JEE Main 2023 — Mathematics Calculus

The length of wire is $20m$20m.

According to the question:

$${\ell }_{1}+{\ell }_{2}=20$$ℓ1​+ℓ2​=20

$$\Rightarrow \frac{d{\ell }_{2}}{d{\ell }_{1}}=-1$$⇒dℓ1​dℓ2​​=−1

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

$$4x={\ell }_{1}$$4x=ℓ1​

$$\Rightarrow x=\frac{{\ell }_{1}}{4}...(1)$$⇒x=4ℓ1​​...(1)

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

$$2\pi y={\ell }_{2}$$2πy=ℓ2​

$$\Rightarrow y=\frac{{\ell }_{2}}{2\pi }...(2)$$⇒y=2πℓ2​​...(2)

Hence,

${A}_{1}={(\frac{{\ell }_{1}}{4})}^{2}$A1​=(4ℓ1​​)2 and ${A}_{2}=\pi {(\frac{{\ell }_{2}}{2\pi })}^{2}$A2​=π(2πℓ2​​)2

Let $S=2{A}_{1}+3{A}_{2}$S=2A1​+3A2​

$$\Rightarrow S=\frac{({\ell }_{1}{)}^{2}}{8}+\frac{3({\ell }_{2}{)}^{2}}{4\pi }$$⇒S=8(ℓ1​)2​+4π3(ℓ2​)2​

$$\Rightarrow \frac{dS}{d{\ell }_{1}}=\frac{2{\ell }_{1}}{8}+\frac{6{\ell }_{2}}{4\pi }\cdot (\frac{d{\ell }_{2}}{d{\ell }_{1}})$$⇒dℓ1​dS​=82ℓ1​​+4π6ℓ2​​⋅(dℓ1​dℓ2​​)

For $S$S to be minimum,

$$\frac{ds}{d{\ell }_{1}}=0$$dℓ1​ds​=0

$$\Rightarrow \frac{2{\ell }_{1}}{8}+\frac{6{\ell }_{2}}{4\pi }\cdot (\frac{d{\ell }_{2}}{d{\ell }_{1}})=0$$⇒82ℓ1​​+4π6ℓ2​​⋅(dℓ1​dℓ2​​)=0

$$\Rightarrow \frac{2{\ell }_{1}}{8}-\frac{6{\ell }_{2}}{4\pi }=0(\because \frac{d{\ell }_{2}}{d{\ell }_{1}}=-1)$$⇒82ℓ1​​−4π6ℓ2​​=0(∵dℓ1​dℓ2​​=−1)

$$\Rightarrow \frac{{\ell }_{1}}{4}=\frac{6{\ell }_{2}}{4\pi }$$⇒4ℓ1​​=4π6ℓ2​​

$$\Rightarrow \frac{\pi {\ell }_{1}}{{\ell }_{2}}=6$$⇒ℓ2​πℓ1​​=6