Question

The length $ 'x' $ of a rectangle is decreasing at the rate of $ 6 $ cm/min and the width y is increasing at the rate of $ 4\, cm/\min $ . When $ x = 8 \, cm$ and $ y = 4\, cm $ , the rate of change of the area of the rectangle is

• A. 8
• B. 16
• C. 24
• D. 32

Answer 1

Answer: (A)

We have,
$\frac{d x}{d t}=-6, \frac{d y}{d t}=4$
Area of rectangle, $A=x y$
$\Rightarrow \frac{d A}{d t}=x \frac{d y}{d t}+y \cdot \frac{d x}{d t}$
$\Rightarrow \frac{d A}{d t}=x(4)+(-6) y$
$\Rightarrow \frac{d A}{d t}=4 x-6 y$
$\because$ Rate of change of area $(A)$ of rectangle i.e. $\frac{d A}{d t}$
$\therefore\left(\frac{d A}{d t}\right)_{(8,4)}=4(8)-6(4)=8$