Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\dfrac{8}{27}$ of the volume of the sphere. **OR** Show that the semi-vertical angle of the right circular cone of the maximum volume and of given slant height is $\tan^{-1}\sqrt{2}$.

> > [!TLDR] > Express the cone's volume in terms of a single variable (its height $h$) using the sphere's geometric constraint, maximize with calculus, then compare to the sphere's own volume. > > **Setup.** Let a cone of height $h$ and base radius $r$ be inscribed in a sphere of radius $R$. If the cone's apex and the centre of its base lie on a diameter, standard geometry (Pythagoras on the right triangle formed by $R$, $r$, and $(h-R)$) gives: > $$r^2 = R^2-(h-R)^2 = 2Rh-h^2, \qquad 0 > **Step 1: Volume as a function of $h$.** > $$V = \frac13\pi r^2h = \frac13\pi(2Rh-h^2)h = \frac{\pi}{3}\left(2Rh^2-h^3\right)$$ > > **Step 2: Differentiate and find critical points.** > $$\frac{dV}{dh} = \frac{\pi}{3}\left(4Rh-3h^2\right) = \frac{\pi}{3}h(4R-3h)$$ > > Setting $\dfrac{dV}{dh}=0$: $h=0$ (rejected, degenerate) or $h=\dfrac{4R}{3}$. > > **Step 3: Second derivative test.** …

Prove that the volume of the largest cone that can

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Meghalaya MboseMBOSE Meghalaya Intermediate Board 2026Subjective· 5mImportance★★★★★est

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For a cone inscribed in a sphere of radius $R$ , $r^2=2Rh-h^2$ ; maximizing $V=\frac13\pi r^2h$ gi …

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