Consider a rectangle whose length is increasing at the uniform rate of 2 m/sec, breadth is decreasing at the uniform rate of 3 m/sec and the area is decreasing at the uniform rate of 5 m2/sec. If after some time the breadth of the rectangle is 2 m then the length of the rectangle is

Question

Consider a rectangle whose length is increasing at the uniform rate of 2 m/sec, breadth is decreasing at the uniform rate of 3 m/sec and the area is decreasing at the uniform rate of 5 m2/sec. If after some time the breadth of the rectangle is 2 m then the length of the rectangle is (A) 2 m (B) 4 m (C) 1 m (D) 3 m

Tardigrade Admin · Accepted Answer

Let A be the area, b be the breadth and

ℓ

be the length of the rectangle.
Given :

\frac{d A}{d t} = - 5, \frac{d ℓ}{d t} = 2, \frac{d b}{d t} = - 3

We know,

A = ℓ \times b

\Rightarrow \frac{d A}{d t} = ℓ . \frac{d b}{d t} + b . \frac{d ℓ}{d t} = 3 ℓ + 2 b

\Rightarrow - 5 = — - 3 ℓ + 2 b .

When

b = 2

, we have

- 5 = - 3 ℓ + 4 \Rightarrow ℓ = \frac{9}{3} = 3 m

$2 m$

$4 m$

$1 m$

$3 m$

Q. Consider a rectangle whose length is increasing at the uniform rate of 2m/sec, breadth is decreasing at the uniform rate of 3m/sec and the area is decreasing at the uniform rate of 5 m2/sec. If after some time the breadth of the rectangle is 2m then the length of the rectangle is

2m

4m

1m

3m

Let A be the area, b be the breadth and ℓ be the length of the rectangle. Given : dtdA​=−5,dtdℓ​=2,dtdb​=−3 We know, A=ℓ×b ⇒dtdA​=ℓ.dtdb​+b.dtdℓ​=3ℓ+2b ⇒−5=—−3ℓ+2b. When b=2, we have −5=−3ℓ+4⇒ℓ=39​=3m

$2 m$

$4 m$

$1 m$

$3 m$