A wire of length 20m is to be cut into two pieces. A piece of...

The length of wire is $20m$.

According to the question:

${ℓ}_{1} + {ℓ}_{2} = 20$

$⇒ \dfrac{d{ℓ}_{2}}{d{ℓ}_{1}} = - 1$

Let the side of the square be $x$, then

$4x = {ℓ}_{1}$

$⇒ x = \dfrac{{ℓ}_{1}}{4} . . . \left(1\right)$

Let the radius of the circle be $y$, then

$2\piy = {ℓ}_{2}$

$⇒ y = \dfrac{{ℓ}_{2}}{2\pi} . . . \left(2\right)$

Hence,

${A}_{1} = {\left(\dfrac{{ℓ}_{1}}{4}\right)}^{2}$ and ${A}_{2} = \pi{\left(\dfrac{{ℓ}_{2}}{2\pi}\right)}^{2}$

Let $S = 2{A}_{1} + 3{A}_{2}$

$⇒ S = \dfrac{( {ℓ}_{1}{)}^{2}}{8} + \dfrac{3 ( {ℓ}_{2}{)}^{2}}{4\pi}$

$⇒ \dfrac{dS}{d{ℓ}_{1}} = \dfrac{2{ℓ}_{1}}{8} + \dfrac{6{ℓ}_{2}}{4\pi} \cdot \left(\dfrac{d{ℓ}_{2}}{d{ℓ}_{1}}\right)$

For $S$ to be minimum,

$\dfrac{ds}{d{ℓ}_{1}} = 0$

$⇒ \dfrac{2{ℓ}_{1}}{8} + \dfrac{6{ℓ}_{2}}{4\pi} \cdot \left(\dfrac{d{ℓ}_{2}}{d{ℓ}_{1}}\right) = 0$

$⇒ \dfrac{2{ℓ}_{1}}{8} - \dfrac{6{ℓ}_{2}}{4\pi} = 0\left(∵ \dfrac{d{ℓ}_{2}}{d{ℓ}_{1}} = - 1\right)$

$⇒ \dfrac{{ℓ}_{1}}{4} = \dfrac{6{ℓ}_{2}}{4\pi}$

$⇒ \dfrac{\pi{ℓ}_{1}}{{ℓ}_{2}} = 6$