Question

The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x=8 cm and y=6 cm, find the rate of change of the area of the rectangle.

Answer 1

At any instant of time t, let length, breadth, perimeter and area of rectangle are x, y, p and A respectively, then
$P=2(x+y)andA=xy...\left(i\right)Perimeterofrectangle=(length+breadth)andarea=length\times breadth$
It is given that $\frac{dx}{dt}=-5cm/minand\frac{dy}{dt}=4cm/min$.
(-ve sign shows that the length is decreasing)

Here, area of rectangle A = xy. On differentiating w.r.t t, we get
Rate of change $\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}=8\times 4+6\times (-5)$
$(\because \frac{dx}{dt}=-5and\frac{dy}{dt}=4)=32-30=2c{m}^2/min$
Hence, area of the rectangle is increasing at the rate of $2c{m}^2/min$.
Note If rate of change is increasing, we take positive sign and if rate of change is decreasing, then we take negative sign.