum of the areas so formed is

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