Question

You have a wire that is 20 cm long. You wish to cut it into two pieces. One-piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?

Answer 1

The circumference of the circle is  $C =\dfrac{20\pi}{4+\pi}$  when A is a minimum.

What is circumference?

The circumference of the circle is the length of the curve traced out from the fixed point which is equidistant from the locus of the curve.

Let the side of the square be s

and the radius of the circle be r

The perimeter of the square is = 4s

The circumference of the circle is =2πr

Given that the length of the wire is 20 cm.

According to the problem,

4s + 2πr =20

⇒2s+πr =10

$s=\dfrac{10-\pi r}{2}$

The area of the circle is = πr²

The area of the square is = s²

A represents the total area of the square and circle.

A=πr²+s²

Putting the value of s

$A =\pi r^2 +(\dfrac{10-\pi r}{2})^2$

$A =\pi r^2 +(\dfrac{10}{2})^2-2\times \dfrac{10}{2}\times \dfrac{\pi r}{2}+(\dfrac{\pi r}{2})^2$

$A = { \pi r^2}(\dfrac{4+\pi}{4})-5\pi r+25$

For maximum or minimum $\dfrac{dA}{dr}=0$

Differentiating with respect to r

$\dfrac{dA}{dr}=\dfrac{2\pi r(4+\pi )}{4}-5 \pi$

Again differentiating with respect to r

$\dfrac{d^2A}{dr^2}=\dfrac{2\pi (4+\pi )}{4}$    > 0

For maximum or minimum

$\dfrac{dA}{dr}=0$

$\dfrac{2\pi r(4+\pi )}{4}-5 \pi=0$

$r = \dfrac{10\pi}{\pi (4+\pi)}$

$r=\dfrac{10}{4+\pi}$

Therefore the circumference of the circle is

C = 2 π r

$C = \dfrac{20 \pi }{4+\pi }$

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