एक आयत की लम्बाई x, 5 cm / min की दर से घट रही है और चौड़ाई y, 4 cm / min कि दर से बढ़ रही है जब x=8 cm और y= 6 cm है तब आयत के (a) परिमाप (b) क्षेत्रफल की परिवर्तन की दर ज्ञात कीजिए

Answer

### Step-by-Step Solution:

Given:
- Length of the rectangle = \( x \) cm (decreasing at a rate of 5 cm/min)
- Width of the rectangle = \( y \) cm (increasing at a rate of 4 cm/min)
- \( x = 8 \) cm
- \( y = 6 \) cm

We need to find the rates of change of:
(a) Perimeter of the rectangle
(b) Area of the rectangle

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

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

2. **Differentiate with respect to time \( t \)**:
\[ \frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) \]

3. **Substituting the values**:
Given \( \frac{dx}{dt} = -5 \) cm/min (since length is decreasing) and \( \frac{dy}{dt} = 4 \) cm/min (since width is increasing):
\[ \frac{dP}{dt} = 2\left(-5 + 4\right) = 2(-1) = -2 \text{ cm/min} \]

Thus, the rate of change of the perimeter is \( -2 \) cm/min.

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

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

2. **Differentiate with respect to time \( t \)**:
\[ \frac{dA}{dt} = x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt} \]

3. **Substituting the values**:
Substitute \( x = 8 \) cm, \( y = 6 \) cm, \( \frac{dx}{dt} = -5 \) cm/min, and \( \frac{dy}{dt} = 4 \) cm/min:
\[ \frac{dA}{dt} = 8 \cdot 4 + 6 \cdot (-5) \]
\[ = 32 - 30 = 2 \text{ cm}^2/\text{min} \]

Thus, the rate of change of the area is \( 2 \) cm²/min.

### Final Answers:
(a) Rate of change of perimeter: \( -2 \) cm/min
(b) Rate of change of area: \( 2 \) cm²/min