The radii of the circualr ends of a bucket of height 15 cm are 14 cm and r cm `(r lt 14)`. If the volume of the bucket is `5390 cm^(3)`, find the value of r.

Answer

To solve the problem, we need to find the value of \( r \) given the volume of a bucket with two circular ends (a frustum of a cone). The parameters given are:

- Height of the bucket \( h = 15 \) cm
- Radius of the top circular end \( R = 14 \) cm
- Radius of the bottom circular end \( r \) (where \( r < 14 \))
- Volume of the bucket \( V = 5390 \) cm³

### Step-by-step Solution:

1. **Volume Formula for Frustum of a Cone**:
The volume \( V \) of a frustum of a cone is given by the formula:
\[ V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) \]

2. **Substituting Known Values**:
We know \( V = 5390 \) cm³, \( h = 15 \) cm, and \( R = 14 \) cm. We can substitute these values into the volume formula:
\[ 5390 = \frac{1}{3} \pi (15) (14^2 + r^2 + 14r) \]

3. **Calculating Constants**:
First, calculate \( 14^2 \):
\[ 14^2 = 196 \]
Now substitute this back into the equation:
\[ 5390 = \frac{1}{3} \pi (15) (196 + r^2 + 14r) \]

4. **Simplifying the Equation**:
Multiply both sides by \( 3 \) to eliminate the fraction:
\[ 16170 = \pi (15) (196 + r^2 + 14r) \]
Now divide both sides by \( 15\pi \):
\[ \frac{16170}{15\pi} = 196 + r^2 + 14r \]

5. **Calculating \( \frac{16170}{15\pi} \)**:
Using \( \pi \approx \frac{22}{7} \):
\[ 15\pi \approx 15 \times \frac{22}{7} = \frac{330}{7} \]
Now calculate \( \frac{16170 \times 7}{330} \):
\[ \frac{16170 \times 7}{330} = \frac{113190}{330} = 343 \]

6. **Setting Up the Quadratic Equation**:
Now we have:
\[ 343 = 196 + r^2 + 14r \]
Rearranging gives:
\[ r^2 + 14r + 196 - 343 = 0 \]
Simplifying further:
\[ r^2 + 14r - 147 = 0 \]

7. **Factoring the Quadratic**:
To factor \( r^2 + 14r - 147 \), we look for two numbers that multiply to \(-147\) and add to \(14\). These numbers are \(21\) and \(-7\):
\[ (r - 7)(r + 21) = 0 \]

8. **Finding the Roots**:
Setting each factor to zero gives:
\[ r - 7 = 0 \quad \Rightarrow \quad r = 7 \]
\[ r + 21 = 0 \quad \Rightarrow \quad r = -21 \quad (\text{not valid since } r < 14) \]

9. **Conclusion**:
The valid solution is:
\[ r = 7 \text{ cm} \]