The semi vertical angle of a right circular cone of maximum volume of a given slant height is

Let the semi-vertical angle be $\alpha$ (the angle between the height and the slant height).

Given that slant height $= l$ (fixed/constant)

From trigonometry in the right-angled triangle:

Height: $h = l \cos \alpha$

Radius: $r = l \sin \alpha$

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Volume of cone $= \frac{1}{3}\pi r^2 h$

Substituting $r$ and $h$:

$V = \frac{1}{3}\pi (l\sin\alpha)^2 (l\cos\alpha)$

$V = \frac{1}{3}\pi l^3 \sin^2\alpha \cos\alpha$

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To find maximum volume, differentiate with respect to $\alpha$:

$\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3 \left[2\sin\alpha\cos\alpha \cdot \cos\alpha + \sin^2\alpha \cdot (-\sin\alpha)\right]$

$\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3 \left[2\sin\alpha\cos^2\alpha - \sin^3\alpha\right]$

$\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3 \sin\alpha\left[2\cos^2\alpha - \sin^2\alpha\right]$

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Setting $\frac{dV}{d\alpha} = 0$:

$\sin\alpha\left[2\cos^2\alpha - \sin^2\alpha\right] = 0$

Since $\alpha \neq 0$:

$2\cos^2\alpha - \sin^2\alpha = 0$

$2\cos^2\alpha = \sin^2\alpha$

Using $\sin^2\alpha = 1 - \cos^2\alpha$:

$2\cos^2\alpha = 1 - \cos^2\alpha$

$3\cos^2\alpha = 1$

$\cos^2\alpha = \frac{1}{3}$

$\cos\alpha = \frac{1}{\sqrt{3}}$

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Therefore, the semi-vertical angle is:

$\alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$