Consider a rectangle whose length is increasing at the uniform rate of $2 \mathrm{~m} / \mathrm{sec}$,…
Consider a rectangle whose length is increasing at the uniform rate of $2 \mathrm{~m} / \mathrm{sec}$, breadth is decreasing at the uniform rate of $3 \mathrm{~m} / \mathrm{sec}$ and the area is decreasing at the uniform rate of $5 \mathrm{~m}^2 / \mathrm{sec}$. If after some time the breadth of the rectangle is $2 \mathrm{~m}$ then the length of the rectangle is
$2 \mathrm{~m}$
$4 \mathrm{~m}$
$1 \mathrm{~m}$
$3 \mathrm{~m}$
Solution
Let $A$ be the area, $b$ be the breadth and $\ell$ be the length of the rectangle.
Given: $\frac{d A}{d t}=-5, \frac{d \ell}{d t}=2, \frac{d b}{d t}=-3$
We know, $A=\ell \times b$
$
\begin{aligned}
& \Rightarrow \frac{d A}{d t}=\ell \cdot \frac{d b}{d t}+b \cdot \frac{d \ell}{d t}=-3 \ell+2 b \\
& \Rightarrow-5=-3 \ell+2 b .
\end{aligned}
$
When $b=2$, we have
$
-5=-3 \ell+4 \Rightarrow \ell=\frac{9}{3}=3 m
$