Minimum primeter

L = Rectangle length

W = Rectangle width

Area of rectangle $=256 \ m^2 = L \times W$

perimeter $P = 2(L + W) = 2(L + \frac{256}{L} )$

differentiating w.r.t L gives :

$\dfrac{dP}{dL} = 2(1 - \frac{256}{L^2}) = 0$

$\dfrac{256}{L^2} = 1 \\ \ \ L = 16 \ \text{and} \ W = 16$ $\\P = 64$

Let the length of the rectangle be $l$ and the breadth be $b$ . Then it's area is $lb=256$ and the perimeter is $2(l+b)$ . Applying AM-GM inequality we get $2(l+b)\geq {4\sqrt {lb}}$ . So the minimum perimeter is $4\sqrt {256}=\boxed {64}$