Cone VolumeDate: 04/19/99 at 16:59:27 From: SHAMSI Subject: Volume of a cone Would you please answer this question for me? A right circular cone is to be circumscribed about a sphere of radius R cm. Find the ratio of the altitude to the base radius of the cone of largest possible volume. I started this question but got confused because there are no actual numbers to work with. Your help would be greatly appreciated. Thanks, Shamsi Date: 04/19/99 at 17:31:27 From: Doctor Rob Subject: Re: Volume of a cone Thanks for writing to Ask Dr. Math! Let the radius of the sphere be r, the radius of the base of the cone R, and the altitude of the cone h. Draw this cross-sectional picture: C o /|\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / _,,--+--.._ \ / ,-' |h-r `-. \ /,' | `.\ ,' | `. / | \ D / h | _,-'\ , | _,-' . /| | _,-' r |\ / + O+' + \ / | | | \ / . | , \ / \ | / \ / \ |r / \ / `. | ,' \ / `. | ,' \ / `-._ | _,-' \ o--------------------``--+--''--------------------o A R B Using similar right triangles ABC and DOC, you get the equation h/R = CD/r CD = h*r/R Using the Pythagorean Theorem on DOC, you get (h*r/R)^2 + r^2 = (h-r)^2 You are interested in the value of x = h/R, so elminate R from the equation by substituting R = h/x: (r*x)^2 + r^2 = h^2 - 2*h*r + r^2 r^2*x^2 = h^2 - 2*h*r Now you can solve this equation for h, and substitute that into the formula V = Pi*h*r^2/3 for the volume. Then differentiate the result with respect to x, remembering that r is a constant. Set the derivative equal to zero, and solve for x. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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