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Maximizing the Volume of a Cylinder

Date: 7/24/96 at 15:11:4
From: Anonymous
Subject: Maximizing the Volume of a Cylinder

I need to maximize the volume of a right-circular cylinder that fits 
inside a sphere of radius 1 m.

I know that the first derivative of the volume of the cylinder is
2pi*r.  When I set this equal to zero, I get r = O, which is no help. 

Am I approaching this in the wrong way?


Date: 7/24/96 at 18:2:47
From: Doctor Anthony
Subject: Re: Maximizing the Volume of a Cylinder

Let r = radius of sphere (I will put this equal to 1 later)

Now a line drawn from the centre of the sphere to an edge of the 
cylinder where it meets the surface of the sphere makes an angle  
'theta' with the axis of the cylinder, and 'theta' is the variable we 
shall be using to maximize the volume of the cylinder.

In terms of theta the radius of the cylinder = r*sin(theta)
 "   "   "    "    "  height "   "     "     = 2*r*cos(theta)

Volume of cylinder = pi(r*sin(theta))^2*{2*r*cos(theta)}

                   = 2*pi*r^2*sin^2(theta)*cos(theta)  now put r=1

        Volume = 2*pi*(1 - cos^2(theta))*cos(theta)

               = 2*pi*(cos(theta) - cos^3(theta))

Now find dV/d(theta) and equate this to zero to find value of theta 
giving the maximum volume.

   dV/d(theta) = 2*pi{-sin(theta) - 3*cos^2(theta)*(-sin(theta)) = 0

Divide out by (-sin(theta)).  This put equal to 0 would satisfy the 
equation, and corresponds to value of theta giving MINIMUM volume (= 

Other solution is given by   1 - 3*cos^2(theta) = 0

                                cos^2(theta) = 1/3

                                cos(theta) = 1/sqrt(3)

So max Volume = 2*pi{1/sqrt(3) - (1/3)^(3/2)}

              = 2*pi/sqrt(3){1 - 1/3}

              = 2*pi/sqrt(3){2/3}

              = 4*pi/(3sqrt(3))        

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Calculus

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