A wire of length 20 m is to be cut into two pieces. A piece of length is bent to make a square of area and the other piece of length is made into a circle of area . If is minimum then is equal to

Question

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

Smart Achievers · Accepted Answer

(1)

$Since, given ℓ_{1} + ℓ_{2} = 20 \Rightarrow \frac{d ℓ_{2}}{d ℓ_{1}} = - 1$

$Now, A_{1} = {(\frac{ℓ_{1}}{4})}^{2} and A_{2} = π {(\frac{ℓ_{2}}{2 π})}^{2}$

$Let S = 2 A_{1} + 3 A_{2} = \frac{ℓ_{1}^{2}}{8} + \frac{3 ℓ_{2}^{2}}{4 π}$

$For max or min$

$\frac{d S}{d ℓ_{1}} = 0$

$\Rightarrow \frac{2 ℓ_{1}}{8} + \frac{6 ℓ_{2}}{4 π} \cdot \frac{d ℓ_{2}}{d ℓ_{1}} = 0$

$\Rightarrow \frac{ℓ_{1}}{4} = \frac{6 ℓ_{2}}{4 π}$

$\Rightarrow \frac{π ℓ_{1}}{ℓ_{2}} = 6$

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

1 6 : 1

2 3 : 1

3 1 : 6

4 4 : 1

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