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.

The answer is that there is always a solution whenever $$b\geq 2.$$ One example solution is given by
$$w \equiv \frac{b + \sqrt{b^2-2b}}{2}$$
$$\ell \equiv \frac{b - \sqrt{b^2-2b}}{2}$$
In this case, you can confirm that the sum is $2b$ and the product is $\frac{b}{2}$, exactly as required. The condition $b\geq 2$ ensures that the discriminant is positive so that the solutions are real.

1. You're asking for which values of $b$ there exists a rectangle with perimeter $2b$ and area $\frac{b}{2}$. 



*Let $\ell$ and $w$ be the sides of such a rectangle. Then you're asking for which values of $b$ does there exist a solution $\langle \ell, w\rangle$ to the following system of equations:


$$\ell w =\frac{b}{2} \qquad \ell+w = b.$$


*We can eliminate one of the variables using one of the equations, say $\ell = b - w$. Then the question becomes for which values of $b$ does there exist a solution $\langle w\rangle$ to the equation


$$\begin{align*}(b-w)w &= \frac{b}{2}\\ bw - w^2 &= \frac{b}{2}\\w^2-bw+\frac{b}{2} &= 0\end{align*}$$ 


*This formula is quadratic in $w$, and hence it always has two solutions. In order to be a physical rectangle though, we must have that $w>0$ and also $\ell > 0$ (which requires that $b-w > 0$, i.e. $w<b$.) 
Putting this all together, we ask for what values of $b$ does this quadratic equation have a solution $w$ where $0<w<b$.

*The solution in general is given by the quadratic formula:
$$w = \frac{b\pm\sqrt{b^2-2b}}{2}$$
Hence our question is: for which values of $b$ does at least one of the following conditions hold:
$$0 < \frac{b+\sqrt{b^2-2b}}{2} < b\qquad\text{(positive root)}$$
$$0 < \frac{b-\sqrt{b^2-2b}}{2} < b\qquad\text{(negative root)}$$


*With some arithmetic, we can show that each condition is actually equivalent to:



$$-b < \sqrt{b^2-2b} < b$$


*We are given that $b>0$, so assuming the quantity under the square root is also positive ($b^2 > 2b$, i.e. $b>2$), the condition $-b < \sqrt{b^2-2b}$ is always satisfied, and we only need to worry about whether 


$$\begin{align*}\sqrt{b^2-2b} &< b\\
b^2 - 2b &< b^2 \\
-2b & < 0\\
b &> 0\\
\end{align*}$$


*We originally asked under which conditions a rectangle with the desired physical parameters existed. We converted this into a quadratic formula $w^2-bw + \frac{b}{2} = 0$ and sought solutions $w$ which were physically possible, meaning that $0<w<b$. 
The first condition we imposed was that the discriminant $b^2 -2b$ would be positive so that the solutions $w$ wouldn't be imaginary. This is equivalent to requiring $b>2$. 
It turns out that under that restriction, the quadratic formula always has a positive root less than $b$. (To see this, note that $b^2-2b$ is always less than $b^2$, so $\sqrt{b^2-2b}$ is always less than $b$, so $b-\sqrt{b^2-2b}$ is always positive.)

*Hence the answer is that there is always a solution whenever $$b\geq 2.$$ One example solution is given by
$$w \equiv \frac{b + \sqrt{b^2-2b}}{2}$$
$$\ell \equiv \frac{b - \sqrt{b^2-2b}}{2}$$
You know that each of these quantities is positive because $\sqrt{b^2-2b} < \sqrt{b^2} = b$, and of course their sum is $(2b)/2 = b$ and their product is 
$$\frac{1}{4}[b^2 - (b^2-2b)] = \frac{b}{2}$$
exactly as required.
