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July 29, 2014

July 29, 2014

Posted by **Emma** on Saturday, January 1, 2011 at 8:21pm.

- PreCalculus -
**Reiny**, Saturday, January 1, 2011 at 11:51pmThe key thing is to see that the radius of 4 inches of the original circle shows up as the slant height of the cone.

let the height of the cone be h, and the radius of its base be r

r^2 + h^2 = 16, r^2 = 16-h^2

V = (1/3)πr^2 h

= (1/3)π(16-h^2)h

= 16πh/3 - πh^3/3

16πh/3 - πh^3/3 > 21

divide by -π/3

h^3 - 16h + 20.05 < 0

let's consider the equation

h^3 - 16h + 20.05 = 0

using my trusty cubic equation solver

http://www.1728.com/cubic.htm

I got h = -4.5, h = 3.08 and h = 1.44

looking at the graph of

h^3 - 16h + 20.05 < 0

I saw that it was negative for 1.44 < h < 3.08

so r has to be 2.55 < r < 3.73

Since the circumference of the base of the cone is the arc-length x as defined in your question

sub in the values of r into 2πr

if r = 2.55, x = 16.022

if r = 3.73, x = 23.44

check my arithmetic

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