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This problem involves related rates for a right circular cone. Let's break it down step by step.

The formula for the volume \( V \) of a cone is:
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \( r \) is the radius,
- \( h \) is the height.

Given:
- The rate of change of the radius \( \frac{dr}{dt} = 4 \) inches per second,
- The rate of change of the height \( \frac{dh}{dt} = -3 \) inches per second (it's decreasing),
- At the moment when \( r = 40 \) inches and \( h = 20 \) inches, we need to find the rate of change of the volume \( \frac{dV}{dt} \).

### Step 1: Differentiate the Volume Formula with Respect to Time
Differentiate both sides of the volume formula using the product and chain rules:
\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi r^2 h \right)
\]
Using the product rule:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right)
\]
Now substitute the given values:
- \( r = 40 \),
- \( h = 20 \),
- \( \frac{dr}{dt} = 4 \),
- \( \frac{dh}{dt} = -3 \).

### Step 2: Calculate the Rate of Change of the Volume
Substitute these values into the differentiated formula:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2(40)(4)(20) + (40)^2(-3) \right)
\]
Simplify each term:
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 2(40)(4)(20) + 1600(-3) \right)
\]
\[
\frac{dV}{dt} = \frac{1}{3} \pi \left( 6400 - 4800 \right)
\]
\[
\frac{dV}{dt} = \frac{1}{3} \pi (1600)
\]
\[
\frac{dV}{dt} = \frac{1600 \pi}{3}
\]

So, the rate of change of the volume is \( \frac{1600 \pi}{3} \) cubic inches per second, which is approximately **1675.52 cubic inches per second** when using \( \pi \approx 3.1416 \).

Would you like more details on this, or have any questions? 

Here are some related questions for further practice:
1. How does the rate of change of volume depend on the radius and height?
2. What happens to the rate of change of volume if both the radius and height are increasing?
3. Can you derive the related rates for the surface area of the cone?
4. What if the radius was decreasing at the same rate as the height? How would the volume change?
5. How would the volume change if the height remained constant but the radius changed?

**Tip**: Always keep track of the signs when dealing with related rates, especially when values are increasing or decreasing.

The radius of a right circular cone is increasing at a rate of 4 inches per second and its height is decreasing at a rate of 3 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?

Solve a related rates problem where the radius of a cone is increasing at 4 inches per second and the height is decreasing at 3 inches per second. Learn step-by-step how to find the rate of change of the volume when the radius is 40 inches and the height is 20 inches. Suitable for students in Grades 11-12.

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