Question

A rectangular box has a square base with a edge length of X cm and a height of h cm. The volume of the box is given by V = x^2× H cm^3. Find the rate at which the volume of the box is changing when the edge length of the base is 12 cm, that edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min.

Answer 1

Answer: To find the rate at which the volume of the box is changing, we need to find the derivative of the volume with respect to time, which is the rate of change of the volume with respect to time.

The volume of the box is given by the equation:

V = x^2 * h

To find the rate of change of the volume, we need to find the derivative of this equation with respect to time:

dV/dt = d/dt (x^2 * h)

= 2x * dx/dt * h + x^2 * dh/dt

At the moment when the edge length of the base is 12 cm and the height of the box is 6 cm, we can substitute the values into the equation:

dV/dt = 2 * 12 * (-2) * 6 + 12^2 * 1

= -144 + 144

= 0

So the rate at which the volume of the box is changing is 0 cm^3/min, which means the volume of the box is not changing.

Explanation: