(a) The length $x$ of a rectangle is decreasing at the rate of $5 \mathrm{~cm} / \mathrm{min} . \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-5 \mathrm{~cm} \min \quad \ldots(i)$
The width $\mathrm{y}$ is increasing at the rate of $4 \mathrm{~cm} / \mathrm{min}$.
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=4 \mathrm{~cm} / \mathrm{min} \text {. } \quad \ldots(ii)$
The perimeter $\mathrm{p}$ of the rectangle is
$\mathrm{p}=2(\mathrm{x}+\mathrm{y}), \frac{\mathrm{dp}}{\mathrm{dt}}=2\left(\frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dy}}{\mathrm{dt}}\right)$
from (i) \& (ii); $\frac{d p}{d t}=2(-5+4)=-2 \mathrm{~cm} / \mathrm{min}$.
$\therefore$ This shows the perimeter decreases at the rate of $2 \mathrm{~cm} / \mathrm{min}$.
(b) $\mathrm{A}=\mathrm{xy} \therefore \frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{dx}}{\mathrm{dt}} \mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dt}} \quad \ldots (iii)$
But $\frac{\mathrm{dx}}{\mathrm{dt}}=-5 \mathrm{~cm} / \mathrm{min}, \frac{\mathrm{dy}}{\mathrm{dt}}=4 \mathrm{~cm} / \mathrm{min}$. and $x=8, y=6$; Put these values in (iii)
$\frac{\mathrm{dA}}{\mathrm{dt}}=(-5 \times 6)+8 \times 4=-30+32=2 \mathrm{~cm}^2 / \mathrm{min}$
$\Rightarrow$ Area is increasing at the rate of $2 \mathrm{~cm}^2 / \mathrm{min}$