Given a right circular cone, you put an upside-down cone inside it so that its vertex is at the center of the base of the larger cone, and its base is parallel to the base of the larger cone. if you choose the upside-down cone to have the largest possible volume, what fraction of the volume of the larger cone does it occupy?

Question

Pittman

Mathematics

31 August, 08:47

Given a right circular cone, you put an upside-down cone inside it so that its vertex is at the center of the base of the larger cone, and its base is parallel to the base of the larger cone. if you choose the upside-down cone to have the largest possible volume, what fraction of the volume of the larger cone does it occupy?

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Answers (1)

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Aiden · Answer 1

OK, letting x be height and b radius of smaller,

then x/b=h/r, and

volumeof smaller is 1/3pi (h-x) b^2, or

1/3pi (h-x) (x^2r^2/h^2=

1/3pi ((x^2r^2/h) - (x^3r^2/h^2).

Now differinate with respect to x,

and set to 0, so

((2xr^2/h) - (3x^2r^2/h^2) = 0,

so you get x=2/3h.

So, fraction is ((1/3pi (h-2/3h) (4/9r^2) / (1/3pir^2h) = 4/27