A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then:

Length of wire = 2 units
Length of side of square = x units
Radius of circle = r units
Total area = x² + πr²
Also, 4x + 2πr = 2
So, x = (1 - πr)/2
Total area = ((1 - πr)/2)² + πr²
dA/dr = 2(1/4)(1 - πr)(-π) + 2πr = 0 for minimum.
⇒ -π/2(1 - πr) + 2πr = 0
⇒ -π/2 + π²/2r + 2πr = 0
⇒ π²/2r + 2πr = π/2
⇒ r(π² + 4π) = π
⇒ r = 1/(π + 4)
x = (1 - πr)/2 = (1 - π/(π + 4))/2 = (4 + π - π)/(2(4 + π)) = 4/(2(4 + π)) = 2/(4 + π)
Therefore, 2x = 4/(4 + π)
2x = 4/(4+π)
And r = 1/(4 + π)
Then 2x = 4r
2x = (π + 4)r