Let a right circular cone be inscribed in a sphere of radius $r$. Let $h$ be the altitude of the cone and $x$ be the radius of its base. Let the sphere center be $O$ and the cone's vertex be at the top of the sphere. Let $c$ be the distance from $O$ to the base center.

The altitude $h = r + c$, where $c$ is the distance from sphere center to base center. For the base circle to be inscribed in the sphere:

$$x^2 + c^2 = r^2$$

Therefore,

$$x^2 = r^2 - c^2 = r^2 - (h - r)^2 = 2rh - h^2 = h(2r - h)$$

The volume of the cone is:

$$V = \frac{1}{3}\pi x^2 h = \frac{1}{3}\pi h^2(2r - h)$$

Differentiating $V$ with respect to $h$:

$$\frac{dV}{dh} = \frac{1}{3}\pi[2h(2r - h) - h^2] = \frac{1}{3}\pi(4rh - 3h^2)$$

Setting $\dfrac{dV}{dh} = 0$:

$$h(4r - 3h) = 0$$

giving $h = 0$ or $h = \dfrac{4r}{3}$.

The second derivative:

$$\frac{d^2V}{dh^2} = \frac{1}{3}\pi(4r - 6h)$$

At $h = \dfrac{4r}{3}$:

$$\frac{d^2V}{dh^2} = \frac{1}{3}\pi(4r - 8r) < 0$$

Hence, the altitude $h = \dfrac{4r}{3}$ gives maximum volume.

∴ The altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\dfrac{4r}{3}$.