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 May 2nd, 2009, 12:13 AM #1 Newbie   Joined: Apr 2009 Posts: 13 Thanks: 0 geometry has been an area of the ball 27pi cm^2. Sphere is designed with a cylinder. Find the radius. Find the cylinder of the radius of the bottom r height h through. Find the volume of a cylinder height of the cylinder h through. How big should be the height of the cylinder, the cylinder volume shall be the maximum. Please give some explanations too !
 May 2nd, 2009, 12:25 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: geometry Post again in the original language. Don't translate. :P Hints: for a sphere, $V=\frac43\pi r^3, \qquad S=4\pi r^2$
 May 3rd, 2009, 04:41 AM #3 Newbie   Joined: Apr 2009 Posts: 13 Thanks: 0 Re: geometry there is sphere and sphere area is S=27Pi . In the sphere is cylinder. Find cylinder bottom radius . Find cylinder volume. how big should be the height of the cylinder, the cylinder volume is a maximum !
 May 3rd, 2009, 08:21 AM #4 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: geometry The surface area gives you the radius of the sphere. That can be used to express the height of the cylinder, h, in terms of the radius of the cylinder, r [Use Pythagoras]. Express the volume of the cylinder in terms of "h" and "r". Substitute for "r", then differentiate to find the max. You could just as well use "h", subbing "h", appropriately, for "r". This is the 3D equivalent of finding the max area of a rectangle within a circle.
 May 4th, 2009, 09:44 PM #5 Global Moderator   Joined: Dec 2006 Posts: 17,912 Thanks: 1382 Let R be the radius of the sphere, so that 4?R² = 27?, and so R = (3/2)?3. Let r, 2h and V be the radius, height and volume, respectively, of the inscribed cylinder. By Pythagoras's theorem, R² = r² + h², so V = 2?r²h = 2?(R² - h²)h = 2?R²h - 2?h³. dV/dh = 2?R² - 6?h², which is zero if h = R/?3 = 3/2, so V is maximized when the height of the cylinder = 3. It's now easy to calculate the cylinder's radius and volume.

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