Inscribed Truncated Cone.

$m = r\cos(\theta)$ and $h = r\sin(\theta)$

$\implies$ the volume of the truncated cone $V = \dfrac{\pi}{3}(r^2 + r^2\cos(\theta) + r^2\cos^2(\theta))r\sin(\theta) =$

$\dfrac{\pi}{3}r^3(2\sin(\theta) + \dfrac{1}{2}\sin(2\theta) -\sin^3(\theta))$

$\implies \dfrac{dV}{d\theta} = \dfrac{\pi}{3}r^3(2\cos(\theta) + \cos(2\theta) - 3\sin^2(\theta)\cos(\theta)) =$

$\dfrac{\pi}{3}r^3(2\cos(\theta) + 2\cos^2(\theta) - 1 - 3(1 - \cos^2(\theta))\cos(\theta))$

$\implies 3\cos^3(\theta) + 2\cos^2(\theta) - \cos(\theta) - 1 = 0$

Let $u = \cos(\theta) \implies 3u^3 + 2u^2 - u - 1 = 0$

To solve the cubic:

To eliminate the square term let $u = w - \dfrac{2}{9}$

$\implies 729w^3 - 351w - 173 = 0 \implies w^3 - \dfrac{13}{27}w - \dfrac{173}{729} = 0$

Let $w = z + \dfrac{13}{81z} \implies 531441z^6 - 1261172z^3 + 2197 = 0 \implies$

$z^3 = \dfrac{173 + 27\sqrt{29}}{1458} = R = R(\cos(2\pi n) + i\sin(2\pi n)) \implies$

$z = R^{\frac{1}{3}}(\cos(\dfrac{2\pi n}{3}) + i\sin(\dfrac{2\pi n}{3}) \implies$

$z = R^{\frac{1}{3}}, R^{\frac{1}{3}}(-\dfrac{1}{2} \pm i\sqrt{\dfrac{3}{2}})$

In this case we have exactly one real root for $3u^3 + 2u^2 - u - 1 = 0$

Using $z = R^{\frac{1}{3}} \implies w = R^{\frac{1}{3}} + \dfrac{13}{81 R^{\frac{1}{3}}} \implies$

$u = \dfrac{R^{\frac{2}{3}} - \dfrac{2}{9}R^{\frac{1}{3}} + \dfrac{13}{81}}{R^{\frac{1}{3}}}$ $= \cos(\theta) \approx 0.646488$

$m = r\cos(\theta) \implies \dfrac{m}{r} = \boxed{0.646488}$ .

Checking for relative max at $\theta = \theta_{0}$ .

Incidentally $\theta_{0} \approx 0.867824$ radians.

$\cos(\theta) = 0.646488 \implies \sin(\theta) = \sqrt{1 - \cos^2(\theta)} = 0.762924$

and

$\dfrac{d^{2}V}{d\theta^{2}}|_{\theta = \theta_{0}} =$ $\dfrac{\pi}{3}r^3\sin(\theta)(-9\cos^2(\theta) - 4\cos(\theta) + 1)|_{\theta = \theta_{0}} = -4.79715 < 0$

$\implies$ relative max at $\theta = \theta_{0}$ .

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Quick Note: In solving the cubic above I used the following method:

Given: $ax^3 + bx^2 + cx + d = 0$ we let $x = u - \dfrac{b}{3a}$ to eliminate the square term.

This results in the cubic: $u^3 + Pu + Q$ .

Letting $u = z - \dfrac{P}{3z}$ we obtain a quadratic in $z^3$ as was the case above.