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A wire of length 22 mt 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 square and equilateral triangle is minimum is

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

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

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

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

answer is B.

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Detailed Solution

Side of the $Δ^{l e}$ be x and side of the square be y then $3 x + 4 y = 22$
Sum of Areas A = $y^{2} + \frac{\sqrt{3}}{4} x^{2} = {(\frac{22 - 3 x}{4})}^{2} + \frac{\sqrt{3}}{4} x^{2}$ $\frac{d A}{d x} = 0 \Rightarrow x = \frac{66}{9 + 4 \sqrt{3}}$

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