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.
To find the rate at which the volume of the box is changing, we need to use the concept of derivatives.
Given the formula for the volume of the rectangular box V = x^2h, we can differentiate with respect to time (t).
dV/dt = d(x^2h)/dt
Now, let's substitute the given values:
x = 4 cm (edge length of the base)
dx/dt = 2 cm/min (rate at which the edge length is increasing)
h = 15 cm (height of the box)
dh/dt = -3 cm/min (rate at which the height is decreasing)
Now, differentiate V = x^2h with respect to t:
dV/dt = d(x^2h)/dt
= 2x(dx/dt)h + x^2(dh/dt)
Substituting the given values:
dV/dt = 2(4)(2)(15) + (4^2)(-3)
= 120 + 48(-3)
= 120 - 144
= -24 cm^3/min
Therefore, the volume of the box is changing at a rate of -24 cm^3/min. Since the volume is decreasing, the negative sign indicates that the volume is decreasing.