A rectangle with perimeter of 100 has area at least of 500, within what bounds must the length of the rectangle lie? Problem
The problem states that there is a rectangle that has a perimeter of $100$ and an area of at least $500$ and it asks for the bounds of the length 
which can be given in interval notation or in the <> (greater than or less than) signs
My steps and thought process
So I set some few equations
1)$2x+2y=100$ which becomes $x+y=50$
2)$xy \geq 500$
3)I then made $y = 50-x$
so that I can substitute it into equation #2 to get:
$50x-x^2 \geq 500$
which eventually got me $0 \geq x^2-50x+500$
and this is where I got stuck.
 A: You are correct that $x + y = 50$ and that $xy \geq 500$.  We can solve the inequality by completing the square.
\begin{align*}
xy & \geq 500\\
x(50 - x) & \geq 500\\
50x - x^2 & \geq 500\\
0 & \geq x^2 - 50x + 500\\
0 & \geq (x^2 - 50x) + 500\\
0 & \geq (x^2 - 50x + 625) - 625 + 500\\
0 & \geq (x - 25)^2 - 125\\
125 & \geq (x - 25)^2\\
\sqrt{125} & \geq |x - 25|\\
5\sqrt{5} & \geq |x - 25|
\end{align*}
Hence,
\begin{align*}
-5\sqrt{5} & \leq x - 25 \leq 5\sqrt{5}\\
25 - 5\sqrt{5} & \leq x \leq 25 + 5\sqrt{5}
\end{align*}
Alternatively, note that since the perimeter of the rectangle is $100$ units, the average side length is $100/4 = 25$ units.  Hence, we can express the lengths of adjacent sides as $25 + k$ and $25 - k$.  Hence, the area is 
$$A(k) = (25 + k)(25 - k) = 625 - k^2$$
The requirement that the area must be at least $500$ square units means
\begin{align*}
625 - k^2 & \geq 500\\
125 & \geq k^2\\
5\sqrt{5} & \geq |k|\\
5\sqrt{5} & \geq k \geq -5\sqrt{5}
\end{align*}
Thus, for the area of the rectangle to be at least $500$ square units, the length of the longer side of the rectangle must be at most $25 + 5\sqrt{5}$ units and the length of the shorter side of the rectangle must be at least $25 - 5\sqrt{5}$ units. Also, note that the maximum area of $625$ square units occurs when $k = 0$, that is, when the rectangle is a square.
A: Since $x+y= 100$ and $xy \geq 500$, we have $x(50-x) \geq 500$. Thus $x^2 -50x + 500 \leq 0$. Consequently, $x$ must lie between the roots of the quadratic $x^2-50x+500 = 0$. The roots are $25 \pm 5\sqrt{5}$. Thus $25-5\sqrt{5} \leq x \leq 25+5\sqrt{5}$.
