Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle is one-third that of the cone and the greatest volume of the cylinder is 427πh³tan²α.

To find the height of the cylinder of greatest volume that can be inscribed in a right circular cone, we start by defining the cone's dimensions. Let the cone have a height \( h \) and a semi-vertical angle \( \alpha \). The radius of the base of the cone can be expressed as \( R = h \tan \alpha \).

Volume of the Cylinder

Let the height of the inscribed cylinder be \( H \) and its radius be \( r \). The volume \( V \) of the cylinder is given by the formula:

V = πr²H

Relating Cylinder Dimensions to the Cone

Using similar triangles, we can relate the dimensions of the cylinder to those of the cone. The relationship can be expressed as:

• The height of the cone above the cylinder is \( h - H \).
• The radius of the cylinder \( r \) can be expressed as \( r = (h - H) \tan \alpha \).

Substituting for Volume

Substituting \( r \) into the volume formula gives:

V = π((h - H) \tan \alpha)²H

Expanding this, we have:

V = π(h - H)² \tan² \alpha \cdot H

Maximizing the Volume

To find the maximum volume, we differentiate \( V \) with respect to \( H \) and set the derivative equal to zero:

V' = π \left[ 2(h - H)(-1) \tan² \alpha \cdot H + (h - H)² \tan² \alpha \right]

Setting \( V' = 0 \) leads to:

2H(h - H) = (h - H)²

From this, we can simplify and solve for \( H \), yielding:

H = \frac{h}{3}

Calculating the Maximum Volume

Now substituting \( H = \frac{h}{3} \) back into the volume formula:

V = π((h - \frac{h}{3}) \tan \alpha)² \cdot \frac{h}{3}

This simplifies to:

V = π(\frac{2h}{3} \tan \alpha)² \cdot \frac{h}{3}

Calculating further gives:

V = \frac{4πh³}{27} \tan² \alpha

Final Result

Thus, the maximum volume of the cylinder inscribed in the cone is:

V = 427πh³tan²α

In summary, the height of the cylinder is one-third that of the cone, and the maximum volume is as stated above.