1. Let the circle use length $k$ meters.

   Then the remaining wire used for the square is: $36-k$

2. Area of the circle in terms of $k$

   Since circumference of the circle is $k$, $2\pi r = k \implies r = \frac{k}{2\pi}$

   So its area is: $A_c = \pi r^2 = \pi\left(\frac{k}{2\pi}\right)^2 = \frac{k^2}{4\pi}$

3. Area of the square in terms of $k$

   The perimeter of the square is $36-k$, so side length is: $s = \frac{36-k}{4}$

   Therefore area of the square is: $A_s = s^2 = \left(\frac{36-k}{4}\right)^2 = \frac{(36-k)^2}{16}$

4. Total area

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

5. Minimize $A(k)$

   Differentiate with respect to $k$: $A'(k) = \frac{k}{2\pi} - \frac{36-k}{8}$

   For minimum, $A'(k)=0$

   So, $\frac{k}{2\pi} = \frac{36-k}{8}$

   Multiply by $8\pi$: $4k = \pi(36-k)$

   $4k = 36\pi - \pi k$

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

   $k = \frac{36\pi}{4+\pi}$

6. Compute the required value

   We need: $\left(\frac{4}{\pi}+1\right)k = \frac{4+\pi}{\pi}\cdot k$

   Substitute $k$: $\frac{4+\pi}{\pi}\cdot \frac{36\pi}{4+\pi} = 36$

7. Check minimum condition

   $A''(k)=\frac{1}{2\pi}+\frac{1}{8}>0$ Hence the value indeed gives a minimum.

Therefore, the required integer is: $\boxed{36}$