Answer to Question #341049 in Calculus for boyyy

Question

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

radius 9 inches.

Answer 1

Let $R=$ the radius of the sphere.

Let $AD=r, BD=h, \angle ABC=\theta.$

$\triangle ABC$

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

The Law of Cosines

$$(2r)^2=R^2+R^2-2R^2\cos(2\theta)$$ $$4r^2=2R^2(2\sin^2\theta)$$ $$r=R\sin \theta$$ $$h=r\tan(\theta/2)=R\sin \theta\tan(\theta/2)$$ $$V_{cone}=\dfrac{1}{3}\pi r^2 h$$ $$V_{cone}=V_{cone}(\theta)=\dfrac{1}{3}\pi R^3\sin^3 \theta\tan(\theta/2)$$ $$(V_{cone})'_{\theta}=\dfrac{1}{3}\pi R^3\bigg(3\sin^2 \theta\cos(\theta)\tan(\theta/2)$$ $$+\dfrac{1}{2}\sin^3\theta(\dfrac{1}{\cos ^2(\theta/2)})\bigg)$$

Find the critical numbr(s)

$$(V_{cone})'_{\theta}=0$$ $$\dfrac{1}{3}\pi R^3\bigg(3\sin^2 \theta\cos\theta\tan(\theta/2)$$ $$+\dfrac{\sin^3\theta}{2\cos ^2(\theta/2)}\bigg)=0$$ $$\dfrac{\sin^3 \theta(3\cos\theta+1)}{2\cos ^2(\theta/2)}=0$$ $$\cos \theta=-\dfrac{1}{3}$$

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

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

The volume of inscribed cone has the absolute maximum at

$$\theta=\pi-\cos^{-1}(1/3)$$