Question

Find the height of the right circular cone of maximum volume, which is inscribed in a sphere of radius 12 cm.

Hint:

In optimization problems involving volumes, use calculus to find critical points by setting the derivative equal to zero.

Answer 1

Step 1: Understanding the geometry of the problem.
Consider a sphere with radius \( R = 12 \, \text{cm} \) and a right circular cone inscribed in it. Let the height of the cone be \( h \) and the radius of its base be \( r \). The vertex of the cone is at the center of the sphere, and the base lies on the surface of the sphere. The relationship between the height \( h \), the radius \( r \), and the radius of the sphere \( R \) can be derived using the Pythagorean theorem: \[ r^2 + \left( \frac{h}{2} \right)^2 = R^2 \] Substitute \( R = 12 \, \text{cm} \): \[ r^2 + \left( \frac{h}{2} \right)^2 = 12^2 = 144 \] Thus, the radius of the base is: \[ r = \sqrt{144 - \left( \frac{h}{2} \right)^2} \] Step 2: Volume of the cone.
The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] Substitute \( r^2 \) from the previous equation: \[ V = \frac{1}{3} \pi \left( 144 - \left( \frac{h}{2} \right)^2 \right) h \] Simplify the expression for the volume: \[ V = \frac{1}{3} \pi h \left( 144 - \frac{h^2}{4} \right) \] \[ V = \frac{1}{3} \pi \left( 144h - \frac{h^3}{4} \right) \] Step 3: Maximizing the volume.
To find the value of \( h \) that maximizes the volume, we take the derivative of \( V \) with respect to \( h \) and set it equal to zero: \[ \frac{dV}{dh} = \frac{1}{3} \pi \left( 144 - \frac{3h^2}{4} \right) \] Set the derivative equal to zero to find the critical points: \[ 144 - \frac{3h^2}{4} = 0 \] Solving for \( h \): \[ \frac{3h^2}{4} = 144 \] \[ h^2 = \frac{4 \times 144}{3} = 192 \] \[ h = \sqrt{192} = 8\sqrt{3} \, \text{cm} \] Step 4: Conclusion.
The height of the cone that maximizes the volume is \( h = 8\sqrt{3} \, \text{cm} \).