Water is leaking out of an inverted conical tank at a rate of 6 liters per minute. At
the same time that water is being pumped into the tank at a constant rate. The tank has a
height of 18 meters and the diameter at the top is 12 meters. If the water level is rising at a
rate of 15 centimeters per minute when the height of the water is 2 meters, find the rate at
which water is being pumped into the tank in milliliters per minute.

We are dealing with a problem involving related rates, specifically in the context of an inverted conical tank. Let's solve it step by step:

### Step 1: Understanding the geometry of the tank
The tank is an inverted cone, and we are given:
- The height of the tank: \( h_{\text{tank}} = 18 \) meters.
- The diameter at the top of the tank: \( d = 12 \) meters, which gives a radius \( r_{\text{tank}} = \frac{12}{2} = 6 \) meters.

The relationship between the radius and height for any cross-section of water in the tank is proportional. So, the radius \( r \) at any height \( h \) can be described as:

\[
\frac{r}{h} = \frac{6}{18} = \frac{1}{3}, \quad \text{so} \quad r = \frac{h}{3}.
\]

### Step 2: Volume of the cone
The volume \( V \) of water in the cone at height \( h \) is given by the formula for the volume of a cone:

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

Since \( r = \frac{h}{3} \), we substitute this into the formula:

\[
V = \frac{1}{3} \pi \left( \frac{h}{3} \right)^2 h = \frac{1}{3} \pi \frac{h^3}{9} = \frac{\pi h^3}{27}.
\]

### Step 3: Differentiate the volume with respect to time
We are interested in the rate of change of the volume, so we differentiate both sides with respect to time \( t \):

\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi h^3}{27} \right) = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi h^2}{9} \frac{dh}{dt}.
\]

### Step 4: Applying the given values
We are given:
- The rate at which the water level is rising: \( \frac{dh}{dt} = 15 \) centimeters per minute \( = 0.15 \) meters per minute.
- The height of the water at the moment we are interested in: \( h = 2 \) meters.
- The rate at which water is leaking: \( 6 \) liters per minute \( = 6000 \) milliliters per minute.

Substitute \( h = 2 \) meters and \( \frac{dh}{dt} = 0.15 \) meters per minute into the equation for \( \frac{dV}{dt} \):

\[
\frac{dV}{dt} = \frac{\pi (2)^2}{9} \cdot 0.15 = \frac{\pi \cdot 4}{9} \cdot 0.15 = \frac{0.6\pi}{9} \text{ cubic meters per minute}.
\]

Since 1 cubic meter equals 1,000,000 milliliters, we convert this rate to milliliters per minute:

\[
\frac{dV}{dt} = \frac{0.6\pi}{9} \times 1,000,000 \approx 209,439.51 \text{ milliliters per minute}.
\]

### Step 5: Account for the leakage
The net rate of change of the volume is the rate at which water is being pumped into the tank minus the rate at which water is leaking out. Let \( P \) be the rate at which water is being pumped into the tank in milliliters per minute. We have:

\[
P - 6000 = 209,439.51.
\]

Solving for \( P \):

\[
P = 209,439.51 + 6000 = 215,439.51 \text{ milliliters per minute}.
\]

Thus, the rate at which water is being pumped into the tank is approximately **215,440 milliliters per minute**.

---

Would you like further clarification or more details on any part of the solution? Here are five related questions to expand on this topic:

1. How would the solution change if the diameter of the tank were different?
2. What would happen if the rate of water leaking out was not constant?
3. How does the relationship between radius and height affect the volume calculation?
4. Can we calculate the rate of pumping if the tank were cylindrical instead of conical?
5. How can this approach be applied to non-conical tanks with other shapes?

**Tip:** Always ensure that units are consistent throughout the problem to avoid confusion when converting between volume measurements.

Learn how to solve a related rates problem involving an inverted conical tank. Calculate the rate at which water is pumped into the tank given the dimensions and rates of change. Suitable for high school students studying calculus and related rates.

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