Let the length of pieces of wire are x & 28-x

consider the perimeter of square(side=a) =x & circumference of circle(radius=r)=28-x

so, 4a=x

a=x/4

and, 2*pi*r=(28-x)

r=(28-x)/2*pi

combine area (A) = area of square+area of circle

A = [{(x/4)^2} + {pi*(28-x/2*pi)^2}]

when we solved this equestion ,we get,

A = x^2/16 + (28^2)/(4*pi) + x^2/(4*pi) - 14x/pi --------------------------eqn(1)

A will be minimum ,when dA/dx=0

dA/dx = 2x/16 + 0 + 2x/4pi - 14/pi

0 = x/8 + x/2pi - 14/pi

14/pi =x/2(1/4+1/pi)

(14*2*4pi)/pi*(pi+4) = x

x = 112/(pi+4)

x= 15.68 metre when,pi=22/7

put the value of x in eqn(1)

A (minimum) = 38.12 metre^2