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.

Short answer

The maximum volume V of the cone is calculated from the value of \(r\) which is found by setting the derivative of V with respect to \(r\) to zero. The corresponding value of \(h\) is then determined using \(h = \sqrt{3-r^2}\). The radius, height, and volume values are then used to construct the cone.

Step-by-step solution

Express the radius in terms of height

Let the height of the cone be \(h\) and the radius be \(r\). Using the Pythagorean Theorem, we get \(h^2 + r^2 = (\sqrt{3})^2 \), which simplifies to \(h^2 + r^2 = 3\). Since we will be taking \(r\) as a variable, we express \(h\) in terms of \(r\), i.e., \(h = \sqrt{3-r^2}\).

Express the volume in terms of radius

Now, substitute the value of \(h\) from step 1 into the formula for the volume of a cone \(V = \frac{1}{3}πr^2h\), we get \(V = \frac{1}{3}πr^2(\sqrt{3-r^2})\).

Find the maximum volume

To find the maximum volume, we need to take the derivative of \(V\) with respect to \(r\), set it to zero and solve for \(r\). The value of \(r\) obtained is the radius of the cone that yields the maximum volume when the height is equal to \( \sqrt{3-r^2}\). Use this \(r\) to find the corresponding maximum volume \(V\).

Calculate the height

Finally, substitute \(r\) into the equation \(h = \sqrt{3-r^2}\) to find the value of \(h\). This \(h\) is the height of the cone with the maximum volume.

Key concepts

Pythagorean Theorem in Cone Volume

Understanding the role of the Pythagorean Theorem in determining the volume of a cone is critical when dealing with geometric problems involving right circular cones. In our exercise, we have a fixed hypotenuse length \(\sqrt{3} \text{m}\), and the right triangle is revolved around one of its legs to create our cone. Using the Pythagorean Theorem \( a^2 + b^2 = c^2 \), where \(a\) and \(b\) are the legs of the right triangle (height and radius of the cone, respectively), and \(c\) is the hypotenuse (fixed length \(\sqrt{3}\)), we can express one of the legs in terms of the other.
This relationship is pivotal as it allows us to express the height \(h\) as a function of the radius \(r\), \(h = \sqrt{3-r^2}\). This expression is a foundational step that links our geometrical understanding with calculus tools needed to maximize the volume.

Differentiation to Find Maximum Volume

To optimize and find the maximum volume of a shape, differentiation is an essential mathematical tool. In the context of our cone volume problem, we look to maximize the volume function \( V(r) = \frac{1}{3}\pi r^2\sqrt{3-r^2} \). By taking the derivative of this volume function with respect to the radius \(r\), we can find the critical points of the function, which are the values of \(r\) where the first derivative is zero or undefined.

At these critical points, the function can have a maximum, minimum, or neither, but at least one of these points will give us the maximum volume of the cone. Setting the derivative equal to zero provides us with an equation that, when solved, yields the value of \(r\) that maximizes the cone's volume. The real challenge lies in solving the derivative equation which often requires additional calculus techniques.

Volume Formula for a Cone

The generalized formula to calculate the volume of a right circular cone is \( V = \frac{1}{3}\pi r^2h \), where \( V \) stands for volume, \( r \) for the cone's radius, and \( h \) for its height.
The factor \( \frac{1}{3} \) arises from the integral calculus-based derivation of the volume of a cone, which fundamentally indicates that a cone's volume is one-third that of a cylinder with the same base and height.

It is essential to understand this formula thoroughly as it directly affects how we can manipulate and optimize the cone's dimensions in problems. In the previously mentioned exercise, by using this formula in conjunction with our expression for \( h \) in terms of \( r \) derived from the Pythagorean Theorem, we can create a volume function purely in terms of \( r \) which we then differentiate to find the maximum volume.

Implicit Differentiation in Volume Optimization

Implicit differentiation comes into play when we face an equation that describes a relationship between variables but one is not explicitly solved in terms of the other. In our cone volume problem, once we've expressed the height \( h \) as \( \sqrt{3-r^2} \) and have plugged it into our volume formula, we have our volume function implicitly defined in terms of \( r \).

When we take the derivative of the volume \( V \) with respect to \( r \) to find the optimal radius, we are performing implicit differentiation. Calculus then allows us to differentiate the composite function \( r^2\sqrt{3-r^2} \) found in the volume formula. This process helps us to get to the critical point equation which we need to solve to optimize the cone's volume. Proper application of implicit differentiation is crucial for solving many real-world optimization problems, like maximizing the volume of this cone.