[tex]\begin{gathered} \text{Area =20} \\ \text{Let' sassume the l and b is the length and the width of the rectangle.} \\ So, \\ l\times b=20\Rightarrow b=\frac{20}{l} \\ \text{Now,perimeter } \\ \text{p}=2(l+b) \\ \text{For largest perimeter, differentiate the above equation} \\ \frac{dP}{dl}=\frac{d2(l+b)}{dl} \\ \frac{dP}{dl}=\frac{2d(l+\frac{20}{l})}{dl} \\ 0=2-\frac{40}{l^2} \\ 2=\frac{40}{l^2} \\ l^2=20 \\ l=4.47\text{unit} \end{gathered}[/tex]
