The height of a right circular cone of maximu | vedclass

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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,if $x

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$(2/3)a$

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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,if $x$ is the distance from the center of the sphere to the base of the cone,then $h = R + x = a/2 + x$.The radius of the base of the cone is given by $r^2 = R^2 - x^2 = (a/2)^2 - x^2$.The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi ((a/2)^2 - x^2)(a/2 + x)$.Expanding this,$V = \frac{\pi}{3} (a^2/4 - x^2)(a/2 + x) = \frac{\pi}{3} (a^3/8 + a^2x/4 - ax^2/2 - x^3)$.To maximize volume,find $dV/dx = 0$:$dV/dx = \frac{\pi}{3} (a^2/4 - ax - 3x^2) = 0$.$3x^2 + ax - a^2/4 = 0$.Using the quadratic formula,$x = \frac{-a \pm \sqrt{a^2 - 4(3)(-a^2/4)}}{2(3)} = \frac{-a \pm \sqrt{a^2 + 3a^2}}{6} = \frac{-a \pm 2a}{6}$.Since $x$ must be positive,$x = a/6$.Therefore,the height $h = a/2 + a/6 = (3a + a)/6 = 4a/6 = (2/3)a$.

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

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