Show that the volume of greatest cylinder which can be inscribed in a cone of height h and semi vertcal angle t is 4/27 π h³ tan²t

To find the volume of the largest cylinder that can be inscribed in a cone, we can use some geometry and calculus. The problem involves determining the dimensions of the cylinder that maximize its volume while being confined within the cone. Let's break this down step by step.

Understanding the Cone and Cylinder Dimensions

Consider a right circular cone with a height \( h \) and a semi-vertical angle \( t \). This cone will have a base radius \( R \) that we can express based on the height and angle.

• The relationship between the height \( h \) and the radius \( R \) can be described using trigonometric functions: \( R = h \tan(t) \).

Setting Up the Cylinder

Next, we need to think about the cylinder that fits inside this cone. Let’s define the height of the cylinder as \( y \) and its radius as \( r \). The top of the cylinder will be at a height \( y \) from the base of the cone, and we can observe that the radius \( r \) of the cylinder will also depend on the height at which it is located within the cone.

Relating the Cylinder Dimensions to the Cone

From the geometry of the cone, we can derive the radius \( r \) of the cylinder in terms of its height \( y \). At height \( y \), the radius of the cone can be expressed as:

• Radius of the cone at height \( y \): \( r = (h - y) \tan(t) \).

Volume of the Cylinder

The volume \( V \) of the cylinder can be expressed by the formula:

V = πr²y

Substituting for \( r \) gives us:

V = π[(h - y) tan(t)]²y

Expanding this expression yields:

V = π(h - y)² tan²(t) y

Maximizing the Volume

To find the maximum volume, we need to take the derivative of \( V \) with respect to \( y \), set it to zero, and solve for \( y \).

First, let's expand the volume function:

V = π tan²(t) (h² - 2hy + y²) y

This simplifies to:

V = π tan²(t) (hy² - 2hy² + y³)

Now we differentiate with respect to \( y \):

dV/dy = π tan²(t) (3y² - 4hy)

Setting the derivative equal to zero gives:

3y² - 4hy = 0

Factoring out \( y \), we find:

y(3y - 4h) = 0

Thus, \( y = 0 \) or \( y = \frac{4h}{3} \). The height of the cylinder that maximizes the volume is \( \frac{4h}{3} \).

Calculating the Maximum Volume

Substituting \( y = \frac{4h}{3} \) back into the expression for \( r \), we can find the corresponding radius:

r = (h - \frac{4h}{3}) tan(t) = \frac{h}{3} tan(t)

Now substituting \( r \) and \( y \) into the volume formula:

V = π \left( \frac{h}{3} tan(t) \right)² \left( \frac{4h}{3} \right)

Calculating this gives:

V = π \frac{h²}{9} tan²(t) \frac{4h}{3} = \frac{4}{27} π h³ tan²(t)

Final Result

Thus, we've shown that the volume of the largest cylinder that can be inscribed in a cone of height \( h \) and semi-vertical angle \( t \) is:

V = \frac{4}{27} π h³ tan²(t)

This result illustrates the interplay between geometry and calculus in optimizing shapes within constraints.