Question

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 $2 A _1+3 A _2$ is minimum then $\left(\pi \ell_1\right): \ell_2$ is equal to:

• A. $6: 1$
• B. $3: 1$
• C. $4: 1$
• D. $1: 6$

Answer 1

Answer: (A)

$ \ell_1+\ell_2=20 \Rightarrow \frac{ d \ell_2}{ d \ell_1}=-1$
$ A _1=\left(\frac{\ell_1}{4}\right)^2 \text { and } A _2=\pi\left(\frac{\ell_2}{2 \pi}\right)^2$
Let $S =2 A _1+3 A _2=\frac{\ell_1^2}{8}+\frac{3 \ell_2^2}{4 \pi}$
$ \frac{ ds }{ d \ell}=0 \Rightarrow \frac{2 \ell_1}{8}+\frac{6 \ell_2}{4 \pi} \cdot \frac{ d \ell_2}{ d \ell_1}=0$
$\Rightarrow \frac{\ell_1}{4}=\frac{6 \ell_2}{4 \pi} \Rightarrow \frac{\pi \ell_1}{\ell_2}=6$