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