Answer
$\frac{32\pi}{3}$
Work Step by Step
Step 1. Using the figure given in the Exercise, the volume of the cone can be written as $V=\frac{1}{3}\pi x^2(y+3)$ where $x$ is the radius of the base and $y+3$ is the height.
Step 2. Use the Pythagorean Theorem in the right triangle shown in the figure, $x^2+y^2=3^2$ which gives $x^2=9-y^2$
Step 3. Combining the above relations, we have $V=\frac{1}{3}\pi (9-y^2)(y+3)=\frac{1}{3}\pi (-y^3-3y^2+9y+27)$
Step 4. To find the maximum of $V$, take its derivative to get $V'=\frac{1}{3}\pi (-3y^2-6y+9)=(3-2y-y^2)\pi=(3+y)(1-y)\pi$
Step 5. Let $V'=0$ and consider $y\gt0$; we get $y=1$ and $x=\sqrt {9-y^2}=2\sqrt 2$ and the volume is $V=\frac{1}{3}\pi (8)(1+3)=\frac{32\pi}{3}$
Step 6. We can confirm the above is a maximum because $V'$ changes signs from positive to negative when passing through $y=1$ (concave down or $V''=(-2-2y)\pi\lt0$)
Step 7. We conclude that the maximum volume is $V=\frac{32\pi}{3}$ cubic units.
