Exercise 6.3 · Q23: Largest cone in a sphere is $\tfrac{8}{27}$ of the sphere

Basics you need

Solution

1. The cross-section and the variable

   Let O be the centre of the sphere, A the apex of the cone, D the centre of its base and C a point on the rim of the base. Cut through the axis AD: the sphere gives a circle of radius $R$ and the cone gives the triangle ABC inscribed in it.

   Let the cone have height AD $= h$ and base radius DC $= r > 0$⁠. The axis AD passes through O (see the Why below), so it meets the sphere again at A′, the point diametrically opposite A, and AA′ $= 2R$⁠. The base centre D lies strictly inside the sphere, because $\text{OD}^2 = R^2 - r^2 < R^2$ (Step 2). So D is on the diameter AA′ but at neither end, and

   $$0 < h < 2R.$$

   Conversely, for every $h$ in this range the plane at distance $h$ from A, perpendicular to the diameter through A, cuts the sphere in a circle, and that circle with apex A is an inscribed cone. So every $h \in (0, 2R)$ is allowed.

   The sphere's centre lies on the cone's axis: every rim point C is at distance $R$ from O, so the foot of the perpendicular from O to the base plane is equally far from all of them, which makes it the centre D of the base circle. The cone is right circular, so its apex is also on the line through D perpendicular to the base: A, O and D are on one line. At $h = 2R$ the base would shrink to a point, so there is no cone.
   An open range: the cone must have a positive height and a real base. Basics: interval notation

2. The base radius from triangle ODC

   Triangle ODC has a right angle at D, with OC $= R$ and OD $= |h - R|$ (the base is below O when $h > R$ and above it when $h < R$⁠). By Pythagoras' theorem,

   $$\begin{aligned} r^2 &= R^2 - (h - R)^2 \\ &= R^2 - h^2 + 2Rh - R^2 \\ &= 2Rh - h^2. \end{aligned}$$
   This is the link between the cone's height and its width imposed by the sphere. Because OD appears squared, the formula is the same whether the base lies below or above the centre, so one formula covers every $h$ in $(0, 2R)$⁠. Check: $h = R$ gives $r = R$ (the base passes through O), and $h \to 2R$ gives $r \to 0$⁠.
   $(h - R)^2 = h^2 - 2Rh + R^2$⁠; and $|h - R|^2 = (h - R)^2$⁠.

3. The volume as a function of $h$

   $$\begin{aligned} V(h) &= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(2Rh - h^2)h \\ &= \frac{\pi}{3}\left(2Rh^2 - h^3\right), \quad 0 < h < 2R. \end{aligned}$$
   This is the quantity to maximise. Choosing $h$ as the variable avoids square roots: $r$ appears only as $r^2$⁠, which Step 2 gives as a polynomial in $h$⁠.
   Volume of a cone (Basic 1). Basics: building the function to optimise

4. Differentiate and find the critical point

   $$V'(h) = \frac{\pi}{3}\left(4Rh - 3h^2\right) = \frac{\pi}{3}\,h\,(4R - 3h).$$

   On $(0, 2R)$ we have $h \ne 0$⁠, so $V'(h) = 0$ exactly when $4R - 3h = 0$⁠:

   $$h = \frac{4R}{3}.$$

   Since $0 < \frac{4R}{3} < 2R$⁠, this point is in the range, and it is the only critical point there ($V$ is a polynomial).

   Factoring out $h$ shows the two zeros of $V'$⁠: $h = 0$ (not a cone) and $h = \frac{4R}{3}$⁠. Only the second is in the allowed range.
   Power rule: $\frac{d}{dh}(2Rh^2) = 4Rh$ and $\frac{d}{dh}(h^3) = 3h^2$⁠; $R$ is a constant.

5. Second derivative test

   $$\begin{aligned} V''(h) &= \frac{\pi}{3}(4R - 6h), \\ V''\!\left(\tfrac{4R}{3}\right) &= \frac{\pi}{3}(4R - 8R) = -\frac{4\pi R}{3} < 0. \end{aligned}$$

   By the second derivative test, $h = \frac{4R}{3}$ is a point of local maxima of $V$⁠.

   A negative second derivative means the graph bends downwards: the flat point is the top of a hill.
   Theorem 4 (i), Section 6.4. Lesson: first and second derivative tests

6. It is the greatest volume of all inscribed cones

   In $V'(h) = \frac{\pi}{3}h(4R - 3h)$ the factors $\frac{\pi}{3}$ and $h$ are positive on $(0, 2R)$⁠, so $V'$ has the sign of $4R - 3h$⁠: positive for $0 < h < \frac{4R}{3}$ and negative for $\frac{4R}{3} < h < 2R$⁠. Hence $V$ increases on $\left(0, \frac{4R}{3}\right]$ and decreases on $\left[\frac{4R}{3}, 2R\right)$⁠, and

   $$V(h) < V\!\left(\tfrac{4R}{3}\right) \quad\text{for every other } h \in (0, 2R).$$

   At both ends of the range the volume shrinks to 0: $V(h) \to 0$ as $h \to 0^+$ and as $h \to 2R^-$⁠.

   "Largest cone" means largest among all inscribed cones, so a local maximum is not enough. One critical point with $V$ rising before it and falling after it is the highest point of the whole graph.
   Theorem 1, Section 6.3 (positive derivative: increasing; negative: decreasing). Lesson: increasing and decreasing functions

7. The greatest volume

   Put $h = \frac{4R}{3}$ into $V(h)$⁠:

   $$\begin{aligned} V_{\max} &= \frac{\pi}{3}\left(2R \cdot \frac{16R^2}{9} - \frac{64R^3}{27}\right) \\ &= \frac{\pi}{3}\left(\frac{96R^3}{27} - \frac{64R^3}{27}\right) \\ &= \frac{\pi}{3} \cdot \frac{32R^3}{27} = \frac{32\pi R^3}{81}. \end{aligned}$$

   (The base radius is then given by $r^2 = \frac{8R^2}{3} - \frac{16R^2}{9}$ $= \frac{8R^2}{9}$⁠, so $r = \frac{2\sqrt{2}R}{3}$⁠.)

   Substitute only after differentiating. Writing $\frac{32R^3}{9}$ as $\frac{96R^3}{27}$ puts both terms over the same denominator.
   $\left(\frac{4R}{3}\right)^2 = \frac{16R^2}{9}$⁠, $\left(\frac{4R}{3}\right)^3 = \frac{64R^3}{27}$⁠.

8. Compare with the sphere

   $$\frac{V_{\max}}{\frac{4}{3}\pi R^3} = \frac{32\pi R^3}{81} \cdot \frac{3}{4\pi R^3} = \frac{96}{324} = \frac{8}{27}.$$

   So the volume of the largest inscribed cone is $\frac{8}{27}$ of the volume of the sphere, as was to be proved.

   The ratio does not depend on $R$⁠: every sphere, large or small, holds a largest cone that fills the same fraction of it, a little under 30%.
   $96 = 12 \times 8$ and $324 = 12 \times 27$⁠.

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