The let the height of the cone be h and radius r inscribed
in the cone.
The vertical semiangle of the cone with
axis , x = arc tan (4/2*12) = arc tan (1/6) .
x = arc
tan (1/6)
Therefore the radius of the cylinder r = (12-h)
tan x = (1-h)/6, where h is the height of cone.
Therefore
the volume of the cylinder v(h) = pi r^2*h =
pi*{(12-h)^2/6^2}*h.
To find maximum v(x) , we find x = c ,
for which v'(c) = 0 and v"(c) < 0.
v'(x) = p^2/36 {
(12-h)^2*h}'
v'(x) = p^2/36{ 144h -
24h^2+h^3}'
v'(x) =
p^2/36{144-48h+3h^2}
v'(x) = 0 gives 144-48h+3h^2 = 0.
Or
Divide by 3:
48-16h+h^2 =
0.
(h-12)(h-4) = 0
So h = 4,
or h= 12.
v"(x) = pi/36{ -48 +6h} < 0 for x= 4 as
-48+6*4 = -24.
Therefore the volume of the inscribed
cylinder is maximum for x = 4 and the maximum volume is
:
v(4) = (pi/36){ (12-4)^2* 4} = (pi/36)(64*4)= 64pi/9 =
22.34. sq units.
The height of the cone = 4 and the radius
of the cone = (12-4)/6 = 4/3.
No comments:
Post a Comment