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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

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

3 comments

  1. Anonymous says:

    The analysis of the problem, shown in the drawing at the beginning of the solution, assumes that the height h that maximizes the volume of the cone is greater than R. But h can be anything in between 0 and 2R. To allow for this, we could choose R^2 = r^2 + (h – R)^2 (or equivalently R^2 = r^2 + (R – h)^2) as the expression that relates R, r, and h. We could then write the volume of the cone as a function of either h or r (h works best), determine the derivative of that function, and so on. Finally, we would find that the height h that maximizes the volume of the cone is indeed greater than R.

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