A wire of length 20 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 ( l 1 ) : l 2 is equal to:

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Solution

Q.
A wire of length 20 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 $(π l_{1}) : l_{2}$ is equal to: [2023]

1 6 : 1

2 3 : 1

3 4 : 1

4 1 : 6

Ans.
(1)

Let $x$ be the side of the square and $r =$ radius of the circle.

Now, $l_{1} = 4 x and l_{2} = 2 π r .$ Then $4 x + 2 π r = l (let) ...(i)$

Also, $A_{1} = x^{2} and A_{2} = π r^{2}$

Let $A = 2 A_{1} + 3 A_{2} = 2 x^{2} + 3 π r^{2}$

$= 2 x^{2} + 3 π {(\frac{l - 4 x}{2 π})}^{2} = 2 x^{2} + \frac{3}{4 π} {(l - 4 x)}^{2} [From (i)]$

Now, $\frac{d A}{d x} = 4 x + \frac{3}{2 π} (l - 4 x) (- 4)$

For minima, $\frac{d A}{d x} = 0;$ $4 x + \frac{3}{2 π} (l - 4 x) \times (- 4) = 0$

$\Rightarrow x = \frac{6 l}{4 π + 24}$

Now, $l_{1} = 4 x = \frac{6 l}{π + 6} and l_{2} = 2 π r = l - 4 x$

$\Rightarrow l_{2} = l - \frac{6 l}{π + 6} = \frac{π l}{π + 6}$

Now,
$\frac{π l_{1}}{l_{2}} = \frac{π (\frac{6 l}{π + 6})}{(\frac{π l}{π + 6})} = 6 : 1$

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