  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

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 