Question

Inscribing a Cone Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.

Short answer

The volume of the largest cone that can be inscribed in a sphere of radius 3 is \( \frac{32}{3} \pi \) cubic units.

Step-by-step solution

Find the relationship between the radius and height

Since the cone is inscribed in a sphere, the radius 'r' of the cone plus its height 'h' is equal to the diameter of the sphere, or twice the given radius. Hence, \( r + h = 2 \times 3 = 6 \) . Also, by the Pythagorean theorem on the right triangle formed by the cone's height, radius, and slant height (which is the radius of the sphere), we have \( r^2 + h^2 = (2 \times 3)^2 = 36 \).

Express 'h' in terms of 'r'

From the equation \( r + h = 6 \), we can express 'h' in terms of 'r' as \( h = 6 - r \).

Substitute 'h' in the volume formula

Substituting \( h = 6 - r \) in the cone volume formula \( V = \frac{1}{3} \pi r^2 h \) gives \( V = \frac{1}{3} \pi r^2 (6 - r) = 2 \pi r^2 - \frac{1}{3} \pi r^3 \).

Find the maximum volume

Next, differentiate the volume expression with respect to 'r' and set it equal to zero to find the radius at the maximum volume. This gives \( \frac{dV}{dr} = 4 \pi r - \pi r^2 = 0 \). Solving for 'r' yields two solutions, \( r = 0 \) and \( r = 4 \). The solution \( r = 0 \) corresponds to a minimum volume (no cone), so the maximum volume is attained when \( r = 4 \).

Calculate the maximum volume

Finally, substitute \( r = 4 \) in the volume expression to get the maximum volume. This gives \( V_{max} = 2 \pi (4)^2 - \frac{1}{3} \pi (4)^3 = 32 \pi - \frac{64}{3} \pi = \frac{32}{3} \pi \) cubic units.