Minimum area of a square and a circle of total 1 unit circumference length We have a 1 unit line and we can can cut it into 2 parts, where 1 part is used at the circumference of the square and the other as the circumference of a circle. How do i find out where to cut the line to get the least area as possible?
 A: Say the part of length $x$ forms the circle. That circle has radius $\frac x{2\pi}$ and area $\frac{x^2}{4\pi}$. The remaining part of length $1-x$ forms a square with side $\frac{1-x}4$ and area $\frac{(1-x)^2}{16}$. So we have to minimise
$$\frac{(x-1)^2}{16}+\frac{x^2}{4\pi}=\left(\frac1{16}+\frac1{4\pi}\right)x^2-\frac x8+\frac1{16}$$
and the minimum occurs where the derivative is zero:
$$\left(\frac18+\frac1{2\pi}\right)x-\frac18=0$$
$$x=\frac{1/8}{1/8+1/2\pi}=0.439900\dots$$
So this length is reserved for the circle, and the remaining part for the square.
A: Let the square and circle parts be partitioned $ (x, 1-x)$ respectively.
$$ x= 4a,\, 2 \pi r = (1-x),\, U= 4 a + 2 \pi r $$
$$ A = a^2 + \pi r^2 $$
EDIT1:
simplifies to 
$$ \frac{4 A}{\pi} = \frac{\pi x^2}{4}+(1-x)^2 = x^2(\pi/4+1)-2x+ 1 $$
By usual differentiation its parabola graph has at 
$$ x=\frac{1}{1+\pi/4} \approx 0.44$$ 
gives a minimum for square/circle.
Also, by Lagrange Multiplier method we gain extra geometrical insight ( I mention this  even if outside pre-calculus scope ):
$$ U= 4a+2\pi r;\, A = a^2+\pi r^2;$$
If each of $U$ and $A$ are partially differentiated with respect to $r$ and $a$ separately, and their ratios equated, we get a simple relation
$$ a = 2r$$
or the side should equal the circle diameter   if laid side by side after dividing the given length of string as sketched below:

