The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute.

Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:

$\frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}$ and $\frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min}$

(a) The perimeter (P) of a rectangle is given by,

P = 2(x + y)

$\therefore \frac{d P}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)=2(-5+4)=-2 \mathrm{~cm} / \mathrm{min}$

Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b) The area (A) of a rectangle is given by,

A = x⋅ y

$\therefore \frac{d \mathrm{~A}}{d t}=\frac{d x}{d t} \cdot y+x \cdot \frac{d y}{d t}=-5 y+4 x$

When $x=8 \mathrm{~cm}$ and $y=6 \mathrm{~cm}, \frac{d A}{d t}=(-5 \times 6+4 \times 8) \mathrm{cm}^{2} / \mathrm{min}=2 \mathrm{~cm}^{2} / \mathrm{min}$

Hence, the area of the rectangle is increasing at the rate of $2 \mathrm{~cm}^{2} / \mathrm{min}$.