Application Of Derivatives Question 8
Question 8 - 31 January - Shift 1
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 $(\pi \ell_1): \ell_2$ is equal to:
(1) $6: 1$
(2) $3: 1$
(3) $1: 6$
(4) $4: 1$
Show Answer
Answer: (1)
Solution:
Formula: Minima of a function
$\ell_1+\ell_2=20 \Rightarrow \frac{d \ell_2}{d \ell_1}=-1$
$A_1=(\frac{\ell_1}{4})^{2}$ and $A_2=\pi(\frac{\ell_2}{2 \pi})^{2}$
Let $S=2 A_1+3 A_2=\frac{\ell_1^{2}}{8}+\frac{3 \ell_2^{2}}{4 \pi}$
$\frac{ds}{d \ell}=0 \Rightarrow \frac{2 \ell_1}{8}+\frac{6 \ell_2}{4 \pi} \cdot \frac{d \ell_2}{d \ell_1}=0$
$\Rightarrow \frac{\ell_1}{4}=\frac{6 \ell_2}{4 \pi} \Rightarrow \frac{\pi \ell_1}{\ell_2}=6$
