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(C) Perimeter of the square $= 4x$. Perimeter of the circle $= 2\pi r$. Given total length of wire is $2$ units,so $4x + 2\pi r = 2$. Dividing by $2$,

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$x = 2r$

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(C) Perimeter of the square $= 4x$. Perimeter of the circle $= 2\pi r$. Given total length of wire is $2$ units,so $4x + 2\pi r = 2$. Dividing by $2$,we get $2x + \pi r = 1$,which implies $r = \frac{1 - 2x}{\pi}$. Sum of the areas $A = x^2 + \pi r^2$. Substituting $r$ in terms of $x$: $A = x^2 + \pi \left( \frac{1 - 2x}{\pi} \right)^2 = x^2 + \frac{(1 - 2x)^2}{\pi}$. Differentiating $A$ with respect to $x$: $\frac{dA}{dx} = 2x + \frac{2(1 - 2x)(-2)}{\pi} = 2x - \frac{4(1 - 2x)}{\pi}$. For minimum area,set $\frac{dA}{dx} = 0$: $2x - \frac{4}{\pi} + \frac{8x}{\pi} = 0$. Multiply by $\pi$: $2\pi x - 4 + 8x = 0 \Rightarrow (2\pi + 8)x = 4 \Rightarrow (\pi + 4)x = 2$. Thus,$x = \frac{2}{\pi + 4}$. Substituting $x$ back into the perimeter equation: $2(\frac{2}{\pi + 4}) + \pi r = 1 \Rightarrow \pi r = 1 - \frac{4}{\pi + 4} = \frac{\pi + 4 - 4}{\pi + 4} = \frac{\pi}{\pi + 4}$. Therefore,$r = \frac{1}{\pi + 4}$. Comparing $x$ and $r$,we see that $x = 2r$.

$A$ wire of length $2$ units is cut into two parts,which are bent respectively to form a square of side $x$ units and a circle of radius $r$ units. If the sum of the areas of the square and the circle so formed is minimum,then:

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