Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the greatest volume that a right circular cylinder can have if it is inscribed in a sphere of radius \(r\).

Step-by-Step Solution

Answer

The maximum volume of the cylinder is \(\frac{4}{3} \pi r^3\).

Step 1: Identify the Problem

The problem requires us to find the greatest (or maximum) volume of a right circular cylinder that can fit inside a sphere with a given radius \(r\). This is an optimization problem involving the geometry of a sphere and a cylinder.

Step 2: Understand the Geometry

Visualize a right circular cylinder inscribed in a sphere. The height of the cylinder is \(h\) and the radius of its base is \(R\). The diameter of the sphere, which is \(2r\), must equal the length of the cylinder's diagonal, dictated by the Pythagorean theorem: \(h^2 + (2R)^2 = (2r)^2\).

Step 3: Formulate the Cylinder Volume

The volume \(V\) of a cylinder is given by \(V = \pi R^2 h\). We need a way to express \(V\) solely in terms of \(r\) for maximal simplification.

Step 4: Substitute for the Cylinder Height

From the relationship \(h^2 + (2R)^2 = (2r)^2\), express \(h^2\) as \(h^2 = 4r^2 - 4R^2\), so \(h = \sqrt{4r^2 - 4R^2}\). Substitute \(h = 2\sqrt{r^2 - R^2}\) into the volume formula: \(V = \pi R^2 (2\sqrt{r^2 - R^2})\).

Step 5: Express Volume as a Single Variable Function

The volume becomes \(V(R) = 2\pi R^2 \sqrt{r^2 - R^2}\). Now, \(V\) is a function of a single variable \(R\).

Step 6: Derive and Find Critical Points

To find the maximum volume, take the derivative of \(V(R)\) with respect to \(R\), and set it to zero: \(V'(R) = 4\pi R \sqrt{r^2 - R^2} - \frac{2\pi R^3}{\sqrt{r^2 - R^2}} = 0\). Solve for \(R\) in terms of \(r\) to find the critical points.

Step 7: Simplify and Solve

Notice that \(4r^2 - 4R^2 - R^2 = 0\) leads to \(r^2 = 3R^2\), therefore \(R = \frac{r}{\sqrt{3}}\). Substituting back, the optimal \(h\) becomes \(h = \frac{2r}{\sqrt{3}}\).

Step 8: Evaluate Volume at Critical Points

Substitute \(R = \frac{r}{\sqrt{3}}\) and \(h = \frac{2r}{\sqrt{3}}\) into the volume equation to find \(V = \frac{4}{3}\pi r^3\).

Step 9: Conclude With Maximum Volume

Thus, the maximum volume of the cylinder inscribed in a sphere of radius \(r\) is \(\frac{4}{3} \pi r^3\).

Key Concepts

Inscribed Cylinder in Sphere

When we think of an inscribed cylinder, imagine a cylinder snugly fitting inside a sphere, like a pencil inside a ball. The sphere's radius is given as \(r\), and our task is to find a way to make this cylinder's volume as large as possible. This involves understanding the spatial relationship between the sphere and the cylinder. Remember, the diameter of the sphere is \(2r\). This diagonal runs through the center of the cylinder, playing a crucial role as it equals the cylinder's diagonal in its profile view.

• The problem is essentially a maximization problem within a confined space – the sphere.
• The diagonal of the cylinder, when expressed in terms of the cylinder's height \(h\) and base radius \(R\), adheres to the Pythagorean theorem: \(h^2 + (2R)^2 = (2r)^2\).

This relationship is vital for setting up equations to solve for maximal volume later on.

Volume Maximization

Maximizing the volume of the cylinder inside the sphere is the ultimate goal. To do this, we first need to express the volume formula in a way that can be easily differentiated and solved for its critical points.

• The basic formula for the volume of a cylinder is \(V = \pi R^2 h\).
• However, due to the spatial restrictions set by the sphere, \(h\) must be expressed in terms of \(R\) and \(r\).

Using the relationship derived from the Pythagorean theorem, the height \(h\) can be rewritten as \(h = 2\sqrt{r^2 - R^2}\). This transformation allows the volume formula to solely depend on \(R\) and \(r\):\[V(R) = 2\pi R^2 \sqrt{r^2 - R^2}\]Here, \(V\) becomes a function in terms of \(R\), which is essential for applying calculus to find the maximum value.

Geometry of Solids

Geometry plays a fundamental role in solving optimization problems involving solid shapes like cylinders and spheres. Understanding their geometric properties and relationships is crucial.

• The sphere in question provides a boundary condition, with its diameter directly influencing the cylinder that can fit within its confines.
• The cylinder's dimensions, height \(h\) and radius \(R\), are bound by the constants of the sphere's radius \(r\), and the equation derived from the Pythagorean theorem translates these geometric constraints into algebraic expressions: \(h^2 + (2R)^2 = (2r)^2\).

Understanding this interplay forms the foundation for translating the geometric constraints into mathematical functions that can be optimized using calculus.

Derivative and Critical Points

To find the maximum volume, we must delve into calculus concepts like derivatives and critical points. By differentiating the volume function \(V(R)\) with respect to \(R\), we prepare ourselves to find where this function increases, decreases, or achieves its maximum.Start by taking the derivative:\[V'(R) = 4\pi R \sqrt{r^2 - R^2} - \frac{2\pi R^3}{\sqrt{r^2 - R^2}}\]Setting this derivative equal to zero helps locate the critical points:

• Simplify the expression \(4r^2 - 4R^2 - R^2 = 0\) to derive implicitly \(r^2 = 3R^2\).
• This simplifies to the critical radius \(R = \frac{r}{\sqrt{3}}\).
• Substituting back, the optimal height \(h\) is then \(h = \frac{2r}{\sqrt{3}}\).

Evaluating the volume function at this critical point ensures that we have indeed found the maximum volume the inscribed cylinder can possess, \(V = \frac{4}{3} \pi r^3\).