Maximizing the Volume of a Cone Inscribed in a Sphere

Imagine a sphere of a given radius. When a cone is inscribed in this sphere, how can we find the height of the cone that maximizes its volume? This question is not only geometrically intriguing but also practically useful in various fields. Let's explore how to calculate the optimal height of such a cone with a sphere having a radius of (frac{3}{2}).

Geometric Relations and Volume Formula

When a cone is inscribed in a sphere, the relationship between the radius of the sphere (R frac{3}{2}), the height of the cone (h), and the radius of the base of the cone (r) can be derived from the geometry of the situation. The volume of a cone is given by the formula:

[V frac{1}{3} pi r^2 h]

The radius of the base of the cone can be expressed in terms of the height and the radius of the sphere using the Pythagorean theorem:

[R^2 r^2 (R - h)^2]

Substituting the given radius of the sphere:

[left(frac{3}{2}right)^2 r^2 left(frac{3}{2} - hright)^2]

Simplifying the above equation:

[frac{9}{4} r^2 left(frac{9}{4} - 3h h^2right)]

Solving for (r^2):

[r^2 3h - h^2]

Substituting (r^2) back into the volume formula:

[V frac{1}{3} pi (3h - h^2) h frac{1}{3} pi (3h^2 - h^3)]

Finding the Maximum Volume

To maximize the volume, we take the derivative of (V) with respect to (h) and set it to zero:

[V' frac{1}{3} pi (6h - 3h^2)]

Setting the first derivative equal to zero:

[6h - 3h^2 0 Rightarrow 3h(2 - h) 0]

This gives solutions (h 0) or (h 2). Checking the second derivative:

Second derivative: (V'' frac{1}{3} pi (6 - 6h))

Evaluating at (h 2):

[V''(2) frac{1}{3} pi (6 - 12) -frac{6}{3} pi -2pi]

Since the second derivative is negative, (h 2) gives a local maximum. Therefore, the height of the cone that maximizes its volume is:

boxed{2}

Additional Insight

Let's also verify this result using a slightly different approach. Consider the sphere with radius (frac{3}{2}). Let the distance from the center of the sphere to the base of the cone be (X). Then:

Radius of the cone: (sqrt{left(frac{3}{2}right)^2 - X^2}) Height of the cone: (H frac{3}{2} - X)

The volume of the cone is given by:

[V frac{1}{3} pi left(left(frac{3}{2}right)^2 - X^2right) left(frac{3}{2} - Xright)]

Simplifying:

[V frac{1}{3} pi left(frac{9}{4} - X^2right) left(frac{3}{2} - Xright)]

Finding the derivative and setting it to zero:

[V' frac{1}{3} pi left[left(frac{9}{4} - X^2right) - 2Xleft(frac{3}{2} - Xright)right]]

Solving:

[0 left(frac{9}{4} - X^2right) - 2Xleft(frac{3}{2} - Xright)]

Solving for (X):

[X frac{1}{2}, -frac{3}{2}]

Ignoring the negative solution, we find:

[H 2]

Conclusion

Thus, the height of the cone with the largest volume inscribed in a sphere with radius (frac{3}{2}) is clearly (2). This result is a beautiful interplay between geometry, calculus, and mathematics in general, showcasing the power of analytical approaches in solving real-world problems.