Finding the smallest rectangle able to contain a rectangle with diagonal of length n is the same as asking what the largest area a rectangle with a diagonal of length n can have is (since this rectangle could contain itself, and all others are smaller).

Let a and b be the sides of this rectangle.

n^2 = a^2 + b^2     (1)
The area of this rectangle will then be
c = a*b             (2)

So we are asking what the largest value for c is, for some constant n.

From (1), we get 
a = sqrt(n^2-b^2)
thus, 
c=sqrt(n^2-b^2)b

We take the first derivative of this to determine the maximum:
c(b)'=(n^2-2b^2)/sqrt(n^2-b^2).
Setting to 0,
0 = n^2-2b^2
2b^2=n^2
b=sqrt(n^2/2)       (3)

Back into (1) gives us:
n^2 = a^2 + sqrt((n^2)/2)^2
n^2 = a^2 + (n^2)/2
a^2 = n^2 - (n^2)/2

So a^2 = (n^2)/2, a=sqrt((n^2)/2), and a=b.

(This is entirely unsurprising if you play with rectangles in your mind: the biggest area occurs at the square.)

Finally, we put a and b back into (2), finding the area of a square with sides of length sqrt(n^2)/2:
c = a*b
c = sqrt((n^2)/2)^2
c = (n^2)/2