1. Set up the length constraint

Let the wire be cut into:

• a part forming a square of side $x$
• a part forming a circle of radius $r$

Since total wire length is $2$ units,

• perimeter of square $=4x$
• circumference of circle $=2\pi r$

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

Equivalently, $2x+\pi r=1$

2. Write the total area

Area of square: $A_s=x^2$

Area of circle: $A_c=\pi r^2$

Hence total area, $A=x^2+\pi r^2$

We must minimize this subject to $2x+\pi r=1$

3. Express one variable in terms of the other

From the constraint, $r=\frac{1-2x}{\pi}$

Substitute into area: $A(x)=x^2+\pi\left(\frac{1-2x}{\pi}\right)^2$

$A(x)=x^2+\frac{(1-2x)^2}{\pi}$

4. Differentiate and find critical point

$A'(x)=2x+\frac{2(1-2x)(-2)}{\pi}$

$A'(x)=2x-\frac{4(1-2x)}{\pi}$

For minimum, set $A'(x)=0$:

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

Multiply by $\pi$: $2\pi x-4+8x=0$

$2\pi x+8x=4$

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

$x=\frac{2}{\pi+4}$

Now from constraint, $2x+\pi r=1$

$\pi r=1-2x=1-\frac{4}{\pi+4}$

$\pi r=\frac{\pi}{\pi+4}$

$r=\frac{1}{\pi+4}$

Therefore, $x=\frac{2}{\pi+4}=2\left(\frac{1}{\pi+4}\right)=2r$

So, $\boxed{x=2r}$

5. Check that it is a minimum

$A''(x)=2+\frac{8}{\pi}>0$

Hence the critical point gives a minimum.

6. Evaluate the options

• A: $x=2r$ ✅ correct
• B: $2x=r$ ❌
• C: $2x=(\pi+4)r$; since $x=2r$, this would give $4r=(\pi+4)r$, false in general ❌
• D: $(4-\pi)x=\pi r$; substituting $x=2r$ gives $2(4-\pi)r=\pi r$, not true in general ❌

Therefore the correct option is $\boxed{\text{A}}$