Answer
$20\leq length \leq 40$
Work Step by Step
Let's assume, that length of the rectangle is $xft$, so width will be $(60-x)ft$
(For a better understanding: sum of width and length is half of a perimeter and we know that perimeter in this case is $120$)
We need area $A$, that is more or equal to $800ft^2$.
$A=x(60-x)$
$60x-x^2\geq 800$
$x^2-60x+800\leq0$
At first, we need to find key points (where it gets to $0$) and then find solution using interval method. We can write it as quadratic equation to find the key points:
$x^2-60x+800=0$
$x_1=20$ ; $x_2=40$
We have intervals:
$(-\infty, 20]$ - Positive
$[20, 40]$ - Negative
$[40, +\infty)$ - Positive
According to the last inequality, we need interval where the equation is less than or equal to $0$, that is the second interval $[20, 40]$
$20\leq x \leq 40$
