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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.

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Solution

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...(i)Perimeter of rectangle=(length+breadth) and area=length×breadth
It is given that dxdt=5 cm/min and dydt=4 cm/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 dAdt=xdydt+ydxdt=8×4+6×(5)
(dxdt=5 and dydt=4)=3230=2cm2/min
Hence, area of the rectangle is increasing at the rate of 2cm2/min.
Note If rate of change is increasing, we take positive sign and if rate of change is decreasing, then we take negative sign.


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