The length x of a rectangle is decreasing at the rate of 5 cm/minute

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.

Solution:

(i) Let $P$ be the perimeter of the rectangle at any time $t$. Then,

$P=2(x+y)$

$\Rightarrow \frac{d P}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)$

$\Rightarrow \frac{d P}{d t}=2(-5+4)$               $\left[\because \frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}\right.$ and $\left.\frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min}\right]$

$\Rightarrow \frac{d P}{d t}=-2 \mathrm{~cm} / \mathrm{min}$

(ii) Let $A$ be the area of the rectangle at any time $t$. Then,

$A=x y$

$\Rightarrow \frac{d A}{d t}=x \frac{d y}{d t}+y \frac{d x}{d t}$

$\Rightarrow \frac{d A}{d t}=8(4)+6(-5)$         $\left[\because x=8 \mathrm{~cm}, y=6 \mathrm{~cm} \frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}\right.$ and $\left.\frac{d y}{d t}=4 \mathrm{~cm} / \mathrm{min}\right]$

$\Rightarrow \frac{d A}{d t}=32-30$

$\Rightarrow \frac{d A}{d t}=2 \mathrm{~cm}^{2} / \mathrm{min}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now