Problem 27 - Calculus Early Transcendentals

Question

Find the dimensions of the right circular cylinder of largest volume that can be inscribed in a sphere of radius \(R .\)

Step-by-Step Solution

Answer

Cylinder dimensions: Radius \(\frac{\sqrt{6}}{3}R\), Height \(\frac{2\sqrt{3}}{3}R\).

Step 1: Understand the Problem Context

We need to find the dimensions of a cylinder (i.e., its radius and height) that give it the largest possible volume, while the cylinder is inscribed in a sphere of radius \(R\). The solution requires maximizing the cylinder’s volume with the constraint imposed by the sphere.

Step 2: Set Up the Mathematical Model

Let the radius of the cylinder be \(r\) and the height be \(h\). The volume of the cylinder is given by \(V = \pi r^2 h\). Since the cylinder is inscribed in a sphere of radius \( R \), the relationship between \(r\), \(h\), and \(R\) can be expressed using the Pythagorean theorem in the right triangle: \[(R)^2 = (h/2)^2 + r^2 \]

Step 3: Express Volume in Terms of a Single Variable

From the equation \((R)^2 = (h/2)^2 + r^2\), solve for \(h\): \[h = 2 \sqrt{R^2 - r^2}\]Substitute \(h\) in the volume formula:\[V = \pi r^2 (2 \sqrt{R^2 - r^2}) = 2\pi r^2 \sqrt{R^2 - r^2}\]

Step 4: Differentiate the Volume Function

Take the derivative of the volume function \(V(r) = 2\pi r^2 \sqrt{R^2 - r^2}\) with respect to \(r\): \[V'(r) = 2\pi \left(r^2 \cdot \frac{-r}{\sqrt{R^2 - r^2}} + 2r\sqrt{R^2-r^2}\right)\]Simplify the expression: \[V'(r) = \frac{2\pi r (2(R^2 - r^2) - r^2)}{\sqrt{R^2 - r^2}} = \frac{2\pi r (2R^2 - 3r^2)}{\sqrt{R^2 - r^2}}\]

Step 5: Find Critical Points

Set \(V'(r) = 0\) to find critical points:\[2R^2 - 3r^2 = 0 \]Solve for \(r\):\[r^2 = \frac{2}{3}R^2\]\[r = \frac{\sqrt{6}}{3}R\]

Step 6: Determine the Optimal Height

Substitute \(r = \frac{\sqrt{6}}{3}R\) back into the equation for \(h\):\[h = 2 \sqrt{R^2 - \left(\frac{\sqrt{6}}{3}R\right)^2} = 2 \sqrt{\frac{R^2}{3}} = \frac{2R}{\sqrt{3}} = \frac{2\sqrt{3}}{3}R\]

Step 7: Conclusion

The dimensions of the maximum volume cylinder inscribed in the sphere are: Radius \(r = \frac{\sqrt{6}}{3}R\), Height \(h = \frac{2\sqrt{3}}{3}R\). These values maximize the volume of the cylinder within the given constraint.

Key Concepts

Maximizing Volume

To find the largest possible volume of a cylinder that fits inside a sphere, the concept of volume maximization is used. This involves determining the optimal dimensions of the cylinder — that is, its radius and height. Since the cylinder is inscribed in the sphere, the problem is constrained by the maximum dimensions that the sphere allows. Maximizing volume means identifying the cylinder shape that takes up the most space within its spherical boundary.

A good approach is to express one dimension in terms of the other using the constraint and then formulate the volume as a function of a single variable. This simplification allows us to use calculus to find the optimal value that maximizes the volume. Here, given that the volume of a cylinder is calculated by the formula \( V = \pi r^2 h \), the challenge is to adjust \( r \) and \( h \) such that this product is as large as possible while respecting the relation provided by the sphere constraint.

Pythagorean Theorem

The Pythagorean theorem plays a crucial role in solving this problem. When dealing with a cylinder inscribed within a sphere, the theorem helps establish a relationship between the dimensions of the cylinder and the sphere’s radius.

Since the sphere envelops the cylinder, imagine a right triangle formed where:

• The hypotenuse is the sphere's radius \( R \).
• The base is half the cylinder's height \( h/2 \).
• The vertical side is the cylinder's radius \( r \).

By the Pythagorean theorem, this relationship is \( R^2 = (h/2)^2 + r^2 \). This equation allows us to express the height \( h \) in terms of the cylinder’s radius \( r \), forming a critical part of setting up the problem so that calculus can be applied to maximize the volume.

Derivative of Volume Function

Once the volume of the cylinder is expressed as a function of one variable, finding the derivative allows us to locate the maximum volume. Calculus is applied here because the derivative gives insight into how changes in \( r \) affect the volume.

The volume, expressed as \( V(r) = 2\pi r^2 \sqrt{R^2 - r^2} \), is a function that depends solely on \( r \). To maximize this function, we need to find \( V'(r) \), the derivative of the volume with respect to \( r \). This new function \( V'(r) \) will provide critical points where potential maxima (or minima) might occur.

• This involves applying the product rule and chain rule of differentiation.
• Setting the derivative \( V'(r) \) equal to zero lets us find these critical points.

Solving \( V'(r) = 0 \) leads to the radius \( r \) that maximizes the volume, and by further calculation, the corresponding height \( h \) can be found. This process is crucial to identifying the cylinder dimensions that offer the largest volume within the sphere.