Question

A thin rope 200 inches length is going to be cut into three pieces to form one circle and two equal squares. Which of the following represents the `r` of the circle, if the rope was finally cut in such a way that the total area of the shapes is minimized?

Answer 1

`r=100/(8+pi)`

Explanation:

Rope is 200 inches long.
Let Length of Rope `L=200`
Now Rope is going to be cut into three pieces to form one circle and two equal squares.
Let radius of circle `=r`
Side of Square `=a`
Perimeter of Circle`=2pir`
perimeter of Square `=4a`
Note- Both squares are same.

`"Rope"="circle"+"square"+"square"`
`L=2pir+4a+4a`
`=>2pir+8a=200`
`=>pir+4a=100`
`=>4a=100-pir`
`=>a=(100-pir)/4`

in the question it mentioned that the total area of shapes are minimum.
We are going to use Differentiation method to Find extreme point and then find maxima and minima at all extreme points.

`Area=A="area of circle"+"area of square"+"area of square"`
`Area=A=pir^2+a^2+a^2=pir^2+2a^2`
Put value of `a` in the above equation, we get
`A=pir^2+2((100-pir)/4)^2=pir^2+2/16(100-pir)^2`
`A=pir^2+1/8(100-pir)^2`

Differentiate with respect to r
`(dA)/(dr)=2pir+2/8(100-pir)(-pi)`
`(dA)/(dr)=2pir-pi/4(100-pir)=pi(2r-1/4(100-pir))` (equation 1)

For Extreme points `(dA)/(dr)=0`
`=>pi(2r-1/4(100-pir))=0`
we know that `pi` is not zero.
Hence
`(2r-1/4(100-pir))=0`
Multiply equation with 4
`=>8r-100+pir=0`
`=>r(8+pi)=100`
`=>r=100/(8+pi)`

Now we have to check whether value of r gives us maximum area or minimum area.

Now again differentiate the (equation 1) with respect to r.
`(d^2A)/(dr^2)=d/(dr)(pi(2r-1/4(100-pir)))`
`(d^2A)/(dr^2)=(pi(2-1/4(-pi)))=pi(2+pi/4)="a Positive number" >0`

We have only one value of `r` which will give us minimum area.

Hence `r=100/(8+pi)`