The length l of a rectangle is decreasing at a rate of 3 cm/sec while the width w is increasing at a rate of 3 cm/sec. When l=5 cm and w=12 cm, find the rates of change of the area, the perimeter, and the lengths of the diagonals of the rectangle. Determine which of these quantities are increasing, decreasing, or constant. a) The rate of changes of the area of the rectangle is cm^2/sec
Is the area increasing, decreasing, or constant?
b) The rate of change of the perimeter of the rectangle is cm/sec.
Is the perimeter increasing, decreasing, or constant?
c) The rate of change of the length of the diagonals of the rectangle is cm/sec.
Is the length of the diagonals increasing, decreasing, or constant.

1) dA/dt = -21 cm²/s and Area is decreasing

2)dP/dt = 0 and perimeter is constant

3) dD/dt = 21/13 cm/s and diagonal is increasing

Step-by-step explanation:

We are given that;

length l of a rectangle is decreasing at a rate of 3 cm/sec.

Thus, dl/dt = -3 cm/sec

Also, the width w is increasing at a rate of 3 cm/sec. Thus;

dw/dt = 3 cm/sec

When l=5 cm and w=12 cm;

A) Area is given by the formula;

A = lw

The rate at which area is increasing is;

dA/dt = l(dw/dt) + w(dl/dt)

Plugging in the relevant values;

dA/dt = 5(3) + 12(-3)

dA/dt = 15 - 36

dA/dt = -21 cm²/s

This is less than 0.thus, A is decreasing.

B) Formula for perimeter is;

P = 2l + 2w

rate of change of perimeter is;

dP/dt = 2(dw/dt) + 2(dl/dt)

Plugging in the relevant values, we have;

dP/dt = 2(-3) + 2(3)

dP/dt = 0

Thus,Perimeter is constant

C) the length of the diagonal of a rectangle is given by;

D = √(w² + l²)

Rate of change of diagonal is;

dD/dt = [2w(dw/dt) + 2l(dl/dt)]/(2√(w² + l²))

2 will cancel out in numerator and denominator to give;

dD/dt = [w(dw/dt) + l(dl/dt)]/(√(w² + l²))

Plugging in the relevant values gives;

dD/dt = [12(3) + 5(-3)]/(√(12² + 5²))

dD/dt = (36 - 15)/13

dD/dt = 21/13 cm/s

This is greater than 0.

Thus, diagonal is increasing.