Question

Show that the height of the right circular cylinder of greatest volume
which can be inscribed in a right circular cone of height h and radius r is
one-third of the height of the come, and the greatest volume of the
cylinder is 4/9 times the volume of the cone.
​

Answer 1

Step-by-step explanation:

The given right circular cone of fixed height (h) and semi-vertical angle (α) can be drawn as: Here, a cylinder of radius R and height H is inscribed in the cone. Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Answer 2

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)

= 4/9πh³tan²α