Given a positive number $b$, under what conditions does there exist a rectangle with perimeter $2b$ and area $\frac{b}{2}$? Given a positive number $b$, under what conditions does there exist a rectangle with perimeter $2b$ and area $\frac{b}{2}$?
$$2l + 2w = 2b$$ $$l+w=b$$ 
and $$lw = \frac{b}{2}$$
I tried to use AGM inequality 
$$lw = \frac{l+w}{2}$$
For that to be a AGM inequality the right side had to be squared and the equality will only hold if $l = w$. I don't know how to get the conditions from here.

$2w + 2l = 2b$
$l = b-w$.
$wl = w(b-w) = bw -w^2 = \frac b2$
$w^2 - bw + \frac b2 = 0$
$w = \frac {b \pm\sqrt{b^2 - 4*\frac b2}}2 = \frac {b \pm \sqrt {b^2 - 2b}}2$
This can be done whenever $b^2 \ge 2b$ which (as $b > 0$) mean $b\ge 2$.
If $b = 2$ this is possible with square of $w=l = 1$ and $P = 4=2*2$ and $A = 1^2 = 1 = \frac 22$.
If $b > 2$ then this is possible with $w = \frac {b + \sqrt {b^2 - 2b}}2$ and $l = \frac {b - \sqrt {b^2 - 2b}}2$.  The $P= 2w + 2l = b +  \sqrt {b^2 - 2b} + b -  \sqrt {b^2 - 2b} = 2b$ and $A = wl =  \frac {b + \sqrt {b^2 - 2b}}2*\frac {b - \sqrt {b^2 - 2b}}2 = \frac {b^2 - (b^2 - 2b)}{4} = \frac b2$.
If $b < 2$ well.... 
$\frac {l + w}2 \ge \sqrt {lw}$ by AM-GM
But $\frac {l + w}2 = $ perimeter/$4$ $= \frac b2$
$\sqrt {lw} = \sqrt{\text {area}} = \sqrt {\frac b2}$
So $\frac b2 \ge \sqrt {\frac b2}$
So $\frac b2 \ge 1$ and $b \ge 2$.
So it can't be done.
