Mahesh Godavarti
1

For a given perimeter, a circle has greater area than a square. For a given, P the area of a square is P^2/16 , whereas that of a circle is P^2/(2 \pi) .


Given this, I would say allocate the entire 1 meter length to the circle to get an area of 1 / (2 \pi) m^2 .


However, let's do this formally.


Let l be the length of one side of the square and r be the radius of the circle. Then, we have 4 l + 2 \pi r = 1 \implies l = (1 - 2 \pi r)/4 .


Total area A = l^2 + \pi r^2 = (1 - 2 \pi r)^2/16 + \pi r^2 = \frac{1}{16} - \frac{\pi}{4} r + \frac{5 pi^2}{4} r^2 .


Now, A(r) is an upward facing parabola. So, it has a minimum. The maximums occur at the end of the domain at either r = 0 or at r = \frac{1}{2 \pi} . Therefore, allocating the entire length to the circle is the way to go.


Another interesting thing is the minimum. If we wanted to achieve the minimum, we could say allocate the entire length to the square. Let's see if that is true. The minimum occurs at r = -\frac{\pi}{4} \frac{2}{5 \pi^2} = -\frac{1}{10 \pi} (using the fact that the vertex for y = ax^2 + bx + c occurs at x = - \frac{b}{2a} ) which is less than r= 0 . Therefore, the minimum area occurs by allocating all the length to the square!