A wire of length 22 ; m is to be cut into two | vedclass

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(B) Let the length of the wire used for the equilateral triangle be $x \; m$. Then the length of the wire used for the square is $(22-x) \; m$.For the

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$\frac{66}{9+4 \sqrt{3}}$

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(B) Let the length of the wire used for the equilateral triangle be $x \; m$. Then the length of the wire used for the square is $(22-x) \; m$.For the equilateral triangle,the side length $a = \frac{x}{3}$. The area is $A_1 = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left(\frac{x}{3}\right)^2 = \frac{\sqrt{3} x^2}{36}$.For the square,the side length $b = \frac{22-x}{4}$. The area is $A_2 = b^2 = \left(\frac{22-x}{4}\right)^2 = \frac{(22-x)^2}{16}$.The total area $A = A_1 + A_2 = \frac{\sqrt{3} x^2}{36} + \frac{(22-x)^2}{16}$.To find the minimum area,differentiate $A$ with respect to $x$ and set it to zero:$\frac{dA}{dx} = \frac{2 \sqrt{3} x}{36} + \frac{2(22-x)(-1)}{16} = 0$$\frac{\sqrt{3} x}{18} - \frac{22-x}{8} = 0$$\frac{\sqrt{3} x}{18} = \frac{22-x}{8}$$8 \sqrt{3} x = 18(22-x) = 396 - 18x$$x(8 \sqrt{3} + 18) = 396$$x = \frac{396}{18 + 8 \sqrt{3}} = \frac{198}{9 + 4 \sqrt{3}}$.The side of the equilateral triangle is $a = \frac{x}{3} = \frac{198}{3(9 + 4 \sqrt{3})} = \frac{66}{9 + 4 \sqrt{3}} \; m$.

$A$ wire of length $22 \; m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then,the length of the side of the equilateral triangle,so that the combined area of the square and the equilateral triangle is minimum,is

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