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 $2A_{1}+3A_{2}$ is minimum then $(\pi \ell_{1}): \ell_{2}$ is equal to:

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Application of Derivatives

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:

Answer: (A)

With

ℓ_{1} + ℓ_{2} = 20

, the square of perimeter

ℓ_{1}

has area

A_{1} = \frac{ℓ _{1}^{2}}{16}

, and the circle of circumference

ℓ_{2}

has area

A_{2} = \frac{ℓ _{2}^{2}}{4 π}

. Then

2 A_{1} + 3 A_{2} = \frac{ℓ _{1}^{2}}{8} + \frac{3 ℓ _{2}^{2}}{4 π}

. Substituting

ℓ_{2} = 20 - ℓ_{1}

and setting the derivative with respect to

ℓ_{1}

to zero gives

\frac{ℓ _{1}}{4} = \frac{3 ℓ _{2}}{2 π}

, that is

\frac{π ℓ _{1}}{ℓ _{2}} = 6

. Hence

(π ℓ_{1}) : ℓ_{2} = 6 : 1

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option