A piece of wire 20 cm long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. Atwhat length should the wire be cut to minimize the total area enclosed?...

A piece of wire 20 cm long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. At
what length should the wire be cut to minimize the total area enclosed?

kindly help me in solving this problem?

wanderista | Student

The pieces of wire should be 11.20cm (2 d. p.) and 8.80cm (2 d. p).

hinaje | Student

thank you so very much...............

aruv | Student

Let length of one piece of wire be x, which bent into circle.

Thus circumference of cirle be x. Let radius of the circle be r, so

`2pir=x`

`r=x/(2pi)`

Thus area of the circle be

`A_c=pi(x/(2pi))^2=x^2/(4pi)`              (i)

Perimeter of the square be

P=20-x

Thus side of the square be (20-x)/4

Thus area of the square be

`A_s=(5-x/4)^2`

Total enclosed area be

`A=A_c+A_s`

`A=x^2/(4pi)+(5-x/4)^2`

`(dA)/(dx)=x/(2pi)-2(5-x/4)(1/4)`

`=x/(2pi)-5/2+x/8`

For maximum / minimum, we have

`((4+pi)x)/(8pi)-5/2=0`

`x=(5/2)((8pi)/(4+pi))`

`x=(20pi)/(4+pi)`                (ii)

`(d^2A)/(dx^2)=1/(2pi)+1/8>0AAx.`

`Thus`

x=8.798 cm.

Thus pieces of wire should be  11.202 cm. and 8.798 cm.