Question

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

Answer 1

Given that length of wire = 50 m
let the length of one piece for the shape of the square be x m
∴ Length of other pieces for the shape of circle = (50 - x) m
Now, perimeter of square = 4a = x

⇒ a = `(x)/(4)`

and circumference of circle = 2πr = 50 - x

⇒ r =  `(50 - x)/(2π)`

Combined area = a² + π r²

= `x^2/(16) + π  ( (50 - x)/(2π))^2`

= `x^2/(16) + π   (50 - x)^2/(4π^2)`

A = `x^2/(16) +   (50 - x)^2/(4π)`

Differentiate w.r.t. x, we have 

`(d"A")/(dx) = (2x)/(16) + (2 (50 - x) (-1))/(4π)`

`(d"A")/(dx) = (x)/(8) +  (( x- 50 ))/(2π)`

= `(πx + 4x - 200)/( 8π)`

= `(x ( 4 + π) - 200)/ (8π)`

For maximum or minimum put `(d"A")/(dx)` = 0

∴ x (4 + π) - 200 = 0

x = `(200)/(4 + π)`

`(d^2"A")/(dx^2)` is positive ( >0)

For x = `(200)/(4 + π)`, because

`(d^2"A")/(dx^2) = (4 + π)/(8π)` ( independent of x)

So, length of wire for square shape is x = `(200)/(4 + π)` m 

and length of wire for circle shape = 50 - x 

= 50 - `(200)/(4 + π)`

= `(50π)/(4 + π)`m