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 4/27πh3 tan2 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 R2H
= π R2 / tan α (h tan α - R)
= π R2h - π R3 / tan α
Therefore,
dV/dR = 2π RH - 3π R2 / tan α
Now,
dV/dR = 0
⇒ 2 π R H - 3 π R2 / tan α = 0
⇒ 2 π R H = 3 π R2 / tan α
⇒ 2h tanα = 3R
⇒ R = 2h / 3 tan α
Also,
d2V/dR2 = 2 π H - 6π R / tan α
For R = 2h / 3 tanα, we have:
d2V/dR2 = 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 (4h2 / 9 tan2α) (h / 3)
= 4/27 π h3 tan2α
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 4/27πh3 tan2 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 4/27πh3 tan2 α