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A wire of length 20 m is to be cut into two pieces. A piece of length $l_1$ is bent to make a square of area
$A_1$ and the other piece of length $l_2$ is made into a circle of area $A_2$ . If $2A1 + 3A2$ is minimum then
$(\pi l_1)$ : $l_2$ is equal to :

Option: 1
1:6

Option: 2
6: 1

Option: 3
3: 1

Option: 4
4: 1

Answers (1)

best_answer

Total length of wire = 20 m

area of square $(A_1)$ = $\left(\frac{\ell_1}{4}\right)^2$

area of circle $(A_2)$ = $\left(\frac{\ell_2}{2\pi}\right)^2$

Let S = $2A_1 + 3A_2$

$\mathrm{S}=\frac{\ell_1^2}{8}+\frac{3 \ell_2^2}{4 \pi}$

$\because \ell_1+\ell_2=20$ then

$\begin{aligned} & 1+\frac{\mathrm{d} \ell_2}{\mathrm{~d} \ell_1}=0 \\ & \frac{\mathrm{d} \ell_2}{\mathrm{~d} \ell_1}=-1 \end{aligned}$

$\begin{aligned} & \frac{\mathrm{ds}}{\mathrm{d} \ell_1}=\frac{\ell_1}{4}+\frac{6 \ell_2}{4 \pi} \cdot \frac{\mathrm{d} \ell_2}{\mathrm{~d} \ell_1}=0 \\ & =\frac{\ell_1}{4}=\frac{6 \ell_2}{4 \pi} \\ & =\frac{\pi \ell_1}{\ell_2}=\frac{6}{1} \end{aligned}$

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