A wire of length 50 m is to be cut into two pieces. One piece is bent in the shape of a square and the other in the shape of a circle. What should be the lengths of each piece so that the combined area of the two shapes is minimum?

Length for circle = 22 m, Length for square = 28 m

The correct answer is Option (2) → Length for circle = 22 m, Length for square = 28 m

Let r metres be the radius of the circle and d metres be a side of the square formed by the two pieces of the wire, then

$2πr+4d=50$   ...(i)

Let A sq. metres be the combined area of the circle and the square, then

$A=πr^2+d^2=πr^2+\left(\frac{50 - 2πr}{4}\right)^2$   (using (i))

Differentiating it w.r.t. r, we get

$\frac{dA}{dr}=2πr+\frac{1}{16}.2(50-2πr)(-2π)=2πr+\frac{1}{2}π^2r-\frac{25}{2}π$ and $\frac{d^2A}{dr^2}=2π+\frac{1}{2}π^2$.

Now $\frac{dA}{dr}=0⇒2πr+\frac{1}{2}π^2r-\frac{25}{2}π=0$

$⇒4r + πr=25⇒ \left(4+\frac{22}{7}\right) r=25⇒r=\frac{7}{2}$.

For $r=\frac{7}{2},\frac{d^2A}{dr^2}= 2π+\frac{1}{2}π^2>0$

⇒ A is minimum when $r =\frac{7}{2}$.

∴ The length of the piece of wire bent into the form of circle = $2πr = 2.\frac{22}{7}.\frac{7}{2}\,m=22\,m$ and the length of the piece of wire bent into the form of square = $50\, m - 22\, m = 28\, m$.