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)=32−30=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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