Problem 27 - Thomas Calculus

Question

Constructing cones \(A\) right triangle whose hypotenuse is \(\sqrt{3} \mathrm{m}\) long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.

Step-by-Step Solution

Answer

The cone's radius and height are both \( \sqrt{1.5} \); the greatest volume is \( \sqrt{1.5}^3 \times \frac{\pi}{3}.\)

Step 1: Understand the Problem

We need to find a configuration of a right triangle with hypotenuse \( \sqrt{3} \) that, when revolved about one of its legs, forms a cone with the greatest possible volume. The two legs of the triangle will form the radius and height of the cone.

Step 2: Recall Volume Formula for Cone

The volume \( V \) of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius and \( h \) is the height of the cone. We aim to maximize this volume.

Step 3: Set Up Equations for the Triangle

Consider the two legs of the right triangle as \( a \) and \( b \), and the hypotenuse is \( c = \sqrt{3} \). By Pythagoras' theorem, \( a^2 + b^2 = 3 \). The radius \( r \) and height \( h \) of the cone can be either \( a \) or \( b \).

Step 4: Express Volume in Terms of Triangle Legs

The volume expression in terms of \( a \) and \( b \) can be: \[ V(a) = \frac{1}{3} \pi a^2 b \] or \[ V(b) = \frac{1}{3} \pi b^2 a \] depending on which leg is used as the radius.

Step 5: Differentiate to Find Maximum Volume

Differentiate the volume expression \( V(a) = \frac{1}{3} \pi a^2 (\sqrt{3}^2 - a^2)^{1/2} \) with respect to \( a \). Set the derivative equal to zero and solve to find the critical points.

Step 6: Solve for Maximum Volume

Use \( a^2 + b^2 = 3 \) to find that \( a = \sqrt{1.5} \) and \( b = \sqrt{1.5} \). Therefore, when \( a = b \), the triangle is isosceles, and this gives the maximum volume configuration.

Step 7: Verify Optimal Configuration

Check that \( V = \frac{1}{3} \pi a^2 b \) and \( V = \frac{1}{3} \pi b^2 a \) give the same value when \( a = b = \sqrt{1.5} \), which is indeed the maximum value.

Key Concepts

Right Circular Cone

A right circular cone is a 3-dimensional geometric shape that looks like an ice cream cone. It is defined by a circle at the base and a pointed tip or vertex at the top. The key characteristic of a right circular cone is that the vertex is directly above the center of the circular base, meaning the line segment from the vertex to the center is perpendicular to the base.
This property makes the cone symmetrical around its central axis, which is crucial when analyzing the cone's geometry, such as its volume and height. In our exercise, a right triangle is revolved around one of its legs to form this kind of cone. The radius of the base and the height of the cone come from the triangle's legs.

Hypotenuse in Triangles

In geometry, the hypotenuse is the longest side of a right triangle, directly opposite the right angle. In our problem, the hypotenuse is given as \(\sqrt{3}\) meters, which provides a vital piece of data to connect the triangle to the cone.
When the triangle is revolved around one of its legs, this hypotenuse doesn't directly contribute to the dimensions of the cone but helps in calculating the other two sides of the triangle using the Pythagorean theorem. These sides become the radius and the height of the cone.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that relates the lengths of sides in a right triangle. It states that the square of the length of the hypotenuse \(c\) is equal to the sum of the squares of the other two sides \(a\) and \(b\): \[a^2 + b^2 = c^2\]
In our exercise, the hypotenuse \(c\) is \(\sqrt{3}\). This relationship allows us to express \(a^2 + b^2 = 3\). We use this to determine possible values for \(a\) and \(b\), which ultimately define the ideal dimensions of the cone to maximize its volume.

Differentiation in Calculus

Differentiation is a key concept in calculus that involves finding the rate at which a function is changing at any given point. In the context of maximizing the volume of the cone, differentiation helps us identify when the rate of change of the volume becomes zero, indicating a maximum point.

• We start by expressing the volume of the cone as a function of one side of the triangle, say \(a\): \(V(a) = \frac{1}{3} \pi a^2 (\sqrt{3}^2 - a^2)^{1/2}\).
• Next, we differentiate this volume function with respect to \(a\) and solve for \(a\) when the derivative is zero.
• This provides the critical points where the volume is maximized.

By solving \(a = b = \sqrt{1.5}\), we find the optimal dimensions for the triangle that ensure the maximum volume for the cone.