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Question

The length $$x$$ of a rectangle is decreasing at a rate of $$3cm/min$$ and width $$y$$ is increasing at a rate of $$ 2 cm/min$$. When $$x=10cm$$ and $$y=6cm$$, find the rates of change of (i) the perimeter, (ii) the area of the rectangle.


Solution

Let length of rectangle $$=x\ cm$$
and width of rectangle $$=y\ cm$$

Given the length is decreasing at the rate of $$3\ cm$$/ minutes i.e. is decreasing w.r.t time 

$$\dfrac{dx}{dt}=-3\ cm/min.......(1)$$ as $$x$$ is decreasing and the width $$y$$ is increasing at the rate of $$2\ cm/min$$
i.e. $$y$$ is increasing w.r.t time
$$dy/dt=2\ cm/min .........(2)$$

(1) Let $$P$$ be the perimeter of rectangle
$$=2(l+width)$$
$$P=2(x+y)$$
$$dp/dt=2, (dx/dt+dy/dt)$$
Given : $$dx/dt=-3, dy/dt=2$$
$$\therefore dp/dt=2(-3+2)=2\times -1=-2\ cm/min$$
$$\therefore$$ perimeter is decreasing at the rate of $$2\ cm/min$$

(2) Let $$A$$ be the area of the rectangle 
$$A=l\times width=xy\Rightarrow dA/dt=xdy/dt+y\dfrac{dx}{dt}$$
$$\Rightarrow dA/dt=-3y+2x$$ from $$(1)$$ & $$(2)$$
$$dA/dt \left.\dfrac{}{}\right|_{x=10, y=6}=-3\times 6+2\times 10=-18+20=2$$

Since are is $$m\ cm^{2} dA/dt=2\ cm^{2}/min$$

Hence are is increasing at the rate of $$2\ cm^{2}/min$$

Mathematics

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