The cone’s height is twice its radius. How can you substitute $h = 2r$ into the cone’s volume formula so everything is in terms of $r$ only?

Add the cone’s volume and the hemisphere’s volume, and set this total equal to $4{,}500\pi$. After you simplify, what equation do you get involving $r^3$?

Once you have an equation of the form $\dfrac{4}{3}r^3 = 4{,}500$, isolate $r^3$ and then take the cube root to find $r$. Which answer choice matches this value?

In Desmos, type the equation (4/3)*pi*r^3 = 4500*pi. Desmos will interpret this as a relationship between r and the constant volumes.

Click on r to create a slider, then adjust the slider until the left-hand side (4/3)*pi*r^3 is equal to 4500*pi. The value of r at this point is the radius that satisfies the volume condition.

Identify the two parts of the solid:

• A right circular cone of radius $r$ and height $h$
• A hemisphere of radius $r$

Their volumes are:

• Cone: $V_{\text{cone}} = \dfrac{1}{3}\pi r^2 h$
• Hemisphere: $V_{\text{hemisphere}} = \dfrac{1}{2} \cdot \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3$

We also know the cone’s height is twice its radius, so $h = 2r$. Substitute this into the cone volume.

Use $h = 2r$ in the cone volume:

So both parts have volumes in terms of $r^3$:

• Cone: $\dfrac{2}{3}\pi r^3$
• Hemisphere: $\dfrac{2}{3}\pi r^3$

Add them to get the total volume in terms of $r$.

Add the two volumes and set the sum equal to the given total volume $4{,}500\pi$:

So the equation is

Now solve this equation for $r$.

First, divide both sides of

by $\pi$:

Multiply both sides by $\dfrac{3}{4}$:

Now take the cube root of both sides:

So the radius of the hemisphere is $15$ centimeters.