# Re: Hint?

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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?

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.

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