# Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is 4/27πh^{3} tan^{2} α

**Solution:**

The given right circular cone of fixed height h and semi-vertical angle a can be drawn as:

Here, a cylinder of radius R and height H is inscribed in the cone.

Then, ∠GAO = a, OG = r, OA = h, OE = R and CE = H

We have, r = h tan α

Now, since ΔAOG is similar to ΔCEG, we have:

AO / OG = CE / EG

⇒ h / r = H / (r - R)

⇒ H = h / tan α (h tan α - R)

⇒ H = 1 / tan α (h tan α - R)

Now, the volume (V) of the cylinder is given by,

V = π R^{2}H

= π R^{2 }/ tan α (h tan α - R)

= π R^{2}h - π R^{3 }/ tan α

Therefore,

dV/dR = 2π RH - 3π R^{2 }/ tan α

Now,

dV/dR = 0

⇒ 2 π R H - 3 π R^{2 }/ tan α = 0

⇒ 2 π R H = 3 π R^{2 }/ tan α

⇒ 2h tanα = 3R

⇒ R = 2h / 3 tan α

Also,

d^{2}V/dR^{2} = 2 π H - 6π R / tan α

For R = 2h / 3 tanα, we have:

d^{2}V/dR^{2} = 2 π h - 6π / tan α (2h / 3 tanα)

= 2πh - 4πh.

= - 2πh < 0

By second derivative test,

the volume of the cylinder is the greatest when

R = 2h/3 tanα.

when R = 2h / 3 tan α

Then,

H = 1 / tan α (h tanα - 2h/3 tanα)

= 1 / tanα (h tan α/3)

= h / 3

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Thus, the maximum volume of the cylinder can be obtained as:

π (2h / 3 tan α)^{2} (h / 3)

= π (4h^{2 }/ 9 tan^{2}α) (h / 3)

= 4/27 π h^{3} tan^{2}α

Hence, the given result is proved

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 18

## Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is 4/27πh^{3} tan^{2} α.

**Summary:**

Hence we have shown 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 α is one-third that of the cone and the greatest volume of cylinder is 4/27πh^{3} tan^{2} α

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