Question #341049

Find the angle of the largest right circular cone which can be inscribed in a sphere of

radius 9 inches.


1
Expert's answer
2022-05-17T14:43:28-0400


Let R=R= the radius of the sphere.

Let AD=r,BD=h,ABC=θ.AD=r, BD=h, \angle ABC=\theta.


ABC\triangle ABC

AC=2r,AO=OC=R,AOC=2ABC=2θAC=2r, AO=OC=R, \angle AOC=2\angle ABC=2\theta

The Law of Cosines



(2r)2=R2+R22R2cos(2θ)(2r)^2=R^2+R^2-2R^2\cos(2\theta)4r2=2R2(2sin2θ)4r^2=2R^2(2\sin^2\theta)r=Rsinθr=R\sin \thetah=rtan(θ/2)=Rsinθtan(θ/2)h=r\tan(\theta/2)=R\sin \theta\tan(\theta/2)Vcone=13πr2hV_{cone}=\dfrac{1}{3}\pi r^2 hVcone=Vcone(θ)=13πR3sin3θtan(θ/2)V_{cone}=V_{cone}(\theta)=\dfrac{1}{3}\pi R^3\sin^3 \theta\tan(\theta/2)(Vcone)θ=13πR3(3sin2θcos(θ)tan(θ/2)(V_{cone})'_{\theta}=\dfrac{1}{3}\pi R^3\bigg(3\sin^2 \theta\cos(\theta)\tan(\theta/2)+12sin3θ(1cos2(θ/2)))+\dfrac{1}{2}\sin^3\theta(\dfrac{1}{\cos ^2(\theta/2)})\bigg)

Find the critical numbr(s)



(Vcone)θ=0(V_{cone})'_{\theta}=013πR3(3sin2θcosθtan(θ/2)\dfrac{1}{3}\pi R^3\bigg(3\sin^2 \theta\cos\theta\tan(\theta/2)+sin3θ2cos2(θ/2))=0+\dfrac{\sin^3\theta}{2\cos ^2(\theta/2)}\bigg)=0sin3θ(3cosθ+1)2cos2(θ/2)=0\dfrac{\sin^3 \theta(3\cos\theta+1)}{2\cos ^2(\theta/2)}=0cosθ=13\cos \theta=-\dfrac{1}{3}

If 0<θ<πcos1(1/3),(Vcone)θ>0,Vcone0< \theta<\pi-\cos^{-1}(1/3), (V_{cone})'_{\theta}>0, V_{cone} increases.


If πcos1(1/3)<θ<π,(Vcone)θ<0,Vcone\pi-\cos^{-1}(1/3)<\theta<\pi, (V_{cone})'_{\theta}<0, V_{cone} decreases.


The volume of inscribed cone has the absolute maximum at


θ=πcos1(1/3)\theta=\pi-\cos^{-1}(1/3)

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