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.
 A: As a broad hint: let the two sides be $l, w$ as you've written, and consider $(l-w)^2$. This must be a non-negative quantity, and it's zero iff $l=w$. Now, can you see how to write $(l-w)^2$ as an expression in $l+w$ and $lw$? (Followup hint: what is $(l+w)^2$?)
A: We have $$l+w=b\quad \&\quad lw=\frac b2\implies w+\frac b{2w}=b$$
Now, the minimum of $f(x)=x+\frac b{2x}$ is realized when $x=\sqrt {\frac b2}$ and it yields a value of $2\sqrt {\frac b2}$  Thus you get a solution if and only if
$$2\sqrt {\frac b2}≤b \iff 2b≤b^2\iff 2≤b$$
Note:  The limiting case, $b=2$ corresponds to the unit square.
A: 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.
A: Perimeter $2b$ implies sides $x$ and $b-x$ for $0<x<b$. Area $b/2$ implies $x(b-x)=b/2$.
So $x^2-bx+b/2=0$, or $x = \frac{b\pm\sqrt{b^2-2b}}{2}$. 
This gives a pair of real solutions  (which just swap $x$ and $b-x$, so only one unique solution) with $0<x<b$ when $b^2-2b\geq 0$, which is when $b\geq2$.
A: Suppose we have a right triangle of sides a1,a2,a3 and a3 is diagonal
Given a1+a2+a3=2b & (a1*a2)/2=b/2 => a1*a2=b                                                   $applying$ $Arithmetic$ $geometric$ $inequality$  
=>   $(a1+a2)/2>=\sqrt(a1.a2)$ => $(a1+a2)/2>=\sqrt(b)$=>$(2b-a3)/2>=\sqrt(b)$=> $a3<=2(b-\sqrt(b))$=>$to$ $have$ $this$ $inequality$ $since$ $a3$ $is$ $side(diagonal)$ $so$ $b>\sqrt(b)$ $we$ $get$ $b^2-b>0$=>
$b(b-1)>0$ which gives that $b<0$ $or$ $b>1$=> $so$ $b>1$ $true$  so the conditions are its diagonal must less than 2(b-√b) and b>1
A: $2w + 2l = 2b$
$l = b-w$.
$wl = w(b-w) = bw -w^2 = \frac b2$
$w^2 - bw + \frac b2 = 0$
$w = \frac {b \pm\sqrt{b^2 - 4*\frac b2}}2 = \frac {b \pm \sqrt {b^2 - 2b}}2$
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.
A: Solution from previous exam solution:

