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(A) Let $R = a/2$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.From the geometry of th

Vedclass · Accepted Answer

$(2/3)a$

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(A) Let $R = a/2$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.From the geometry of the sphere,if the center of the sphere is at a distance $x$ from the base of the cone,then $h = R + x = a/2 + x$.The radius of the cone base is $r^2 = R^2 - x^2 = (a/2)^2 - x^2$.The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi ((a/2)^2 - x^2)(a/2 + x)$.$V(x) = \frac{1}{3} \pi (a/2 - x)(a/2 + x)(a/2 + x) = \frac{1}{3} \pi (a/2 - x)(a/2 + x)^2$.To find the maximum volume,differentiate $V$ with respect to $x$ and set to $0$:$\frac{dV}{dx} = \frac{1}{3} \pi [(-1)(a/2 + x)^2 + (a/2 - x) \cdot 2(a/2 + x)] = 0$.$(a/2 + x) [-(a/2 + x) + 2(a/2 - x)] = 0$.Since $h 
eq 0$,$a/2 + x 
eq 0$,so $-(a/2 + x) + a - 2x = 0$.$-a/2 - x + a - 2x = 0 \implies a/2 = 3x \implies x = a/6$.The height $h = a/2 + x = a/2 + a/6 = 3a/6 + a/6 = 4a/6 = (2/3)a$.

The height of a right circular cone of maximum volume inscribed in a sphere of diameter $a$ is-

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