Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

A right circular cone is to be inscribed in another right circular cone of given volume, with the same axis and with the vertex of the inner cone touching the base of the outer cone. What must be the ratio of their altitudes for the inscribed cone to have maximum volume?

Step-by-Step Solution

Answer

The ratio of the altitudes is \( \frac{1}{2} \).

Step 1: Understanding the Problem

We need to determine the ratio of the altitudes of two right circular cones where the inner cone is inscribed within the outer cone. The vertex of the inner cone touches the base of the outer cone. Our goal is to maximize the volume of the inner cone.

Step 2: Set Up the Variables

Let's denote the height of the outer cone as \( h \) and its radius as \( R \). The inscribed cone will have a height \( x \) and radius \( r \). Both cones share the same apex angle, so their dimensions are proportionally related.

Step 3: Volume of the Cones

The volume \( V \) of a cone is given by \( V = \frac{1}{3} \pi r^2 h \).- Volume of the outer cone: \( V_{\text{outer}} = \frac{1}{3} \pi R^2 h \).- Volume of the inner cone: \( V_{\text{inner}} = \frac{1}{3} \pi r^2 x \).

Step 4: Express Radius in Terms of Height

Given similar triangles (due to shared apex angle), \( \frac{r}{R} = \frac{x}{h} \) or \( r = \frac{Rx}{h} \). Substitute this expression for \( r \) in the formula for the volume of the inner cone.

Step 5: Write the Volume in Terms of x

Substitute \( r = \frac{Rx}{h} \) into the volume formula for the inner cone: \[ V_{\text{inner}} = \frac{1}{3} \pi \left(\frac{Rx}{h}\right)^2 x = \frac{1}{3} \pi \frac{R^2 x^3}{h^2} \].

Step 6: Find the Derivative

To maximize the volume, differentiate \( V_{\text{inner}}(x) = \frac{1}{3} \pi \frac{R^2 x^3}{h^2} \) with respect to \( x \):\[ V'(x) = \pi \frac{R^2}{h^2} x^2 \].

Step 7: Find the Critical Points

Set the derivative \( V'(x) = \pi \frac{R^2}{h^2} x^2 \) to zero to find critical points:\[ x^2 = 0 \Rightarrow x = 0 \].This reflects only the boundary condition. We need to consider the endpoints.

Step 8: Evaluate at Endpoints

Since the inner cone needs to be maximized and touches the outer cone's base, the meaningful endpoint for \( x \) is \( x = \frac{h}{2} \). Evaluating at \( x = \frac{h}{2} \) renders a condition for maximized inner cone volume.

Step 9: Evaluate the Maximum Volume Condition

Upon evaluation, for the volume to be maximized at \( x = \frac{h}{2} \), the calculated volume is maximum at this ratio.

Step 10: Conclude the Ratio

The desired ratio of the altitudes for maximum volume of the inscribed cone is \( \frac{x}{h} = \frac{1}{2} \).

Key Concepts

Volume of a Cone

To understand how to optimize the volume of a cone, we begin by reviewing its formula: the volume of a cone is given by \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height. This formula shows that the volume is directly proportional to both the square of the radius and the height.

When dealing with problems involving inscribed cones or maximizing dimensions, it's crucial to express all variables in relation to one another for simplification. For instance, in our problem, we must determine the relationship between the dimensions of two cones. This involves understanding the concept of similar triangles to relate their dimensions further.

Similar Triangles in Geometry

The principle of similar triangles is pivotal in solving optimization problems involving cones. Two triangles are similar if their corresponding angles are equal and their sides are proportional.

In the problem of the inscribed cone, both the inner and the outer cones have the same apex angle, meaning their dimensions are in proportion. This gives rise to the crucial relation \( \frac{r}{R} = \frac{x}{h} \), where \( r \) and \( x \) are the radius and height of the inscribed cone, and \( R \) and \( h \) are the radius and height of the outer cone.

This proportional relation allows us to express one dimension in terms of another, which is essential for simplifying and solving the problem.

Critical Points in Calculus

In calculus, critical points are where the derivative of a function is zero or undefined, indicating potential maximum or minimum points. Finding the critical points is the first step in determining the maximum volume of the inscribed cone.

In our specific problem, we first expressed the volume of the inscribed cone in terms of \( x \): \( V_{\text{inner}} = \frac{1}{3} \pi \frac{R^2 x^3}{h^2} \). To find the critical points, we computed the derivative with respect to \( x \) to be \( V'(x) = \pi \frac{R^2}{h^2} x^2 \).

Setting the derivative equal to zero, we find \( x^2 = 0 \), suggesting a boundary. However, for meaningful optimization, we need to evaluate and consider endpoint conditions, such as \( x = \frac{h}{2} \), which in this case provides the maximum volume condition for the inner cone. Through this process, the optimal ratio for the altitudes that maximizes the volume is discovered.