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December 28, 2014

December 28, 2014

Posted by **Marty** on Monday, March 9, 2009 at 8:01pm.

- Calculus -
**Reiny**, Monday, March 9, 2009 at 8:16pmMake a diagram.

let the radius of the cone be x,

let the distance from the base of the cone to the centre of the sphere by y,

then the height of the cone is y+10

draw a line from the centre of the sphere to the end of the radius of the cone.

You now have a right-angled triangle where

x^2 + y^2 = 100

and x^2 = 100 - y^2

volume of a cone = (1/3)pi(r^2)h

= (pi/3)(100 - y^2)(y+10)

= (pi/3(100y + 1000 - y^3 - 10y^2)

d(volume)/dy = pi/3(100 - 3y^2 - 20y)

= 0 for a max/min of volume

so 3y^2 + 20y - 100 = 0

(3y-10)(y+10) = 0

then y = 10/3 or y = -10 which would be silly

so y = 10/3

sub that back in x^2 - 100 - y^2

and then into the volume of the cone

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