A wire of length 20 m is to be cut into two | vedclass

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(A) Given the total length of the wire is $\ell_1 + \ell_2 = 20$. The side of the square formed by $\ell_1$ is $s = \frac{\ell_1}{4}$,so the area $A_1

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$6:1$

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(A) Given the total length of the wire is $\ell_1 + \ell_2 = 20$. The side of the square formed by $\ell_1$ is $s = \frac{\ell_1}{4}$,so the area $A_1 = (\frac{\ell_1}{4})^2 = \frac{\ell_1^2}{16}$. The radius of the circle formed by $\ell_2$ is $r = \frac{\ell_2}{2\pi}$,so the area $A_2 = \pi(\frac{\ell_2}{2\pi})^2 = \frac{\ell_2^2}{4\pi}$. Let $S = 2A_1 + 3A_2 = 2(\frac{\ell_1^2}{16}) + 3(\frac{\ell_2^2}{4\pi}) = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$. Substitute $\ell_2 = 20 - \ell_1$,then $S = \frac{\ell_1^2}{8} + \frac{3(20 - \ell_1)^2}{4\pi}$. To find the minimum,differentiate with respect to $\ell_1$ and set to zero: $\frac{dS}{d\ell_1} = \frac{2\ell_1}{8} + \frac{6(20 - \ell_1)(-1)}{4\pi} = 0$. $\frac{\ell_1}{4} = \frac{6(20 - \ell_1)}{4\pi} = \frac{6\ell_2}{4\pi}$. $\frac{\pi \ell_1}{4} = \frac{6\ell_2}{4} \Rightarrow \frac{\pi \ell_1}{\ell_2} = 6$.

$A$ 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:

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