durrani cone height 30cm and base radius of 10cm.we want to draw inside asitonp size V(r)takes the largest possible vaue.keep the cylinder height h...

Th e cone has the vertical height 30 and radius 10. Si the
semi vertical angle x of the cone is given by tanx  radiuus/ height . Or tan x = 10/30 =
1/3.

Let the cylinder describes in side the cone have a
radius r. Let h be the height of the cone. Then r , h and the  semi vertical angle are
related by: r/(30-h) = tanx = 1/3.

So 3r =
30-h.

Or h = 30-3r = 3(10-r)
.

The volume v(r) of the cylinder  is given
by:

v(r) = pi*r^2*h =
pi*r^2*(30-3r)

v(r) is maximum for r= c, for which  v'(c) =
0 and v"(c) < 0.

v'(r) = 0 gives {'pi* r^2(30-3r)}'
=0

=> pi{30r^2--3r^3}' =
0

=Pi(60r- 9r^2) = 0, Or r(60 -9r). So r =c = 20/3, or r =
c= 0.

v"(c) = pi(60r-18r)< 0 for r = c =
20/3.

h = 30-3r = 30 -3(20/3) =
10 

Therefore the maximum volume of the cylinder v(r) =
p*r^2h = pi*r^2(30-3r) = pi*(20/3)^2*(30-3*20/3)= (4000pi)/9 = 1396.2634 sq
units.

In terms of radius the of the cylinder, the maximum
volume of the inscribed cylinder  = pi*(30r-3r^2) =
3pi*r^2*(10-r).

(iii) Therefore when r = 20/3,  h = 10, the
inscribed cylinnder has the maximum volume = 1396.2634 sq units.