Problem 37 - Thomas Calculus

Question

A right circular cone is circumscribed in a sphere of radius \(1 .\) Determine the height \(h\) and radius \(r\) of the cone of maximum volume.

Step-by-Step Solution

Answer

The maximum volume cone has height \( h = \sqrt{2} + 1 \) and radius \( r = \sqrt{2} \).

Step 1: Understand the Problem

We need to find the height \( h \) and base radius \( r \) of a right circular cone that is circumscribed around a sphere of radius 1 such that the cone's volume is maximized. The sphere touches the cone at the base and its vertex, forming a simple geometric relationship.

Step 2: Determine the Geometric Relationship

The sphere, with known radius 1, will have its center at height \( h - 1 \) from the cone's base because the sphere is fully inside the cone. The cone's slant height must be equal to the sphere's diameter (which is 2) to ensure the sphere is circumscribed.

Step 3: Set Up the Volume Formula for the Cone

The volume \( V \) of a cone is given by \( V = \frac{1}{3} \pi r^2 h \). To maximize this volume, we should express \( h \) in terms of \( r \) or vice versa, using the geometric relationships we've identified.

Step 4: Use the Pythagorean Theorem

Since the slant height \( l \) of the cone is the sphere's diameter, \( l = 2 \). By the Pythagorean Theorem in the cross-section of the cone: \[ l^2 = r^2 + (h - 1)^2. \] Substituting \( l = 2 \) gives us: \[ 4 = r^2 + (h - 1)^2. \]

Step 5: Solve for \( h \) in terms of \( r \)

From \( 4 = r^2 + (h - 1)^2 \), we get \( (h - 1)^2 = 4 - r^2 \). Therefore, \( h = 1 + \sqrt{4 - r^2} \).

Step 6: Substitute \( h \) Back into Volume Formula

Substituting \( h = 1 + \sqrt{4 - r^2} \) into the volume formula, we have \[ V = \frac{1}{3} \pi r^2 (1 + \sqrt{4 - r^2}). \]

Step 7: Differentiate the Volume Function

Differentiate \( V \) with respect to \( r \) to find the critical points. Set \( \frac{dV}{dr} = 0 \) and solve for \( r \). This involves applying the product rule and chain rule to \( V = \frac{1}{3} \pi r^2 (1 + \sqrt{4 - r^2}) \).

Step 8: Solve the Derivative Equation

Solving \( \frac{dV}{dr} = 0 \) gives the critical points. Simplifying the derivative, solving the equation, and considering the sphere constraint provides the maximum value of \( r \).

Step 9: Confirm with Second Derivative/Additional Check

Check if the critical points found in Step 8 are maximum points by using the second derivative test or examining the context's natural constraints. Calculate \( h \) using \( h = 1 + \sqrt{4 - r^2} \) for this maximum \( r \).

Step 10: Conclusion

The height of the cone \( h \) and the radius \( r \) that maximize the cone's volume while being circumscribed around the sphere are: \( h = \sqrt{2} + 1 \) and \( r = \sqrt{2} \).

Key Concepts

Right Circular Cone

A right circular cone is one of the most fascinating shapes in geometry. It is basically a three-dimensional shape with a circular base and a pointed tip called the vertex. The axis of the cone is a straight line that passes through the middle of the base and ends at the vertex. This alignment makes the cone 'right,' as it suggests that the axis is perpendicular to the base. Here are some of its unique characteristics:

• Height (h): The distance from the center of the base to the vertex.
• Radius (r): The distance from the center of the base to any point on its edge.
• Slant height (l): The distance along the cone's surface from the base to the vertex.
• Lateral Surface: The curved surface joining the base to the vertex.

Understanding these basic properties is crucial when solving problems related to cones, such as finding volumes or dealing with their orientation in space. In geometric problems, especially in optimization, these features play important roles in deriving formulas that help in calculating needed dimensions.

Maximum Volume

Finding the maximum volume of a geometric shape involves applying calculus. For a right circular cone, the volume is expressed by the formula: \(V = \frac{1}{3} \pi r^2 h\)To maximize this volume given specific constraints, such as a circumscribing sphere, you need to express all variables in terms of one of them (usually the radius or height). This often requires critical thinking and a good understanding of calculus techniques like differentiation. Here’s what you typically do:

• Substitute Known Relationships: Write one variable in terms of others using geometric relationships.
• Differentiate: Find the derivative of the volume with respect to the chosen variable.
• Find Critical Points: Set this derivative equal to zero and solve to find where the volume could be at its maximum.
• Confirm Maximum: Use the second derivative test or examine constraints to ensure that the critical point gives a maximum volume.

Understanding these steps helps students apply calculus to practical geometric problems effectively.

Circumscribed Sphere

In our problem, the cone is circumscribed around a sphere. This means the sphere fits perfectly inside the cone, touching it at the base and vertex. A key aspect of this configuration is that:

• The sphere's diameter equals the slant height of the cone.
• The center of the sphere is located at height \( h-1 \) from the base.
• The sphere ensures a symmetrical and uniform contact along the cone's slant height.

These geometric constraints are essential in relating the sphere’s dimensions to the cone's dimensions. Essentially, drawing such a relationship helps in expressing one variable in terms of another, which is vital for setting up equations when tackling optimization problems. Knowing all necessary measurements and relationships simplifies the complex problem into manageable parts and guides the steps for calculus-based optimization.

Geometric Relationships

Understanding the geometric relationships between different components in this problem is the key to finding the solution. The most crucial relationship is that defined by the cone’s slant height \( l \), the radius \( r \), and the height \( h \), as expressed through the Pythagorean theorem:\[ l^2 = r^2 + (h - 1)^2\]Given that the slant height equals the sphere’s diameter (2 in this case), this relationship becomes:\[ 4 = r^2 + (h - 1)^2\]Such equations let us solve for one variable in terms of others, helping to simplify the volume formula. Unlocking these relationships allows us to focus solely on maximizing the volume. Additionally, applying these relationships requires logic and problem-solving skills which are cultivated by practice and understanding. By mastering the geometry fundamentals and appreciating the elegance of these relationships, students can tackle a wide range of mathematical problems effectively.