Question

The area of an expanding rectangle is increasing at the rate of 48 cm²/s. The length of the rectangle is always square of its breadth. At what rate the length of rectangle increasing at an instant, when breadth = 4.5 cm?

Answer 1

Given that,

`(dA)/(dt) = 48  cm^2//sec`

Let l be the length of the rectangle and b be the breadth of the rectangle.

l = b²

b = 4.5 cm

We know that,

Area = l × b

b² × b

A = b³

Differentiate w.r.t.t, we get

`(dA)/(dt) = 3b^2(db)/(dt)`

Put the value `(dA)/(dt)`

48 = `3 xx 4.5 xx 4.5 (db)/(dt)`

`(db)/(dt) = (48 xx 10 xx 10)/(3 xx 45 xx 45)`

= `(16 xx 4)/81`

= `64/81`

but, l = b²

Now differentiate w.r.t. t,

`(dl)/(dt) = 2b (db)/(dt)`

`(dl)/(dt) = 2 xx 45/10 xx 64/81`

`(dl)/(dt) = 64/9  cm//sec`

= 7.11 cm/sec

The rate at which the length of the rectangle is increasing when the breadth is 4.5 cm is 7.11 cm/s.