A wire of length 20 m is to be cut into two | vedclass

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(D) Let the wire be cut into two pieces of length $x$ and $20-x$.The side of the square is $s_1 = \frac{x}{4}$,so the area of the square is $A_1 = \le

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$\frac{10}{3+2 \sqrt{3}}$

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(D) Let the wire be cut into two pieces of length $x$ and $20-x$.The side of the square is $s_1 = \frac{x}{4}$,so the area of the square is $A_1 = \left(\frac{x}{4}ight)^2 = \frac{x^2}{16}$.The side of the regular hexagon is $s_2 = \frac{20-x}{6}$,so the area of the regular hexagon is $A_2 = 6 	imes \frac{\sqrt{3}}{4} \left(\frac{20-x}{6}ight)^2 = \frac{3\sqrt{3}}{2} \frac{(20-x)^2}{36} = \frac{\sqrt{3}}{24} (20-x)^2$.The total area is $A(x) = \frac{x^2}{16} + \frac{\sqrt{3}}{24} (20-x)^2$.To find the minimum area,we differentiate $A(x)$ with respect to $x$ and set it to $0$:$A'(x) = \frac{2x}{16} + \frac{\sqrt{3}}{24} 	imes 2(20-x)(-1) = \frac{x}{8} - \frac{\sqrt{3}}{12}(20-x) = 0$.Multiplying by $24$ gives $3x - 2\sqrt{3}(20-x) = 0$,which simplifies to $3x - 40\sqrt{3} + 2\sqrt{3}x = 0$.Thus,$x(3 + 2\sqrt{3}) = 40\sqrt{3}$,so $x = \frac{40\sqrt{3}}{3 + 2\sqrt{3}}$.The side of the hexagon is $s_2 = \frac{20-x}{6} = \frac{1}{6} \left( 20 - \frac{40\sqrt{3}}{3 + 2\sqrt{3}} ight)$.$s_2 = \frac{1}{6} \left( \frac{60 + 40\sqrt{3} - 40\sqrt{3}}{3 + 2\sqrt{3}} ight) = \frac{1}{6} \left( \frac{60}{3 + 2\sqrt{3}} ight) = \frac{10}{3 + 2\sqrt{3}}$.

$A$ wire of length $20 \ m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in $meters$) of the hexagon,so that the combined area of the square and the hexagon is minimum,is:

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