The length x of a rectangle is decreasing at the rate of 3 cm/minute and the width y is increasing at the rate of 2 cm/minute. When x = 10 cm and y = 6 cm, find the rate of change of the area of the rectangle.

$2\,cm^2/min$

The correct answer is Option (1) → $2\,cm^2/min$

The area, say A, of the rectangle at any time is given by

$A = xy$, diff. w.r.t. t, we get

$\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}=(x.2+y(-3)) cm^2/min = (2x-3y) cm^2/min$.

When $x = 10 cm, y = 6 cm, \frac{dA}{dt}= (10.2 + 6.(-3)) cm^2/min$

$= 2\, cm^2/min$.

Hence, the area is increasing at the rate of $2\, cm^2/min$ when x = 10 cm and y = 6 cm.