Question

Show that the largest circular cylinder that can be inscribed in a sphere is one in which the ratio of diameter to altitude is \(\sqrt{2}: 1\) (Kepler).

Short answer

Answer: The ratio of diameter to altitude of the largest possible cylinder that can fit inside a sphere is \(\sqrt{2}:1\).

Step-by-step solution

Express the problem using variables

Let the radius of the sphere be \(r\). Let the radius and the height of the inscribed cylinder be denoted by \(x\) and \(h\), respectively. The volume of the cylinder, \(V_{cylinder}\), can be expressed as: $$ V_{cylinder} = \pi x^2 h $$

Determine the relationship between the variables

We can use the Pythagorean theorem to determine the relationship between the radius \(r\) of the sphere, the radius \(x\) of the cylinder, and the height \(h\): $$ r^2 = x^2 + \left(\frac{h}{2}\right)^2 $$

Express the volume in terms of one variable

Rearranging the equation obtained from the Pythagorean theorem: $$ h = 2\sqrt{r^2 - x^2} $$ Substitute the expression for \(h\) into the volume equation of the cylinder: $$ V_{cylinder} = \pi x^2 (2\sqrt{r^2 - x^2}) = 2\pi x^2\sqrt{r^2 - x^2} $$

Maximize the volume of the cylinder

To maximize the volume of the cylinder, we must find the value of \(x\) that maximizes the function: $$ f(x) = 2\pi x^2\sqrt{r^2 - x^2} $$ To do this, calculate the derivative of \(f(x)\) with respect to \(x\) and set it equal to zero: $$ \frac{d}{dx} (2\pi x^2\sqrt{r^2 - x^2}) = 0 $$ The derivative is: $$ \frac{d}{dx} \Bigl(2\pi x^2\sqrt{r^2 - x^2}\Bigr) = 2\pi \Bigl(2x\sqrt{r^2 - x^2} - x^3\bigl(-\frac{2x}{2\sqrt{r^2 - x^2}}\bigr)\Bigr) $$ Setting the derivative equal to zero: $$ 2\pi\Bigl(2x\sqrt{r^2 - x^2} + x^3 \frac{x}{\sqrt{r^2 - x^2}}\Bigr) = 0 $$ Solving for \(x\): $$ 2x\sqrt{r^2 - x^2} + x^3\frac{x}{\sqrt{r^2 - x^2}} = 0 $$

Find the optimal relationship between the dimensions

Since we are only trying to find the ratio of diameter to altitude, we can make use of the equation: $$ h = 2\sqrt{r^2 - x^2} $$ Solving for the relationship between \(x\) and \(r\): $$ \frac{x}{r} = \frac{1}{\sqrt{2}} $$ Therefore, the relationship between the diameter and the altitude is \(\sqrt{2}:1\), as required.

Key concepts

Cylinder in a Sphere

Understanding the problem of inscribing a cylinder in a sphere involves visualizing the cylinder snugly fitting within the sphere. The cylinder touches the inner wall of the sphere perfectly.
Imagine the sphere has a radius, denoted as \( r \), while the cylinder's base (circle) has a smaller radius \( x \), and its height is indicated by \( h \). Our goal is to determine the optimal size of this inscribed cylinder.
By applying kepler's proposition in this scenario, we aim to achieve the largest possible cylinder, maximizing its volume under such constraints.

Optimization in Geometry

Optimization in geometry often involves maximizing or minimizing certain parameters, like area or volume, under specific constraints.
Here, we are interested in maximizing the volume of the cylinder, which requires us to adjust its dimensions within the sphere.
The volume \( V \) is given by the formula: \[ V = \pi x^2 h \]
Using optimization techniques, we take derivatives to identify critical points, focusing on maximizing the function.

• Set the derivative equal to zero to find points where the function may have maximum volume.
• Check if these points indeed fall within the permissible solution space.

Optimization facilitates efficient design, where resources or space are limited.

Pythagorean Theorem

The Pythagorean Theorem plays a key role in relating the dimensions of the sphere and the cylinder.
It provides a fundamental relationship that simplifies our mathematical modeling.
In this context, the theorem establishes:\[ r^2 = x^2 + \left(\frac{h}{2}\right)^2 \]
This equation implies that the sum of the squares of the cylinder's radius and half its height equals the square of the sphere's radius.

• This forms the basis for further calculations.
• Helps in reducing the variables to solve for the optimum dimensions of the cylinder.

The theorem is crucial, as it consistently defines relationships in Euclidean geometry.

Derivative in Mathematics

Derivatives are essential in finding optimal points, like maxima or minima of functions commonly encountered in calculus.
Here, they help in determining the cylinder dimensions to maximize volume efficiently.
Consider the function \( f(x) = 2 \pi x^2 \sqrt{r^2 - x^2} \), representing the volume.

• Derivative \( \frac{d}{dx} \) is taken.
• It's set to zero to find potential maximum volume configurations.
• It also helps confirm the nature of these points - whether they represent maxima, minima, or saddle points.

Through differentiation, mathematicians can solve complex geometry problems by studying the rate of changes and relationships amongst involved variables.