★ 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