The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 4 cm/s. when the length is 9 cm and the width is 4 cm, how fast is the area of the rectangle increasing?

The area is the product of length and width. That product can be differentiated:
  A = L·W
  dA/dt = dL/dt·W + L·dW/dt
  dA/dt = (6 cm/s)·(4 cm) + (9 cm)·(4 cm/s)
  dA/dt = 24 cm²/s + 36 cm²/s
  dA/dt = 60 cm²/s

The area of the rectangle is increasing at 60 cm²/s at that moment.

Given :

The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 4 cm/s. When the length is 9 cm and the width is 4 cm .

To find :-

how fast is the area of the rectangle increasing?

Solution :-

As we know that :-

A = lb

To find the rate :-

d(A)/dt = d(lb)/dt .

Differenciate :-

dA/dt = l (db/dt ) + b (dl/dt )

Substitute :-

dA/dt = 9*4 + 4*6

dA/dt = 36+ 24 cm²/s

dA/dt = 60 cm²/s

Ithink its 126 hope its