Water is leaking out of an inverted conical tank at a rate of 8,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)

To solve this problem, we need to find the rate at which water is being pumped into the tank. Let's use related rates and the geometry of the tank to set up the necessary equations.

### Step 1: Understand the Problem and Define Variables
1. **Tank Dimensions**:
   - The tank is an inverted cone with a height of \( h = 600 \) cm (since 6 m = 600 cm).
   - The diameter of the top of the tank is 400 cm, giving a radius \( R = 200 \) cm.

2. **Water Dimensions**:
   - Let \( r \) be the radius of the water surface at a given height \( y \).
   - Let \( V \) represent the volume of water in the tank at height \( y \).
   
3. **Given Rates**:
   - Water is leaking out at a rate of \( \frac{dV}{dt} = -8500 \, \text{cm}^3/\text{min} \).
   - The height of the water level is increasing at a rate of \( \frac{dy}{dt} = 20 \, \text{cm/min} \) when \( y = 200 \, \text{cm} \).

### Step 2: Relate Variables Using Geometry
Since the tank is conical, the radius \( r \) of the water at height \( y \) is proportional to \( y \). Using similar triangles, we have:
\[
\frac{r}{y} = \frac{200}{600} = \frac{1}{3} \Rightarrow r = \frac{y}{3}
\]

### Step 3: Volume of Water in the Tank
The volume \( V \) of a cone is given by:
\[
V = \frac{1}{3} \pi r^2 y
\]
Substitute \( r = \frac{y}{3} \):
\[
V = \frac{1}{3} \pi \left( \frac{y}{3} \right)^2 y = \frac{1}{3} \pi \frac{y^2}{9} y = \frac{\pi}{27} y^3
\]

### Step 4: Differentiate \( V \) with Respect to \( t \)
Differentiate both sides with respect to \( t \):
\[
\frac{dV}{dt} = \frac{\pi}{27} \cdot 3y^2 \cdot \frac{dy}{dt} = \frac{\pi}{9} y^2 \frac{dy}{dt}
\]

### Step 5: Substitute Known Values
We need to find \( \frac{dV}{dt} \) when \( y = 200 \) cm and \( \frac{dy}{dt} = 20 \) cm/min. Substitute these values into the equation:
\[
\frac{dV}{dt} = \frac{\pi}{9} \cdot (200)^2 \cdot 20
\]
Calculate \( (200)^2 = 40000 \):
\[
\frac{dV}{dt} = \frac{\pi}{9} \cdot 40000 \cdot 20 = \frac{\pi}{9} \cdot 800000 = \frac{800000\pi}{9} \approx 279252 \, \text{cm}^3/\text{min}
\]

### Step 6: Account for the Leak
Since water is leaking at \( 8500 \, \text{cm}^3/\text{min} \), the rate at which water is pumped into the tank \( P \) must account for this leak:
\[
P = 279252 + 8500 \approx 287752 \, \text{cm}^3/\text{min}
\]

### Answer
The rate at which water is being pumped into the tank is approximately **287752 cm³/min**.

---

Would you like more details or have any questions about the solution?

Here are some related questions:
1. How would the solution change if the tank were cylindrical instead of conical?
2. How can we adapt this method if the diameter of the tank varied along its height?
3. What if the rate of water leaking changed over time instead of being constant?
4. How would the answer change if we measured in liters instead of cm³?
5. Could we solve for the rate at different water heights in the tank?

**Tip:** When dealing with geometric shapes and changing quantities, always check if similar triangles can simplify relationships!

Water is leaking out of an inverted conical tank at a rate of 8,500 cm³/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)

Solve a related rates problem involving an inverted conical tank with water leaking and being pumped in simultaneously. Learn how to use the volume formula for a cone and apply differentiation to find the rate at which water is being pumped in. Suitable for Grades 11-12 and early college calculus students.

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