Answer
Length: 8 meters
Width: $\dfrac{5}{2}$ meters
Work Step by Step
Let's note:
$l$=the length of the rectangle
$w$=the width of the rectangle
We write the system of equations using the area and the perimeter of the rectangle:
$\begin{cases}
lw=20\\
2l+2w=21
\end{cases}$
Use substitution:
$\begin{cases}
l=\dfrac{20}{w}\\
2\left(\dfrac{20}{w}\right)+2w=21
\end{cases}$
Determine $w$:
$2\left(\dfrac{20}{w}\right)+2w=21$
$\dfrac{40}{w}+2w=21$
$40+2w^2=21w$
$2w^2-21w+40=0$
$2w^2-16w-5w+40=0$
$2w(w-8)-5(w-8)=0$
$(w-8)(2w-5)=0$
$w-8=0\Rightarrow w_1=8$
$2w-5\Rightarrow w_2=\dfrac{5}{2}$
Determine $l$:
$l=\dfrac{20}{w}$
$w_1=8\Rightarrow l_1=\dfrac{20}{8}=\dfrac{5}{2}$
$w_2=\dfrac{5}{2}\Rightarrow l_2=\dfrac{20}{\dfrac{5}{2}}=8$
The solutions are:
$\left(\dfrac{5}{2},8\right),\left(8,\dfrac{5}{2}\right)$
As $l\geq w$ the solution is:
Length: 8 meters
Width: $\dfrac{5}{2}$ meters
