Question

1) The length $x$ of a rectangle is decreasing at the rate of $5\text{cm/minute}$ & the width $y$ is increasing at the rate of $4\text{cm/minute}$ . When $x=8\text{cm}$ and $y=6\text{cm}$ ,
find the rate of change of the perimeter of the rectangle.

2) The length $x$ of a rectangle is decreasing at the rate of $5\text{cm/minute}$ & the width $y$ is increasing at the rate of $4\text{cm/minute}$ . When $x=8\text{cm}$ and $y=6\text{cm}$ ,
find the rate of change of the area of the rectangle.

Answer 1

1) Given: Length $x$, width $y$
Also, length of a rectangle is decreasing at the rate of $5$ cm/min
And the width is increasing at the rate of $4$cm/min
So,

$\frac{dx}{dt}=-5\text{cm/min}\cdots \left(i\right)$

$\frac{dy}{dt}=4\text{cm/min}\cdots \left(ii\right)$

Now, the perimeter of the rectangle is,

$P=2x+2y$

Differentiating w.r.t. $t$, we get

$\frac{dP}{dt}=\frac{d}{dt}(2x+2y)$

$=2(\frac{dx}{dt}+\frac{dy}{dt})$

From $\left(i\right)$ and $\left(ii\right)$, we get

$\Rightarrow \frac{dP}{dt}=2(-5+4)=-2\text{cm/min}$

Hence, perimeter is decreasing at the rate of $2\text{cm/min}$.

2) Given: Length $x$, width $y$
Also, length of a rectangle is decreasing at the rate of $5\text{cm/min}$
And the width is increasing at the rate of $4\text{cm/min}$
So,

$\frac{dx}{dt}=-5\text{cm/min}\cdots \left(i\right)$

$\frac{dy}{dt}=4\text{cm/min}\cdots \left(ii\right)$

Now, the perimeter of the rectangle is,

$A=xy$

Differentiating w.r.t. $t$, we get

$\frac{dA}{dt}=\frac{d}{dt}\left(xy\right)$

$=x\frac{dy}{dt}+y\frac{dx}{dt}$

From $\left(i\right)$ and $\left(ii\right)$, we get

$\Rightarrow \frac{dA}{dt}=x\left(4\right)+y(-5)$

$\Rightarrow \frac{dA}{dt}=4x-5y$

When $x=8\text{cm and}y=6\text{cm}$

$\Rightarrow \frac{dA}{dt}=32-30=2{\text{cm}}^2/\text{min}$

Hence, the area is increasing at the rate of $2{\text{cm}}^2/\text{min},$ when $x=8$ cm and $y=6$ cm.