The sides of rectangle are (6-x) cm and (x-3) cm. If its area is maximum then x =

$\begin{array}{c}Given sides of rec\tan gle are (6-x) and (x-3)\\ Now \mathrm{Area} \mathrm{of} \mathrm{rectangle}=(6-\mathrm{x})(\mathrm{x}-3)\\ let f\left(x\right)=-{\mathrm{x}}^2+9\mathrm{x}-18\\ diff w.r.to x on both sides\\ f\text{'}\left(x\right)=-2x+9\\ for max or min f\text{'}\left(x\right)=0\\ -2x+9=0\Rightarrow x=\frac{9}{2}\\ clearly f\text{'}\text{'}\left(x\right)=-2<0\end{array}$

$\therefore$Area is maximum at $\mathrm{x}=\frac{9}{2}=4.5$