# some hints...

[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]

Posted by T.Gracken on August 29, 2002 at 13:28:27:

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,

Start with volume of a right circular cone is V = (1/3)pi*r2h

now, (I'll use T for angle):

sin(T) = r/3, so r = 3sin(T)

and

cos(T) = h/3, so h = 3cos(T)

This allows us to write the volume as: V = 9pi[cos(T) - cos3(T)] ...with some substitution and simplifying.

Now, find V' and its critical values (keep in mind that T must be between 0o and 180o)

...I got two critical numbers and ended up with a max volume when T = arccos[-sqrt(3)/3] which is appx. 125.3o. But I did the work very quickly and you should not rely on my final answer. The set up should be sound and is just one way to approach this.

hope that helps.

Name:
E-Mail:

Subject: