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Step-by-Step Solution
1. Define Variables
Let the total length of the wire be 20 m. Suppose we cut the wire into two parts of lengths
$x$ (used for the square) and $20 - x$ (used for the regular hexagon).
2. Express the Areas
(i) Square:
If $x$ is the total perimeter for the square, then each side of the square is
$x/4$. Hence, its area is
$$
A_{\text{square}} = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}.
$$
(ii) Regular Hexagon:
If $20 - x$ is the total perimeter for the hexagon, then each side of the hexagon is
$$
\frac{20 - x}{6}.
$$
The area of a regular hexagon with side $a$ is given by
$$
A_{\text{hexagon}} = \frac{3\sqrt{3}}{2}\,a^2.
$$
Substituting $a = \frac{20 - x}{6}$, we get
$$
A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} \left(\frac{20 - x}{6}\right)^2.
$$
3. Write the Total Area Function
The combined area, $A(x)$, is
$$
A(x)
= A_{\text{square}} + A_{\text{hexagon}}
= \frac{x^2}{16}
+ \frac{3\sqrt{3}}{2} \left(\frac{20 - x}{6}\right)^2.
$$
4. Simplify the Total Area
Simplifying $A(x)$,
$$
A(x)
= \frac{x^2}{16}
+ \frac{3\sqrt{3}}{2} \cdot \frac{(20 - x)^2}{36}
= \frac{x^2}{16}
+ \frac{3\sqrt{3}}{72}(20 - x)^2.
$$
5. Differentiate and Find Critical Point
To find the minimum of $A(x)$, we compute the derivative $A'(x)$ and set it to zero:
$$
A'(x)
= \frac{d}{dx}\left(\frac{x^2}{16}\right)
+ \frac{d}{dx}\left(\frac{3\sqrt{3}}{72}(20 - x)^2\right).
$$
$\frac{d}{dx}\left(\frac{x^2}{16}\right) = \frac{2x}{16} = \frac{x}{8}.$
For the second term: let $u = (20 - x)$, so $(20 - x)^2 = u^2$. Then
$$
\frac{d}{dx}\left(\frac{3\sqrt{3}}{72}u^2\right)
= \frac{3\sqrt{3}}{72} \cdot 2u \cdot \frac{du}{dx}.
$$
Since $u = 20 - x$, we have $\frac{du}{dx} = -1$. Hence,
$$
\frac{d}{dx}\left(\frac{3\sqrt{3}}{72}u^2\right)
= \frac{3\sqrt{3}}{72} \cdot 2(20 - x)(-1)
= -\frac{3\sqrt{3}}{36}(20 - x).
$$
Therefore,
$$
A'(x) = \frac{x}{8} - \frac{3\sqrt{3}}{36}(20 - x).
$$
Set $A'(x) = 0$:
$$
\frac{x}{8} = \frac{3\sqrt{3}}{36}(20 - x).
$$
6. Solve for $x$
Multiply both sides by 36 to clear the denominator on the right:
$$
36 \cdot \frac{x}{8}
= 36 \cdot \frac{3\sqrt{3}}{36}(20 - x),
$$
$$
\frac{36}{8}x
= 3\sqrt{3}(20 - x).
$$
This simplifies to
$$
\frac{36}{8}x = 3\sqrt{3}(20 - x).
$$
Or
$$
\frac{9}{2}x = 3\sqrt{3}(20 - x).
$$
$$
9x = 6\sqrt{3}(20 - x).
$$
$$
9x = 120\sqrt{3} - 6\sqrt{3}\,x.
$$
Rearranging terms, we get
$$
9x + 6\sqrt{3}\,x = 120\sqrt{3},
$$
$$
x(9 + 6\sqrt{3}) = 120\sqrt{3}.
$$
Hence
$$
x = \frac{120\sqrt{3}}{9 + 6\sqrt{3}}.
$$
We can rationalize the denominator if needed, but the main quantity we want is the side of the hexagon.
7. Find the Side of the Regular Hexagon
The entire perimeter for the hexagon is $20 - x$, so each side of the hexagon is
$$
\frac{20 - x}{6}.
$$
Using
$x = \frac{120\sqrt{3}}{9 + 6\sqrt{3}},
$
we get
$$
20 - x = 20 - \frac{120\sqrt{3}}{9 + 6\sqrt{3}}.
$$
Upon rationalizing or working algebraically, this leads to
$$
\frac{20 - x}{6} = \frac{10}{3 + 2\sqrt{3}}.
$$
This is precisely the required side length of the hexagon.
8. Conclusion
Therefore, the length of each side of the regular hexagon that minimizes the combined area is
$$
\boxed{\frac{10}{3 + 2\sqrt{3}}}.
$$
