Question

A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x^2h cm^3. Find the rate at which the volume of the box is changing when the edge length of the base is 4 cm, the edge length of the base is increasing at a rate of 2 cm/min, the height of the box is 15 cm, and the height is decreasing at a rate of 3 cm/min.

The volume of the box is decreasing at a rate of 192 cm^3/min.
The volume of the box is increasing at a rate of 288 cm^3/min.
The volume of the box is decreasing at a rate of 288 cm^3/min.
The volume of the box is increasing at a rate of 192 cm^3/min.

Answer 1

option 4 is correct i.e. The volume of the box is increasing at a rate of 192 cm^3/min.

Given : Volume of the rectangular box  = x²h 
where x is edge and h is height.
The edge and the height are varying with time, therefore, we write,x = x(t)
h = h(t)
dh/dt = -3 and we shall calculate when x = 4, dx/dt = 2 and when h=15 
V = x²h dV/dt = (2x × dx/dt × h) + (x² × dh/dt) dV/dt = 2×4×2×15 + (4)^2 ×(-3) dV/dt = 240 - 48
dV/dt = 192 
Because dV/dt is positive, hence the volume is increasing