We have a cone of height 30 cm and radius 10 cm. A cylinder of height h with the radius of the base equal to r is drawn in the cone.
Now we want to maximize the volume of the cone. The volume of the cylinder is pi*r^2*h.
To find the optimum values of h and r we take a two dimensional figure with a rectangle of sides h and 2r in a triangle with height 30 cm and base 20 cm.
Now equating the ratios for deriving the tan of the angle we have:
30/10 = h/(10-r)
=> 3(10-r) = h
=> 30 - 3r = h
The area of the rectangle is 2r*h
=> 2*r*(30 - 3r)
=> 60r - 6r^2
The derivative of this with respect to r is 60 - 12r
60 - 12r = 0
=> 12r = 60
=> r = 5
h = 30 - 15 = 15
For maximum volume the height is 15 and the radius is 5. And we see that 15 = 3*( 10 - 5) = 30 - 15 = 15.
The required height of the cylinder is 15 cm and the radius of its base is 5 cm.