A rectangular box has a width of 3cm and a length of 2cm. The volume of the box is decreasing at a rate of 2 cm 3/min, with the width and the length being held constant. What is the rate of change of the height, in cm/min, when the height is 5cm

dh/dt  = 0,111 cm/min

Step-by-step explanation:

Volume of the rectangular box is:

V = V(b) = Area of the rectanglar base * height

Area of the base is :  3*2  = 6 cm²

V(b)  = 6*h

Differentiating in relation to time  in both sides of the equation give:

d(V(b))/ dt  =  6* dh/dt  (1)

According to problem statement volume of the box is decreasing at the rate of  2 cm each 3 min then by rule of three

2 cm    ⇒   3 min

 x        ⇒    1 min

x  = 2/3  = 0,67 cm/min

As  dimensions of the box ( length and width ) will be constants  decreasing in the volume of the cube does not depend on the level of the height, and we know dV(b) / dt  = 0.67 cm/min, then in equation 1

0.67 = 6*dh/dt

dh/dt  = 0.67/6

dh/dt  = 0,111 cm/min