The length 'x' of a rectangle is decreasing at the rate of 3 cm/m and the width 'y' is increasing at the rate of 2 cm/m. Find the rates of change of :
(a) the perimeter (b) the area of the rectangle when x = 8 cm and y = 6 cm.

Answer

To solve the problem step by step, we will first find the rates of change of the perimeter and then the area of the rectangle.

### Given:
- Length of the rectangle, \( x \) (cm) is decreasing at the rate of \( \frac{dx}{dt} = -3 \) cm/min (negative because it is decreasing).
- Width of the rectangle, \( y \) (cm) is increasing at the rate of \( \frac{dy}{dt} = 2 \) cm/min.
- At the time of interest, \( x = 8 \) cm and \( y = 6 \) cm.

### (a) Finding the rate of change of the perimeter:

1. **Formula for Perimeter**:
The perimeter \( P \) of a rectangle is given by:
\[ P = 2x + 2y \]

2. **Differentiate with respect to time \( t \)**:
To find the rate of change of the perimeter, we differentiate \( P \):
\[ \frac{dP}{dt} = 2 \frac{dx}{dt} + 2 \frac{dy}{dt} \]

3. **Substitute the known values**:
Substitute \( \frac{dx}{dt} = -3 \) cm/min and \( \frac{dy}{dt} = 2 \) cm/min:
\[ \frac{dP}{dt} = 2(-3) + 2(2) \]
\[ \frac{dP}{dt} = -6 + 4 = -2 \text{ cm/min} \]
Thus, the rate of change of the perimeter is \( -2 \) cm/min, indicating that the perimeter is decreasing.

### (b) Finding the rate of change of the area:

1. **Formula for Area**:
The area \( A \) of a rectangle is given by:
\[ A = xy \]

2. **Differentiate with respect to time \( t \)**:
To find the rate of change of the area, we differentiate \( A \):
\[ \frac{dA}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt} \]

3. **Substitute the known values**:
Substitute \( x = 8 \) cm, \( y = 6 \) cm, \( \frac{dy}{dt} = 2 \) cm/min, and \( \frac{dx}{dt} = -3 \) cm/min:
\[ \frac{dA}{dt} = 8(2) + 6(-3) \]
\[ \frac{dA}{dt} = 16 - 18 = -2 \text{ cm}^2/\text{min} \]
Thus, the rate of change of the area is \( -2 \) cm²/min, indicating that the area is decreasing.

### Summary of Results:
- The rate of change of the perimeter is \( -2 \) cm/min.
- The rate of change of the area is \( -2 \) cm²/min.