Question

A wire of length $20m$ 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 $2A_1+3A_2$ 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 $S=2A_1+3A_2=\frac{{\ell }_1^2}{8}+\frac{3{\ell }_2^2}{4\pi }$
$\frac{ds}{d\ell }=0\Rightarrow \frac{2{\ell }_1}{8}+\frac{6{\ell }_2}{4\pi }\cdot \frac{d{\ell }_2}{d{\ell }_1}=0$
$\Rightarrow \frac{{\ell }_1}{4}=\frac{6{\ell }_2}{4\pi }\Rightarrow \frac{\pi {\ell }_1}{{\ell }_2}=6$