# The length l of a rectangle is decreasing at the rate of 2 cm/sec while the width w is increasing at a rate of 2 cm/sec. When l = 12 cm and w=5 cm, find the rate of change of the area, the...

The length l of a rectangle is decreasing at the rate of 2 cm/sec while the width w is increasing at a rate of 2 cm/sec. When l = 12 cm and w=5 cm, find the rate of change of the area, the perimeter and the length of the diagonal.

justaguide | Certified Educator

The length l of a rectangle is decreasing at the rate of 2 cm/sec while the width w is increasing at a rate of 2 cm/sec.

The area of the rectangle is `A = l*w` , perimeter `P = 2*(l + w)` and the diagonal D is related by `D^2 = l^2 + w^2`

When l and w are changing with respect to time,

`(dA)/(dt) = l*(dw)/(dt) + w*(dl)/(dt)` , `(dP)/(dt) = 2*((dl)/(dt) + (dw)/(dt)) ` and `2*D*((dD)/(dt)) = 2*l*(dl)/(dt) + 2*w*(dw)/(dt)`

When l = 12 cm and w = 5 cm,

the rate of change of the area is: 12*(-2) + 5*(2) = -24 + 10 = -14 cm^2/sec

the rate of change of the perimeter is: 2*(-2 + 2) = 0

and the length of the diagonal: `(12*(-2) + 5*2)/(sqrt(12^2 + 5^2))`

=> `(-24 + 10)/13`

=> `-14/13` cm/sec

The required rate of change of area is -14 cm^2/sec, the rate of change of the perimeter is 0 and the rate of change of the diagonal is `-14/13` cm/sec.