# 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 and the radius of its base r (cm) 1-Proved...

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 *

and the radius of its base **r (cm)**

**1-Proved that:h=3(10-r)**

**2-Express V(r) in terms of volume of a cylinder r**

**3-Concluded that the values of h and r takes up the sise V(r) is the largest possible value **

*print*Print*list*Cite

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.