448. The sides $a$ and $b$ of a rectangle change according to the laws $a=(2 t+1)$ cm, $b=(3 t+2)$ cm. At what rate is its area $S$ changing at the moment $t=4$ s?
Solution. We find $S=a b=(2 t+1)(3 t+2)=6 t^{2}+7 t+2, \quad v=$ $=S_{t}^{\prime}=\left(6 t^{2}+7 t+2\right)^{\prime}=12 t+7$. For $t=4$ we get $v_{t=4}=12 \cdot 4+7=$ $=55(\mathrm{~cm} / \mathrm{s})$.