Maximizing the Volume of a CylinderDate: 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? Thanks. 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 (= 0). 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 Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/