Find the maximum volume of a cone inscribed in a sphere of radius R.

Let $R$ be the radius of the sphere.
The volume of the sphere is $V_S = \frac{4}{3}\pi R^3$.

Let the inscribed cone have radius $r$ and height $h$.
The volume of the cone is $V_C = \frac{1}{3}\pi r^2 h$.

Consider a cross-section of the sphere and the inscribed cone. Let the center of the sphere be at the origin $(0,0)$.
Let the height of the cone be $h$. The vertex of the cone is at $(0, R)$.
Let the center of the base of the cone be at $(0, R-h)$.
The radius of the cone's base, $r$, forms a right-angled triangle with the sphere's radius $R$ and the distance from the sphere's center to the cone's base center.
Let $k$ be the distance from the center of the sphere to the center of the cone's base.
Then $h = R + k$.
From the Pythagorean theorem, $r^2 + k^2 = R^2$, so $r^2 = R^2 - k^2$.
Since the cone's vertex is at the top of the sphere, its height $h$ is $R+k$.
The variable $k$ represents the distance from the center of the sphere to the base of the cone. $k$ can range from 0 (base passes through the center) to $R$ (base is a point at the bottom of the sphere).

Substitute $r^2 = R^2 - k^2$ and $h = R+k$ into the volume formula for the cone:

Factor $R^2 - k^2$ as $(R-k)(R+k)$:

We need to maximize $V_C$ with respect to $k$. The domain for $k$ is $0 \le k \le R$.

To find the maximum volume, we differentiate $V_C$ with respect to $k$ and set the derivative to zero.
Let $f(k) = (R-k)(R+k)^2$.
Using the product rule, $\frac{d}{dk}(uv) = u'v + uv'$:

Set $\frac{dV_C}{dk} = 0$:

Since $R > 0$ and $k \ge 0$, $R+k$ cannot be zero.
Therefore, $R-3k = 0$, which implies $k = \frac{R}{3}$.
This value of $k$ is within the valid range $0 \le k \le R$.

We can check the sign of the derivative around $k = \frac{R}{3}$:
If $k < \frac{R}{3}$, then $R-3k > 0$, so $\frac{dV_C}{dk} > 0$ (volume is increasing).
If $k > \frac{R}{3}$, then $R-3k < 0$, so $\frac{dV_C}{dk} < 0$ (volume is decreasing).
This confirms that $k = \frac{R}{3}$ corresponds to a local maximum.

Substitute $k = \frac{R}{3}$ back into the expressions for $h$ and $r^2$:

Now, calculate the maximum volume of the cone:

Now, compare this to the volume of the sphere $V_S = \frac{4}{3}\pi R^3$:

Thus, the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.