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

Answer:

Therefore the circumference of the circle is $=\frac{20\pi}{4+\pi}$

Step-by-step explanation:

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

$\Rightarrow s=\frac{10-\pi r}{2}$

The area of the circle is = πr²

The area of the square is = s²

A represent the total area of the square and circle.

A=πr²+s²

Putting the value of s

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

$\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2$

$\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}$

$\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25$

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

Differentiating with respect to r

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

Again differentiating with respect to r

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

For maximum or minimum

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

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

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

$\Rightarrow r=\frac{10}{4+\pi}$

$\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0$

Therefore at $r=\frac{10}{4+\pi}$  , A is minimum.

Therefore the circumference of the circle is

$=2 \pi \frac{10}{4+\pi}$

$=\frac{20\pi}{4+\pi}$