In a right circular cone, the semi-vertical angle of which is θ, a cube is placed so that four of its vertices are on the base and four on the curved surface. Prove that as θ varies the maximum of the ratio for the volume of the cube to the volume of the cone occurs when sin θ = 1/3.
Note: The semi-vertical angle of a cone is half the vertex angle.
For a cone with height h and radius r, tan θ = r/h.
First note by simple trig: if
sin θ = 1/3 then tan
θ = sqrt(2)/4
Let the h=1 then r=tan θ
and V(cone)=pi*tan^2(θ)/3
As seen from 45 degrees to the side, the cube of side s reaches s units up but only s/sqrt(2) units to the side, where the upper corner meets the curved surface of the cone.
By similar triangles (or tangent)
s= (sqrt(2)tan(θ)) / (1+sqrt(2)tan(θ))
V(cube)=s^3
https://www.desmos.com/calculator/cfsu8kcidy
We wish to maximize the ratio. Since tangent is an increasing function on the domain (0,pi/2), it will be simpler to replace tan(x) with x.
Here's the algebra to simplify the ratio, take it derivative, and set te derivative to zero.
https://www.desmos.com/calculator/5pnknqhcjx
1-6x^2-4sqrt(2)x^3=0
Desmos got some rounding error, but if you plug in sqrt(2)/4 you do actually get 1 - 3/4 - 1/4 = 0
And finally, since tan(θ)=sqrt(2)/4, sin(θ)=1/3.
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Posted by Jer
on 2024-08-23 12:51:29 |
solution