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

A wire of length 22 m is cut into two pieces. One piece forms a square and the other an equilateral triangle.

First, let the side of the equilateral triangle be $$a$$, so its perimeter is $$3a$$. The remaining wire for the square is $$22 - 3a$$, and hence the side of the square is $$\frac{22 - 3a}{4}$$.

Next, the area of the square is $$\left(\frac{22 - 3a}{4}\right)^2 = \frac{(22-3a)^2}{16}$$, while the area of the equilateral triangle is $$\frac{\sqrt{3}}{4}a^2$$. Therefore, the total area as a function of $$a$$ is $$A(a) = \frac{(22-3a)^2}{16} + \frac{\sqrt{3}}{4}a^2$$.

To minimize this area, we differentiate to obtain $$A'(a) = \frac{2(22-3a)(-3)}{16} + \frac{\sqrt{3}}{2}a = \frac{-3(22-3a)}{8} + \frac{\sqrt{3}}{2}a$$. Setting $$A'(a) = 0$$ gives $$\frac{3(22-3a)}{8} = \frac{\sqrt{3}}{2}a$$, which leads to $$3(22-3a) = 4\sqrt{3}\,a$$ and hence $$66 - 9a = 4\sqrt{3}\,a$$. Combining terms yields $$66 = a(9 + 4\sqrt{3})$$, so $$a = \frac{66}{9 + 4\sqrt{3}}$$.

Therefore, the length of the side of the equilateral triangle that minimizes the total area is $$\frac{66}{9+4\sqrt{3}}$$. The answer is Option B.