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October 31, 2014

October 31, 2014

Posted by **Lee Meyer** on Saturday, November 10, 2012 at 2:03pm.

- Math -
**Steve**, Saturday, November 10, 2012 at 2:31pmGiven an arc of x radians, you should be able to convince yourself that you want the maximum volume

v = 1/3 pi r^2 h

where r = x/2pi * 20 = 10x/pi

and h = √(400-r^2)

so,

v = 1/3 pi (10x/pi)^2 √(400-(10x/pi)^2)

= 1/3 pi (10/pi)^3 x^2 √(4pi^2-x^2)

Note that as expected,

v(0) = 0 (zero radius base)

v(2pi) = 0 (zero height)

so, maximum volume must be somewhere in between.

dv/dx = 1000x/3pi^2 (8pi^2 - 3x^2)/√(4pi^2-x^2)

dv/dx = 0 when x=0 (minimum volume)

or x = √(8pi^2/3) (maximum volume)

so, for that value of x,

r = 10/pi √(8pi^2/3)

h = √(400 - 100pi^2 * 8pi^2/3) = 20/√3

max v is thus pi/3 (100/pi^2)(8pi^2/3) (20/√3) = 2000pi/(9√3)

As usual, check my math.

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