Miscellaneous · Q12: Cone of maximum volume in a sphere of radius $r$ has altitude $\frac{4r}{3}$

Basics you need

Solution

1. The cross-section and the variable

   The rim of the cone's base is a circle on the sphere, and every point of it is at distance $r$ from the sphere's centre $O$⁠. So $O$ lies on the line through the centre of the base perpendicular to it, which is the cone's axis. The vertex $V$ is also on the axis and on the sphere. Cutting through the axis gives a circle of radius $r$ with the isosceles triangle of the cone inside it.

   Let $h$ be the height of the cone (from $V$ to the centre $M$ of the base) and $x$ its base radius.

   The whole problem has one degree of freedom: once the height is chosen, the base plane is fixed, and it cuts the sphere in a circle that must be the base. So everything, including the volume, is a function of $h$⁠.
   All points of a sphere are at distance $r$ from its centre; a right circular cone is symmetric about its axis. Basics: building the function

2. The allowed range of $h$

   $$0 < h < 2r.$$
   The vertex is at one end of a diameter along the axis, and the base plane must cut the sphere in a genuine circle, so $M$ lies strictly between the two ends of that diameter, whose length is $2r$⁠. At $h = 0$ or $h = 2r$ the base shrinks to a point and the "cone" has no volume, so both ends are excluded.
   A plane meets a sphere in a circle only when its distance from the centre is less than $r$⁠: here $|h - r| < r$⁠, which is $0 < h < 2r$⁠. Basics: interval notation

3. Base radius in terms of $h$

   In the right triangle with vertices $O$⁠, $M$ and a point $E$ on the rim of the base: $OE = r$⁠, $ME = x$⁠, $OM = |h - r|$⁠. Pythagoras gives

   $$\begin{aligned} x^2 &= r^2 - (h-r)^2 \\ &= r^2 - h^2 + 2rh - r^2 \\ &= 2rh - h^2. \end{aligned}$$
   $V$ is at distance $r$ above $O$ and $M$ is $h$ below $V$⁠, so $M$ is $h - r$ below $O$ (or above it if $h < r$⁠). Squaring removes the sign, so one formula covers both cases. Note $2rh - h^2 = h(2r - h) > 0$ on the allowed range, as a squared radius should be.
   Pythagoras' theorem in triangle $OME$⁠, right-angled at $M$⁠. Basics: area and volume formulas

4. The volume as a function of $h$

   $$\begin{aligned} V(h) &= \frac{1}{3}\pi x^2 h = \frac{\pi}{3}(2rh - h^2)\,h \\ &= \frac{\pi}{3}\left(2rh^2 - h^3\right), \qquad 0 < h < 2r. \end{aligned}$$
   Substituting $x^2$ from Step 3 removes $x$⁠, leaving a polynomial in $h$ (with $r$ a fixed constant). Polynomials are easy to differentiate.
   $V = \frac13\pi x^2 h$ (Basic 1). Basics: area and volume formulas

5. Critical point

   $$\begin{aligned} V'(h) &= \frac{\pi}{3}\left(4rh - 3h^2\right) \\ &= \frac{\pi}{3}\,h\,(4r - 3h). \end{aligned}$$

   On $0 < h < 2r$ the factor $h$ is never 0, so

   $$V'(h) = 0 \iff 4r - 3h = 0 \iff h = \frac{4r}{3},$$

   and $\frac{4r}{3}$ does lie in $(0, 2r)$⁠.

   $h = 0$ also makes $V'$ zero, but it is outside the range (no cone). $V$ is a polynomial, differentiable everywhere, so $h = \frac{4r}{3}$ is the only critical point.
   Critical point: $V'(c) = 0$ (Note after Theorem 2, Section 6.4). Lesson: maxima and minima

6. Second derivative test

   $$V''(h) = \frac{\pi}{3}(4r - 6h),$$ $$V''\left(\tfrac{4r}{3}\right) = \frac{\pi}{3}(4r - 8r) = -\frac{4\pi r}{3} < 0.$$

   By the Second Derivative Test (Theorem 4 (i)), $h = \frac{4r}{3}$ is a point of local maxima of $V$⁠.

   $V' = 0$ and $V'' < 0$ at the same point: the graph of $V$ is flat there and bending downwards, which is a peak.
   Theorem 4 (Second Derivative Test), Section 6.4. Lesson: derivative tests

7. It is the largest volume on the whole range

   For $0 < h < 2r$⁠, the factors $\frac{\pi}{3}$ and $h$ are positive, so $V'(h)$ has the sign of $4r - 3h$⁠:

   • for $0 < h < \frac{4r}{3}$⁠: $4r - 3h > 0$⁠, so $V' > 0$ and $V$ is increasing;
   • for $\frac{4r}{3} < h < 2r$⁠: $4r - 3h < 0$⁠, so $V' < 0$ and $V$ is decreasing.

   So $V(h) < V\left(\frac{4r}{3}\right)$ for every other $h$ in the range: the cone with altitude $\frac{4r}{3}$ has the maximum volume among all inscribed cones.

   The second derivative test only compares $V\left(\frac{4r}{3}\right)$ with nearby heights. The question is about the largest cone of all, so we use the sign of $V'$ over the entire range: $V$ climbs all the way up to $\frac{4r}{3}$ and falls all the way after it. Near both ends of the range $V$ tends to 0.
   Theorem 1, Section 6.3, and its proof, applied on $\left[h, \frac{4r}{3}\right]$ and $\left[\frac{4r}{3}, h\right]$⁠. Lesson: increasing and decreasing

8. The maximum volume (for completeness)

   At $h = \frac{4r}{3}$⁠: $x^2 = 2r\cdot\frac{4r}{3} - \frac{16r^2}{9} = \frac{8r^2}{9}$⁠, so $x = \frac{2\sqrt2\,r}{3}$⁠, and

   $$V_{\max} = \frac{1}{3}\pi\cdot\frac{8r^2}{9}\cdot\frac{4r}{3} = \frac{32\pi r^3}{81}.$$
   The question only asks for the altitude, but the volume is a useful check: $\frac{32\pi r^3}{81} \div \frac{4}{3}\pi r^3 = \frac{8}{27}$⁠, so the largest cone fills $\frac{8}{27}$ of the sphere, the result of Exercise 6.3, Question 23.
   $\frac{8r}{3} = \frac{24r}{9}$⁠, so $\frac{24r^2}{9} - \frac{16r^2}{9} = \frac{8r^2}{9}$⁠. Volume of a sphere: $\frac43\pi r^3$⁠. Basics: area and volume formulas

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