The radius of a right circular cone is increasing at a rate of 2 inches per second and its height is decreasing at a rate of 2 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 20 inches? NOTE: The volume of a cone with base radius r and height h is given by V=13πr2h.

 To find out how the volume of the cone is changing, we need to calculate the rate of change of the volume with respect to time, which is known as the derivative of the volume with respect to time (dV/dt). The volume V of a right circular cone with base radius r and height h is given by:

\[ V = \frac{1}{3}\pi r^2 h \]

We are given that the radius is increasing at a rate of 2 inches per second (dr/dt = 2 inch/s) and the height is decreasing at a rate of 2 inches per second (dh/dt = -2 inch/s). We want to find dV/dt when r = 40 inches and h = 20 inches.

To find dV/dt, we need to take the derivative of the volume equation with respect to time using the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function:

\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{3}\pi r^2 h\right) \]

\[ \frac{dV}{dt} = \frac{1}{3}\pi\left(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\right) \]

Plugging in the values we have:

\[ \frac{dV}{dt} = \frac{1}{3}\pi\left(2 \cdot 40 \cdot 2 \cdot 20 + 40^2 \cdot (-2)\right) \]

\[ \frac{dV}{dt} = \frac{1}{3}\pi\left(3200 - 3200\right) \]

\[ \frac{dV}{dt} = \frac{1}{3}\pi\left(0\right) \]

\[ \frac{dV}{dt} = 0 \]

So the rate at which the volume of the cone is changing when the radius is 40 inches and the height is 20 inches is 0 cubic inches per second. This means that at the moment when the radius is 40 inches and the height is 20 inches, the volume of the cone is momentarily not changing.