A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Given that length of the wire is 28 m .
Let x be the length of the wire that forms the square, so, the length of second wire will be 28 − x m .
As the perimeter of the square is 4 × side , so,
x = 4 × side side = x 4
As the circumference of the circle is 2 π × r , so,
28 − x = 2 π r r = 28 − x 2 π
The total area of the square and circle will be,
A = ( x 4 ) 2 + π ( 28 − x 2 π ) 2 = x 2 16 + ( 28 − x ) 2 4 π
Differentiate the area with respect to x,
A ′ = 2 x 16 + 2 4 π ( 28 − x ) ( − 1 ) = x 8 − 1 2 π ( 28 − x ) (1)
Put A ′ = 0 ,
x 8 − 1 2 π ( 28 − x ) = 0 x π − 4 ( 28 − x ) 8 π = 0 x ( π + 4 ) − 112 = 0 x = 112 π + 4
Differentiate equation (1) with respect to x,
A ″ = 1 8 + 1 2 π > 0
This shows that A ″ is positive, so, the value of x = 112 π + 4 gives the minimum combined area of the square and the circle.
As x = 112 π + 4 , so the other part will be,
28 − x = 28 − 112 π + 4 = 28 π π + 4
Therefore, the given combined area is minimum when the length of wire of one part is 112 π + 4 cm and the length of second part is 28 π π + 4 cm.
Given that length of the wire is
Let
As the perimeter of the square is
As the circumference of the circle is
The total area of the square and circle will be,
Differentiate the area with respect to x,
Put
Differentiate equation (1) with respect to x,
This shows that
As
Therefore, the given combined area is minimum when the length of wire of one part is
