If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is ______________ .

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$\frac{2}{3}$ \text { Let h, r, V and R be the height, radius of the base, volume of the cone and the radius of the sphere, respectively } . \text { Given:} h = R + $\sqrt$(R^2 - r^2) \Rightarrow h - R = $\sqrt$(R^2 - r^2) \text { Squaring both side, we get } h^2 + R^2 - 2hR = R^2 - r^2 \Rightarrow r^2 = 2hr - h^2 ...........\left( 1 \right) \text { Now,} \text { Volume } = $\frac{1}{3}$$\pi$ r^2 h \Rightarrow V = $\pi$/3\left( 2 h^2 R - h^3 \right) ................\left[\text { From eq. } \left( 1 \right) \right] \Rightarrow dV/dh = $\pi$/3\left( 4hR - 3 h^2 \right) \text { For maximum or minimum values of V, we must have } dV/dh = 0 \Rightarrow $\pi$/3\left( 4hR - 3 h^2 \right) = 0 \Rightarrow 4hR - 3 h^2 = 0 \Rightarrow 4hR = 3 h^2 \Rightarrow h = 4R/3 \text { Now,} d^2 V/d h^2 = $\pi$/3\left( 4R - 6h \right) = $\pi$/3\left( 4R - 6 $\times$ 4R/3 \right) = - 4$\pi$ R/3 < 0 \text { So, volume is maximum when h } = 4R/3 . \Rightarrow h = \frac{2\left( 2R \right)}{3} \Rightarrow h/2R = $\frac{2}{3}$ \therefore \frac{\text { Height }}{\text { Diameter of sphere }} = $\frac{2}{3}$

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