Question

A wire of length 28m is to be cut into 2 pieces one of the piece is to be made into a square and other into a circle. what shout be the length of the 2 pieces so that the it combine the area of the square and the circle is minimum?

Answer 1

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