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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$\frac{22}{9 + 4 \sqrt{3}}$

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

$\frac{22}{4 + 9 \sqrt{3}}$

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

answer is B.

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Detailed Solution

$\begin{array}{l} Let a = length of side of the triangle \\ b = length of side of the square \\ 3 a = x and 4 b = 22 - x \\ \begin{array}{l} Area of triangle + Area of square = A = \frac{\sqrt{3}}{4} a^{2} + b^{2} \\ = \frac{\sqrt{3}}{4} (\frac{x^{2}}{9}) + \frac{(22 - x)^{2}}{16} \\ \frac{dA}{dx} = 0 \Rightarrow x (\frac{\sqrt{3}}{2 \times 9} + \frac{1}{8}) - \frac{22}{8} = 0 \\ \Rightarrow x (\frac{4 \sqrt{3} + 9}{36}) = \frac{11}{2} \\ a = x / 3 \\ a = (\frac{11 / 2}{\frac{4 \sqrt{3} + 9}{36}}) (\frac{1}{3}) = \frac{66}{4 \sqrt{3} + 9} \end{array} \end{array}$