JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 8)
A wire of length $$20 \mathrm{~m}$$ 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 $$2 A_{1}+3 A_{2}$$ is minimum then $$\left(\pi l_{1}\right): l_{2}$$ is equal to :
6 : 1
1 : 6
4 : 1
3 : 1
Explanation
$ \ell_{1}+\ell_{2}=20 \Rightarrow \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=-1$
$\mathrm{A}_{1}=\left(\frac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\frac{\ell_{2}}{2 \pi}\right)^{2}$
Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\frac{\ell_{1}^{2}}{8}+\frac{3 \ell_{2}^{2}}{4 \pi}$
$\frac{\mathrm{ds}}{\mathrm{d} \ell_1}=0 \Rightarrow \frac{2 \ell_{1}}{8}+\frac{6 \ell_{2}}{4 \pi} \cdot \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=0$
$\Rightarrow \frac{\ell_{1}}{4}=\frac{6 \ell_{2}}{4 \pi} $
$\Rightarrow \frac{\pi \ell_{1}}{\ell_{2}}=6$
$\mathrm{A}_{1}=\left(\frac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\frac{\ell_{2}}{2 \pi}\right)^{2}$
Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\frac{\ell_{1}^{2}}{8}+\frac{3 \ell_{2}^{2}}{4 \pi}$
$\frac{\mathrm{ds}}{\mathrm{d} \ell_1}=0 \Rightarrow \frac{2 \ell_{1}}{8}+\frac{6 \ell_{2}}{4 \pi} \cdot \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=0$
$\Rightarrow \frac{\ell_{1}}{4}=\frac{6 \ell_{2}}{4 \pi} $
$\Rightarrow \frac{\pi \ell_{1}}{\ell_{2}}=6$
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