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

(a) perimeter, and

(b) the area of the rectangle.

Let us assume that at any instant of time $t$, length of rectangle be $x$, breadth be $y$, the perimeter $P$ and the area be $A$, then

(a) We have, $\cfrac{{dx}}{{dt}} = - 5{\rm{cm/min}}$ and $\cfrac{{dy}}{{dt}} = 4{\rm{cm/min}}$
(a) $P = 2(x + y)$ …(i)

Differentiating (i) w.r.t. $t$, we get
$\cfrac{{dP}}{{dt}} = 2\left( {\cfrac{{dx}}{{dt}} + \cfrac{{dy}}{{dt}}} \right) = 2( - 5 + 4){\rm{cm/min}} = - 2{\rm{cm/min}}$

Therefore, Perimeter of the rectangle is decreasing at the rate of $2{\rm{cm/min}}$

(b) $A = xy$ …(ii)

Differentiating (i) w.r.t. $t$, we get

$\cfrac{{dA}}{{dt}} = x\cfrac{{dy}}{{dt}} + y\cfrac{{dx}}{{dt}} = \left( {8{\rm{cm}}} \right)\left( {4{\rm{cm/min}}} \right) + \left( {6{\rm{cm}}} \right)\left( { - 5{\rm{cm/min}}} \right)$
$= 2{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/min}}$

Therefore, area of the rectangle is increasing at the rate of $2{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/min}}$.