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Find the largest cone that can be inscribed in a given sphere

Consider a sphere of fixed radius R. Find the right circular cone of maximum volume that can be inscribed in the sphere in terms of R, and the radius and altitude of the cone, r and h, respectively.


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We want to maximize the volume of the cone,

    \[ V = \frac{1}{3} \pi r^2 h. \]

From the diagram we find the following expression for h in terms of R and r,

    \[ h = R + \sqrt{R^2 - r^2}. \]

Thus, our expression for V in terms of r is

    \[ V = \frac{1}{3} \pi r^2 R + \frac{1}{3} \pi r^2 \sqrt{R^2 - r^2}. \]

Taking the derivative of this we have

    \begin{align*}  \frac{dV}{dr} &= \frac{2}{3} \pi r R + \frac{2}{3} \pi r \sqrt{R^2 - r^2} - \frac{1}{3} \pi r^3 \left( \frac{1}{ \sqrt{R^2 - r^2} } \right)\\  &= \frac{2}{3} \pi r (R + \sqrt{R^2 - r^2} ) - \frac{\pi r^3}{3 \sqrt{R^2 - r^2}}.  \end{align*}

Setting this equal to 0 we have

    \begin{align*}  \frac{dV}{dr} = 0 && \implies && \frac{2}{3} \pi r (R + \sqrt{R^2 - r^2} ) - \frac{ \pi r^3}{3 \sqrt{R^2 - r^2}} &= 0 \\  && \implies && R + \sqrt{R^2 - r^2} - \frac{r^2}{2 \sqrt{R^2 - r^2}} &= 0 \\  && \implies && 2R \sqrt{R^2 - r^2} + 2R^2 - 2r^2 - r^2 &= 0 \\  && \implies && 3r^2 - 2R^2 &= 2R\sqrt{R^2 - r^2} \\  && \implies && 9r^2 - 12r^2 R^2 + 4R^4 &= 4R^4 - 4R^2 r^2 \\  && \implies && 9r^2 &= 8r^2 R^2 \\  && \implies && r = \frac{2 \sqrt{2}}{3} R.  \end{align*}

(We used that r \neq 0 a couple of times in the computation, which is fine since the cone does not have radius 0.) This critical point is a maximum since

    \begin{align*}  \frac{dV}{dr} &> 0 & \text{when} && r &< \frac{2 \sqrt{2}}{3} R \\  \frac{dV}{dr} &< 0 & \text{when} && r &> \frac{2 \sqrt{2}}{3} R. \end{align*}

Then, plugging this value of r back into our expression for h we have

    \[ h = R + \sqrt{R^2 - \frac{8}{9} R^2} = \frac{4}{3} R. \]

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