Question

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (i) the perimeter (ii) the area of the rectangle.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}P\mathrm{be}\mathrm{the}\mathrm{perimeter}\mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{at}\mathrm{any}\mathrm{time}t.\mathrm{Then}, \\ P\text{=2}\left(x+y\right) \\ \text{⇒}\frac{dP}{dt}\text{=2}\left(\frac{dx}{dt}+\frac{dy}{dt}\right) \\ \text{⇒}\frac{dP}{dt}\text{=2}\left(-5+4\right)\left[\because \frac{dx}{dt}=-5\mathrm{cm}/\min \mathrm{and}\frac{dy}{dt}=4\mathrm{cm}/\min \right] \\ \text{⇒}\frac{dP}{dt}\text{=-2 cm/min} \\ \left(ii\right)\mathrm{Let}A\mathrm{be}\mathrm{the}\mathrm{a}\text{rea of the rectangle at any time}\mathit{\text{t.}}\text{Then,} \\ A=xy \\ \Rightarrow \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt} \\ \Rightarrow \frac{dA}{dt}=8\left(4\right)+6\left(-5\right)\left[\because x=8\mathrm{cm},y=6\mathrm{cm}\frac{dx}{dt}=-5\mathrm{cm}/\min \mathrm{and}\frac{dy}{dt}=4\mathrm{cm}/\min \right] \\ \Rightarrow \frac{dA}{dt}=32-30 \\ \Rightarrow \frac{dA}{dt}=2{\mathrm{cm}}^2/\min \end{gathered}$$