The maximum volume (in \( \mathrm{cu} . \mathrm{m} \) ) of the right circul cone having slant height 3 m is

To find the **maximum volume** of a **right circular cone** with a fixed **slant height** of \( l = 3 \) meters, we'll follow these steps: 1. **Understand the Relationships:** - **Volume of a Cone:** \( V = \frac{1}{3} \pi r^2 h \) - **Slant Height Relation:** \( l^2 = r^2 + h^2 \) 2. **Express Height in Terms of Radius:** \[ h = \sqrt{l^2 - r^2} = \sqrt{9 - r^2} \] 3. **Express Volume Solely in Terms of Radius:** \[ V = \frac{1}{3} \pi r^2 \sqrt{9 - r^2} \] 4. **Maximize the Volume Function:** - To maximize \( V \), take the derivative with respect to \( r \) and set it to zero. - Alternatively, parameterize using trigonometric substitution: - Let \( r = 3 \sin \theta \), then \( h = 3 \cos \theta \). - Volume becomes: \[ V = \frac{1}{3} \pi (9 \sin^2 \theta)(3 \cos \theta) = 9 \pi \sin^2 \theta \cos \theta \] - Maximize \( \sin^2 \theta \cos \theta \): - Taking the derivative and setting it to zero leads to: \[ \sin \theta = \sqrt{\frac{2}{3}} \] - Thus, \( \cos \theta = \frac{1}{\sqrt{3}} \). 5. **Calculate the Maximum Volume:** \[ V_{\text{max}} = 9 \pi \left(\frac{2}{3}\right) \left(\frac{1}{\sqrt{3}}\right) = 2 \pi \sqrt{3} \, \text{m}^3 \] **Final Answer:** The maximum volume is \( 2 \pi \sqrt{3} \) cubic meters.