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Q.

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 $l_{2}$ is 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:

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a

6 :1

b

4 :1

c

3:1

d

1: 6

answer is A.

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Detailed Solution

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$4 x = l_{1} l_{2} = 2 π r = 20 - l_{1} let S = 2 A_{1} + 3 A_{2} = 2 x^{2} + 3 π r^{2}$
$= 2 {(\frac{l_{1}}{4})}^{2} + 3 π {(\frac{20 - l_{1}}{2 π})}^{2}$
for min of S is $\frac{d s}{d l_{1}} = 0 \Rightarrow \frac{l_{1}}{4} + \frac{3}{4 π} 2 (20 - l_{1}) (- 1) = 0$
$\Rightarrow l_{1} = \frac{120}{π + 6}; l_{2} = \frac{20 π}{π + 6} ∴ π l_{1} : l_{2} = 6 : 1$

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