Question

The length $x$ of a rectangle is decreasing at the rate of $5\, cm / min$ and the width $y$ is increasing at the rate of $4 \,cm / min$. If $x =8 \,cm$ and $y =6 \,cm$, then which of the following is correct?
I. The rate of change of the perimeter is $-2\, cm / min$.
II. The rate of change of the area of the rectangle is $12 \,cm ^{2} / min$.

• A. Only I is correct
• B. Only II is correct
• C. Both I and II are correct
• D. Both I and II are incorrect

Answer 1

Answer: (A)

At any instant time $t$, let length, breadth, perimeter and area of the rectangle are $x , y , P$ and A respectively,
then $P =2( x + y )$ and $A = xy$
It is given that $\frac{ d x }{ dt }=-5 \,cm / min$ and
$\frac{ dy }{ dt }=4 \,cm / min$
(- ve sign shows that the length is decreasing)
I. Now $P=2(x+y)$.
On differentiating w.r.t t, we get
$\frac{ dP }{ dt }=2\left(\frac{ dx }{ dt }+\frac{ dy }{ dt }\right)$
$=2\{-5+4\} cm / min =-2\, cm / min$
Hence, perimeter of the rectangle is decreasing at the rate of $2 \,cm / min$.
II. Here, area of rectangle $A = xy$
On differentiating w.r.t. t, we get
Rate of change of area
$\frac{ d A }{ dt }= x \frac{ dy }{ dt }+ y \frac{ dx }{ dt }$
$=8 \times 4+6 \times(-5)$
$=32-30=2 \,cm ^{2} / min$
Hence, area of the rectangle is increasing at the rate of $2cm ^{2} / min$