Posted by Soroban on August 31, 2002 at 01:14:37:
In Reply to: Hint? posted by Ryan on August 29, 2002 at 09:39:04:
: Could someone give me a hint:
: I have a right circular cone with a slant hieght of 3 a radius of r and a hight of h.
: I want to know what angle "o"(between the perpendicular height and the slant hieght) will maximize the volume.
:
: /|\
: /o| \
: / | \
: I know that O = tan-1 ( r/h) or r / sqrt(9-r^2)
: But I am having trouble relating this to the volume of the cone, or if i am even on the right track using an inverse trig.
: Thank you
Ryan,
Did they ask you to find that angle?
Or was that your idea?
Either way, we can solve it without resorting to
inverse trig functions.
As suggested, begin with the Volume formula:
V = (pi/3)(r^2)h
From your right triangle: r^2 + h^2 = 9.
Solve for r^2: r^2 = 9 - h^2
Substitute that into the V-function:
V = (pi/3)(9 - h^2)h
Differentiate, etc. You'll find the value of h
which maximizes the volume.
If they want r, use r^2 + h^2 = 9, to find r.
If they want O<.i> (half of the vertex angle), then
you can use tan^(-1)(O) = r/h.