Let the length of the side of the equare is a

Area of the square = a²

Let radius of the circle  = r

area of the circle = πr²

Now, sum of area A = a² + πr² ..........1

and 4a + 2πr = 1 ..............2                    {circumference of circle and square}

Differentiate w.r.t a of equation 1 and 2, we get

dA/da = 2a + 2πr (dr/da)  .........3

4 + 2π * (dr/da) = 0  ........4

=> dr/da = -4/2π 

Put this value in equation 3, we get

      dA/da = 2a + 2πr * (-4/2π)

=> dA/da = 2a - 4r

For maixima and minima, dA/da = 0

=> 2a - 4r = 0

=> 4r = 2a

=> r = 2a/4

=> r = a/2

Again, d² A/da² = 2 - 4 * (dr/da)

=> d² A/da² = 2 - 4 * (4/2π)

=> d² A/da² = 2 + 8/π > 0

So, the sum of the areas of the circle and the square is the least, if the radius of the circle is half the side of the square.