A wire of length [20 ~m] is to be cut into two pieces.

Question

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 :

Accepted Answer

$ \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} $

$\Rightarrow \frac{\pi \ell_{1}}{\ell_{2}}=6$

Solutions · 1