Assertion (A): A wire of length 20m 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 $l_{2}$ in made into a circle of area $A_{2}$ . If $2 A_{1} + 3 A_{2}$ is minimum then $(π l_{1}) : l_{2}$ is equal to 6:1
Reason (R): If the length of the side of square is $a$ then its area is $a^{2}$ and if the radius of a circle is $r$ then its area is $π r^{2}$ .

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If Assertion (A) is correct but Reason (R) is incorrect

If Assertion (A) is incorrect but Reason (R) is correct

If both Assertion (A) and Reason (R) are correct and Reason (R) is the correct explanation of the Assertion (A)

If both Assertion (A) and Reason (R) are correct but Reason (R) is not correct explanation of Assertion (A)

answer is A.

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Detailed Solution

Since, given $l_{1} + l_{2} = 20 \Rightarrow \frac{d l_{2}}{d l_{1}} = - 1$
Now, $A_{1} = {(\frac{l_{1}}{4})}^{2}$ and $A_{2} = π {(\frac{l_{2}}{2 π})}^{2}$
Let $S = 2 A_{1} + 3 A_{2} = \frac{l_{1}^{2}}{8} + \frac{3 l_{2}^{2}}{4 π}$
For max or min $\frac{d s}{d l_{1}} = 0 \Rightarrow \frac{2 l 1}{8} + \frac{6 l_{2}}{4 π} . \frac{d l_{2}}{d l_{1}} = 0$