A wire of length 20 m is to be cut into two pieces. A piece of length $$\ell_1$$ is bent to make a square of area $$A_1$$ and the other piece of length $$\ell_2$$ is made into a circle of area $$A_2$$. If $$2A_1 + 3A_2$$ is minimum then $$\pi\ell_1 : \ell_2$$ is equal to:

A wire of length 20 m is cut into two pieces. Piece of length $$\ell_1$$ forms a square, and piece of length $$\ell_2$$ forms a circle, where $$\ell_1 + \ell_2 = 20$$.

Area of the square: $$A_1 = \left(\frac{\ell_1}{4}\right)^2 = \frac{\ell_1^2}{16}$$

Area of the circle: Circumference = $$\ell_2 = 2\pi r$$, so $$r = \frac{\ell_2}{2\pi}$$ and $$A_2 = \pi r^2 = \frac{\ell_2^2}{4\pi}$$

Minimize: $$f = 2A_1 + 3A_2 = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$$

Substituting $$\ell_2 = 20 - \ell_1$$ and differentiating:

$$ f'(\ell_1) = \frac{\ell_1}{4} - \frac{3(20 - \ell_1)}{2\pi} = 0 $$

$$ \frac{\pi\ell_1}{4} = \frac{3(20 - \ell_1)}{2} $$

$$ \frac{\pi\ell_1}{2} = 3(20 - \ell_1) = 60 - 3\ell_1 $$

$$ \pi\ell_1 + 6\ell_1 = 120 $$

$$ \ell_1 = \frac{120}{\pi + 6}, \quad \ell_2 = 20 - \frac{120}{\pi + 6} = \frac{20\pi}{\pi + 6} $$

The ratio:

$$ \frac{\pi\ell_1}{\ell_2} = \frac{\pi \times \frac{120}{\pi+6}}{\frac{20\pi}{\pi+6}} = \frac{120}{20} = 6 $$

Therefore, $$\pi\ell_1 : \ell_2 = 6 : 1$$.