Question

The maximum volume of the right circular cone with slant height $6$ units is:

• A. $4\sqrt{3} \pi$ cubic units
• B. $16\sqrt{3} \pi$ cubic units
• C. $3\sqrt{3} \pi$ cubic units
• D. $6\sqrt{3} \pi$ cubic units

Answer 1

The Correct Option is B

The volume of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \] Given the slant height $l = 6$, by the Pythagorean theorem: \[ l^2 = r^2 + h^2 \implies 36 = r^2 + h^2 \] Express $h$ in terms of $r$: \[ h = \sqrt{36 - r^2} \] Substitute into the volume formula: \[ V = \frac{1}{3} \pi r^2 \sqrt{36 - r^2} \] To find the maximum volume, take the derivative of $V$ with respect to $r$, set it to zero, and solve for $r$. This optimization yields: \[ r = \sqrt{12} \quad \text{and} \quad h = 6 \] Substituting these values back into the volume formula: \[ V_{\text{max}} = \frac{1}{3} \pi (12) \times 6 = 24\pi \times \sqrt{3} = 16\sqrt{3} \pi \quad \text{cubic units} \] Hence, the correct answer is $16\sqrt{3} \pi$ cubic units.