https://file.aiquickdraw.com/mathbot/file/1729918531615x4dwknnz.jpg

To solve this problem, we need to find the length of \( AM \) that maximizes the volume of the cone formed by \( \triangle AMC \) when revolved around \( AM \). Let's go through it step-by-step:

1. **Identify variables and cone properties:**
   - \( AM = y \) is the height of the cone.
   - \( CM = x \) represents the radius of the cone's base.
   - \( AB = \sqrt{3} \), as given in the triangle.

2. **Relationship between \( x \) and \( y \):**
   - From the right triangle \( \triangle ABC \), applying the Pythagorean theorem:
     \[
     AB^2 = AM^2 + CM^2 \Rightarrow (\sqrt{3})^2 = y^2 + x^2
     \]
   - So, \( y^2 + x^2 = 3 \).

3. **Volume of the cone:**
   - The volume \( V \) of a cone is given by:
     \[
     V = \frac{1}{3} \pi r^2 h
     \]
   - In our case, \( r = x \) and \( h = y \), so:
     \[
     V = \frac{1}{3} \pi x^2 y
     \]

4. **Express \( x \) in terms of \( y \):**
   - From \( y^2 + x^2 = 3 \), we can isolate \( x \):
     \[
     x^2 = 3 - y^2 \Rightarrow x = \sqrt{3 - y^2}
     \]

5. **Volume as a function of \( y \):**
   - Substitute \( x = \sqrt{3 - y^2} \) into the volume formula:
     \[
     V = \frac{1}{3} \pi (\sqrt{3 - y^2})^2 y = \frac{1}{3} \pi (3 - y^2) y
     \]
     \[
     V = \frac{1}{3} \pi (3y - y^3)
     \]

6. **Differentiate \( V \) with respect to \( y \) and find critical points:**
   - To maximize \( V \), take the derivative with respect to \( y \):
     \[
     \frac{dV}{dy} = \frac{1}{3} \pi (3 - 3y^2)
     \]
     \[
     \frac{dV}{dy} = \pi (1 - y^2)
     \]
   - Set \( \frac{dV}{dy} = 0 \):
     \[
     1 - y^2 = 0 \Rightarrow y^2 = 1 \Rightarrow y = 1
     \]

7. **Verify that \( y = 1 \) gives a maximum:**
   - For this function, \( y = 1 \) is a maximum point for \( V \).

Thus, the length of \( AM \) that maximizes the volume of the cone is:
\[
\boxed{1}
\]

Would you like further details, or do you have any questions?

### Related Questions:
1. How would the solution change if the length of \( AB \) was different?
2. Can we use a different approach, like Lagrange multipliers, for this problem?
3. What is the significance of finding the maximum volume in geometric optimization problems?
4. How would we solve if the figure was rotated around a different axis?
5. Could this problem be solved using integration?

**Tip:** In geometry problems involving optimization, expressing all quantities in terms of a single variable helps simplify differentiation.

The length of AM that makes the volume of the cone as maximum as possible equals .......... (Options: 1/4, 1/2, 1, 1 1/2).

Discover the optimal length of AM to maximize the volume of a cone formed by rotating a right triangle around its height. This problem applies the Pythagorean theorem and calculus-based optimization, ideal for Grades 11-12 students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation