#### provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 30

Answer: (i) -2 cm/min       (ii) 2 cm2/min

Hint: We know that, $p=2(x+y)$

Given: The length of the rectangle is $x$ cm and width is $y$ cm.

Solution: As per the given criteria, the length is decreasing at the rate of 5 cm/min

(i) $\frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}$

And width is increasing at the rate of 4 cm/min

$\frac{d x}{d t}=4 \mathrm{cm} / \mathrm{min}$

Differentiating both sides,

$\frac{d p}{d x}=\frac{d(2(x+y))}{d t}=2\left[\frac{d x}{d t}+\frac{d y}{d t}\right]$

So, $\frac{d p}{d x}=2[-5+4]=-2\; cm/min$

(ii) Let A  be the area of the rectangle, $A=xy$

\begin{aligned} &\frac{d A}{d t}=\frac{d(x y)}{d t} \\\\ &\frac{d(u v)}{d x}=v \frac{d u}{d x}+u \frac{d v}{d x} \end{aligned}

So the above equation becomes,

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

Substituting the values from equation (i) and (ii)

$\frac{d A}{d t}=y(-5)+x(4)$

When $x=8$ cm and $y=6$ cm the above equation becomes,

$\frac{d A}{d t}=6(-5)+8(4)=-30+32=2\: cm^{2}/min$