In your cross-section, compare the large triangle for the whole cone with the smaller triangle from the tip of the cone down to the top of the cylinder. How can you set up a proportion between their heights and radii to relate $r$ and $h$?

Once you have $r$ in terms of $h$, substitute it into $V = \pi r^2 h$ so that $V$ depends only on $h$. Remember that multiplying by a positive constant does not change where the maximum occurs.

You will get a function like $V(h) = \text{constant} \times h(24 - h)^2$. Think about how to find the value of $h$ that makes this expression as large as possible—either by taking a derivative if you know calculus or by graphing this function and looking for its highest point.

In Desmos, enter the expression r(h) = (10/24)*(24 - h) to represent the cylinder’s radius in terms of its height $h$ (from the similar-triangles relation).

Enter V(h) = pi*(r(h))^2*h and, if you like, restrict the domain with {0 < h < 24} so you only see realistic cylinder heights inside the cone.

Look at the graph of $V(h)$. Use Desmos’s maximum feature (type maximum(V, a, b) for an interval like maximum(V, 0, 24) or tap the highest point on the curve). The $h$-coordinate gives the cylinder’s optimal height, and the corresponding $V(h)$ value is the greatest possible volume.

Compare the numerical value of the maximum $V(h)$ shown in Desmos with the answer choices (all are multiples of $\pi$). Ignore $\pi$ in the calculator’s decimal and match the coefficient to select the correct option.

Consider a vertical cross-section through the axis of the cone and the cylinder. This gives an isosceles triangle (the cone) and a rectangle (the cylinder).

• The cone has height $24$ cm and base radius $10$ cm.
• Let the cylinder have height $h$ and radius $r$.
• The base of the cylinder is on the base of the cone, so the top of the cylinder is at height $h$ above the base of the cone.

Our goal is to express the cylinder’s volume $V = \pi r^2 h$ in terms of a single variable (either $h$ or $r$) so we can maximize it.

Look at the cone above the top of the cylinder.

• The whole cone has height $24$ and base radius $10$.
• The smaller cone on top (from the cone’s tip down to the top of the cylinder) has height $24 - h$ and radius $r$.

These two cones are similar, so their corresponding sides are in proportion:

Solve for $r$:

This gives the cylinder’s radius in terms of its height $h$.

The volume of the cylinder is

Substitute $r = \dfrac{10}{24}(24 - h)$:

Simplify the constant:

so

To find the height that maximizes $V$, we only need to maximize the factor

because $\dfrac{25\pi}{144}$ is just a positive constant multiplier.

We need to maximize $f(h) = h(24 - h)^2$ for $0 < h < 24$.

Graph $y = x(24 - x)^2$ in Desmos (using $x$ for $h$) and use the maximum feature.

The maximum occurs at $h = 8$ cm.

You can verify: at $h = 0$ and $h = 24$, the volume is 0 (no cylinder). The maximum must be somewhere in between, and Desmos confirms it's at $h = 8$.

Now use $h = 8$ to find $r$ and then the volume.

From the similar-triangles relation

substitute $h = 8$:

Now compute the volume:

So the greatest possible volume of such a cylinder is $\boxed{\dfrac{3200\pi}{9}}$, which corresponds to choice B.