A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is $$k$$ (meter), then $$\left(\frac{4}{\pi} + 1\right)k$$ is equal to _________

Let the total length of the wire be 36 m. Suppose $$k$$ metres of the wire are used to form the circle, so the remaining $$36-k$$ metres form the square.

If the side of the square is $$x$$, we have the perimeter-side relation for a square: $$4x = 36-k$$, therefore

$$x = \dfrac{36-k}{4}.$$

The circle whose circumference is $$k$$ has radius $$r$$ given by the circumference formula $$2\pi r = k$$, so

$$r = \dfrac{k}{2\pi}.$$

Now, write the total area $$A$$ of the two figures. The area of the square is $$x^{2}$$ and the area of the circle is $$\pi r^{2}$$. Thus

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

Substituting the expressions for $$x$$ and $$r$$, we get

$$A = \left(\dfrac{36-k}{4}\right)^{2} + \pi\left(\dfrac{k}{2\pi}\right)^{2}.$$

Simplifying the second term first:

$$\pi\left(\dfrac{k}{2\pi}\right)^{2} = \pi \cdot \dfrac{k^{2}}{4\pi^{2}} = \dfrac{k^{2}}{4\pi}.$$

Therefore

$$A(k) = \left(\dfrac{36-k}{4}\right)^{2} + \dfrac{k^{2}}{4\pi}.$$

To minimise the area, differentiate $$A$$ with respect to $$k$$ and set the derivative equal to zero.

First expand the square term and differentiate directly:

$$A(k) = \dfrac{(36-k)^{2}}{16} + \dfrac{k^{2}}{4\pi}.$$

Differentiate term-by-term:

$$\dfrac{dA}{dk} = \dfrac{1}{16}\cdot 2(36-k)(-1) + \dfrac{1}{4\pi}\cdot 2k.$$

Simplify each part:

$$\dfrac{dA}{dk} = -\dfrac{36-k}{8} + \dfrac{k}{2\pi}.$$

Set the derivative to zero for a minimum:

$$-\dfrac{36-k}{8} + \dfrac{k}{2\pi} = 0.$$

Multiply through by the common denominator $$8\pi$$ to clear fractions:

$$-\pi(36-k) + 4k = 0.$$

Expand the first product:

$$-36\pi + \pi k + 4k = 0.$$

Group the $$k$$ terms:

$$(\pi + 4)k = 36\pi.$$

Solve for $$k$$:

$$k = \dfrac{36\pi}{\pi + 4}.$$

The expression asked for in the problem is $$\left(\dfrac{4}{\pi} + 1\right)k$$. Compute it by direct substitution:

$$\left(\dfrac{4}{\pi} + 1\right)k = \left(\dfrac{4 + \pi}{\pi}\right)\left(\dfrac{36\pi}{\pi + 4}\right).$$

The numerator and denominator each contain the common factor $$\pi + 4$$, which cancels neatly with $$4 + \pi$$, leaving

$$\left(\dfrac{4 + \pi}{\pi}\right)\left(\dfrac{36\pi}{\pi + 4}\right) = 36.$$

So, the answer is $$36$$.