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? (Let H and R be the height and radius of the large cone, let h and r be the height and radius of the small cone. Use similar triangles to get an equation relating h and r. The formula for the volume of a cone is V = 1/2pir^2h.)
Okay, 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
I have no idea how to approach this problem, if someone knows just how to relate h, r with H,R, that would be extremely helpful and I can workout the rest! Thank you in advance. Given a right circular cone, you put an upside-down
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