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

Let us denote by $$x$$ the side of the square and by $$r$$ the radius of the circle. The total wire length is 2 units and it is divided between the perimeters of these two figures.

Perimeter formula for a square: $$P_{\text{square}} = 4x.$$

Circumference formula for a circle: $$C_{\text{circle}} = 2\pi r.$$

Since the whole wire is used, the length-constraint equation is

$$4x \;+\; 2\pi r \;=\; 2.$$

The area of the square is $$A_{\text{square}} = x^{2},$$ and the area of the circle is $$A_{\text{circle}} = \pi r^{2}.$$

Hence the total area to be minimised is

$$S \;=\; x^{2} \;+\; \pi r^{2}.$$

From the constraint we can express $$r$$ in terms of $$x$$. Starting with

$$4x + 2\pi r = 2,$$

subtract $$4x$$ from both sides:

$$2\pi r = 2 - 4x.$$

Divide by $$2\pi$$:

$$r = \dfrac{2 - 4x}{2\pi} = \dfrac{1 - 2x}{\pi}.$$

Now substitute this expression for $$r$$ into the formula for $$S$$:

$$\begin{aligned} S(x) &= x^{2} + \pi\left(\dfrac{1 - 2x}{\pi}\right)^{2} \\ &= x^{2} + \pi \cdot \dfrac{(1 - 2x)^{2}}{\pi^{2}} \\ &= x^{2} + \dfrac{(1 - 2x)^{2}}{\pi}. \end{aligned}$$

To find the minimum we differentiate $$S(x)$$ with respect to $$x$$ and set the derivative to zero.

Differentiate term by term:

$$\frac{d}{dx}\bigl[x^{2}\bigr] = 2x,$$

and for the second term we first note the chain rule: $$\frac{d}{dx}(1 - 2x)^{2} = 2(1 - 2x)(-2) = -4(1 - 2x).$$ Therefore

$$\frac{d}{dx}\left[\dfrac{(1 - 2x)^{2}}{\pi}\right] = \dfrac{-4(1 - 2x)}{\pi}.$$

So the derivative of $$S$$ is

$$S'(x) = 2x - \dfrac{4}{\pi}(1 - 2x).$$

Setting $$S'(x) = 0$$ for the extremum gives

$$2x - \dfrac{4}{\pi}(1 - 2x) = 0.$$

Multiply every term by $$\pi$$ to clear the denominator:

$$2\pi x - 4(1 - 2x) = 0.$$

Expand the bracket:

$$2\pi x - 4 + 8x = 0.$$

Group the $$x$$ terms:

$$(2\pi x + 8x) = 4.$$

Factor out $$x$$:

$$x(2\pi + 8) = 4.$$

Divide by $$2\pi + 8$$:

$$x = \dfrac{4}{2\pi + 8} = \dfrac{4}{2(\pi + 4)} = \dfrac{2}{\pi + 4}.$$

Now find the corresponding $$r$$ using $$r = \dfrac{1 - 2x}{\pi}$$:

Substitute $$x = \dfrac{2}{\pi + 4}$$:

$$\begin{aligned} r &= \dfrac{1 - 2\left(\dfrac{2}{\pi + 4}\right)}{\pi} \\ &= \dfrac{1 - \dfrac{4}{\pi + 4}}{\pi} \\ &= \dfrac{\dfrac{\pi + 4 - 4}{\pi + 4}}{\pi} \\ &= \dfrac{\dfrac{\pi}{\pi + 4}}{\pi} \\ &= \dfrac{1}{\pi + 4}. \end{aligned}$$

Comparing $$x$$ and $$r$$:

$$x = \dfrac{2}{\pi + 4}, \qquad r = \dfrac{1}{\pi + 4}.$$ Therefore

$$x = 2r.$$

This relation matches Option A.

Hence, the correct answer is Option A.