Question

If the height of a cone of greatest volume that can be inscribed in a sphere of radius $ R $ is $ kR $, then the ratio of the volume of the cone to the volume of the sphere is:

• A. \( 8:27 \)
• B. \( 27:64 \)
• C. \( 8:125 \)
• D. \( 4:5 \)

Hint:

To maximize the volume of a cone inscribed in a sphere, express the volume in terms of one variable (e.g., \( k \)) and use calculus to find the maximum.

Answer 1

The Correct Option is A

Step 1: Set up the geometry.
Cone’s height \( h = kR \), base radius \( r \). Sphere’s centre at origin, equation \( x^2 + y^2 + z^2 = R^2 \). Base at \( z = -d \), vertex at \( (0, 0, h - d) \).
Base: \( x^2 + y^2 = R^2 - d^2 \), so \( r = R \sqrt{1 - (k - 1)^2} \). Vertex: \( h - d = R \), so \( d = R(k - 1) \). Step 2: Compute the volume ratio.
\[ V_{\text{cone}} = \frac{1}{3} \pi R^3 k (2k - k^2), \quad V_{\text{sphere}} = \frac{4}{3} \pi R^3, \quad \text{Ratio} = \frac{2k^2 - k^3}{4}. \] Step 3: Maximize the volume.
Maximize \( 2k^2 - k^3 \): critical point at \( k = \frac{4}{3} \). Substitute:
\[ \text{Ratio} = \frac{2 \left( \frac{4}{3} \right)^2 - \left( \frac{4}{3} \right)^3}{4} = \frac{8}{27} \quad \Rightarrow \quad 8:27. \]