JEE Main 2023 — Application of Derivatives Question with Solution

The length of wire is $20 m$.

According to the question:

${\ell }_1+{\ell }_2=20$

$\Rightarrow \frac{\mathrm{d}{\ell }_2}{ \mathrm{d}{\ell }_1}=-1$

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

$4x={\ell }_1$

$\Rightarrow x=\frac{{\ell }_1}{4} ...\left(1\right)$

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

 $2\mathrm{\pi }y={\mathrm{\ell }}_2$

$\Rightarrow y=\frac{{\mathrm{\ell }}_2}{2\mathrm{\pi }} ...\left(2\right)$

Hence,

$A_1={\left(\frac{{\ell }_1}{4}\right)}^2$ and $A_2=\pi {\left(\frac{{\ell }_2}{2\pi }\right)}^2$

Let $S=2A_1+3A_2$

$\Rightarrow S=\frac{({\ell }_1{)}^2}{8}+\frac{3({\ell }_2{)}^2}{4\pi }$

$\Rightarrow \frac{dS}{d{\ell }_1}=\frac{2{\ell }_1}{8}+\frac{6{\ell }_2}{4\pi }·\left(\frac{d{\ell }_2}{ d{\ell }_1}\right)$

For $S$ to be minimum,

$\frac{ds}{\mathrm{d}{\ell }_1}=0$

$\Rightarrow \frac{2{\ell }_1}{8}+\frac{6{\ell }_2}{4\pi }·\left(\frac{d{\ell }_2}{ d{\ell }_1}\right)=0$

$\Rightarrow \frac{2{\ell }_1}{8}-\frac{6{\ell }_2}{4\pi }=0 \left(\because \frac{\mathrm{d}{\ell }_2}{ \mathrm{d}{\ell }_1}=-1\right)$

$\Rightarrow \frac{{\ell }_1}{4}=\frac{6{\ell }_2}{4\pi }$

$\Rightarrow \frac{\pi {\ell }_1}{{\ell }_2}=6$