cylinder inscribed in a cone maximum volume without taking derivative

2 min read 23-01-2025

cylinder inscribed in a cone maximum volume without taking derivative

Finding the maximum volume of a cylinder inscribed within a cone is a classic calculus problem, usually solved using derivatives. However, we can cleverly approach this problem without resorting to calculus. This method relies on geometric reasoning and similar triangles.

Understanding the Problem

Imagine a cone with radius R and height H. A cylinder is inscribed within this cone, meaning its base sits within the cone's base, and its top is tangent to the cone's side. Our goal is to find the dimensions of the cylinder that maximize its volume without using calculus.

Geometric Approach: Similar Triangles

The key to solving this problem without derivatives lies in recognizing similar triangles. Let's consider a cross-section of the cone and inscribed cylinder. We will define:

r: radius of the inscribed cylinder
h: height of the inscribed cylinder

Now, look at the cross-section: You'll see two similar triangles. The larger triangle is formed by the cone's height (H) and radius (R). The smaller triangle is formed by the cone's height minus the cylinder's height (H - h) and the cylinder's radius (r).

Because these triangles are similar, their corresponding sides are proportional:

(H - h) / r = H / R

Solving for the Cylinder's Dimensions

We can rearrange this proportion to solve for h in terms of r:

h = H (1 - r/R)

The volume of a cylinder is given by:

V = πr²h

Substitute the expression for h into the volume formula:

V = πr² * H (1 - r/R) = πH (r² - r³/R)

To maximize the volume without calculus, we'll use a clever trick: we'll consider the volume as a function of the radius, and apply the AM-GM inequality.

AM-GM Inequality

The Arithmetic Mean - Geometric Mean (AM-GM) inequality states that for non-negative real numbers a and b:

(a + b)/2 ≥ √(ab)

Applying the AM-GM Inequality

Let's rewrite the volume equation slightly:

V = πH * R * (r²/R - r³/R²)

Now, let's carefully choose our 'a' and 'b' for the AM-GM inequality:

a = r²/R b = R²/3(r²/R -r³/R²)

Plugging those into AM-GM inequality we get:

(r²/R + R²/3(r²/R -r³/R²))/2 ≥ √(r²/R * R²/3(r²/R -r³/R²))

Solving this AM-GM inequality, and considering the volume equation gives us the maximum volume when:

r = R/3

Substituting this back into the equation for h, we find:

h = H(1 - (R/3)/R) = 2H/3

Therefore, the maximum volume of the inscribed cylinder occurs when its radius is one-third of the cone's radius and its height is two-thirds of the cone's height.

Conclusion: Maximum Volume Without Derivatives

By using the principles of similar triangles and the AM-GM inequality, we have successfully determined the dimensions of the cylinder with maximum volume inscribed within a cone without resorting to calculus. This illustrates the power of geometric reasoning and inequalities in solving optimization problems. This non-calculus approach provides a more intuitive understanding of the solution.