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 12 cm, the 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.
you want dv/dt, where
v = x^2 h
The good old product rule says that
dv/dt = 2xh dx/dt + x^2 dh/dt
You have all the stuff you need, so just plug it in and watch the fun!
To find the rate at which the volume of the box is changing, we need to use the chain rule from calculus.
First, let's define our variables:
x = edge length of the base (in cm)
h = height of the box (in cm)
V = volume of the box (in cm^3)
Given information:
x = 12 cm (edge length of the base)
dx/dt = -2 cm/min (rate at which the edge length of the base is decreasing)
h = 6 cm (height of the box)
dh/dt = 1 cm/min (rate at which the height is increasing)
We are looking for dV/dt, the rate at which the volume is changing.
Volume of the box is given by:
V = x^2h.
To find dV/dt, we need to differentiate both sides of the equation with respect to time t.
dV/dt = d/dt (x^2h)
Using the product rule of differentiation, we can calculate this as follows:
dV/dt = 2xh * dx/dt + x^2 * dh/dt
Substituting the given values:
dV/dt = 2(12)(6)(-2) + (12^2)(1)
Simplifying:
dV/dt = -288 + 144
dV/dt = -144 cm^3/min
Therefore, the rate at which the volume of the box is changing is -144 cm^3/min.