Show that the height of the cylinder of greatest volume which can be inscribed in a circular cone of height 'h' and having semi-vertical angle @ is one-third that of the cone and the greatest volume of cylinder is
Answers
★ concept :
Finding local maxima and local minima
• first derivative test
• second derivative test : let f(x) be a function defined interval I and c belongs to I .
let f be twice diffferentiable at c .then
↪if f' (c) = 0 and f" < 0, then x = c is a point of local maxima.
↪ if f' (c) = 0 and f" > 0, then x = c is a point of local minima .
★ Solution :
Let h be the height of cylinder and r be it's radius .therefore ,
the volume of cone =π r²h
Let H be the height of cone .and ß as semi- vertical angle from ᐃ APE
refer to the attachment ;
I hope it helps you!
HELLO DEAR,
The given right circular cone of fixed height (h)and semi-vertical angle (α) is in attachment.
now, a cylinder of radius R and height H is inscribed in the cone.
Then, ∠GAO = α, OG = r, OA = h, OE = R, and CE = H.
We have,
r = h tan α
Now, ΔAOG is similar to ΔCEG,
AO/OG = CE/EG
h/r = H/(r - R)
[EG = OG - OE]
H = h(r - R)/r
H = h(h tanα - R)/htanα
H = (htanα - R)/tanα
now, the volume(V) of cylinder is
V = πR²H = πR² * (htanα - R)/tanα
V = πR²h - πR³/tanα
dV/dR = 2πRh - 3πR²/tanα
NOW, dV/dR = 0
2πRh = 3πR²/tanα
2htanα = 3R
R = 2htanα/3
NOW, d²V/dR² = 2πh - 6π/tanα(2htanα/3)
2πh - 4πh = -2πh
-2πh < 0
By second derivative , the volume of the cylinder is the greatest when R = 2htanα/3.
where, R = 2htanα/3,
H = 1/tanα(htanα - 2htanα/3)
H = 1/tanα(htanα)/3
H = h/3
hence, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.
Now, the maximum volume of the cylinder can be
π(2htanα/3)²(h/3)
= π(4h²tan²α/9)(h/3)
= 4/27πh³tan²α
hence,proved.
I HOPE IT'S HELP YOU DEAR,
THANKS