Key Concepts and Formulas

• Optimization using Derivatives: Finding maxima/minima of a function by setting its derivative to zero.
• Volume of a Cylinder: $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height.
• Trigonometric Identity: $\sin^2\theta + \cos^2\theta = 1$

Step-by-Step Solution

Step 1: Visualize and Define Variables

• We are given a sphere of radius $R=3$. We need to find the height $h$ of a cylinder with radius $r$ inscribed in this sphere such that the cylinder's volume is maximized.
• Consider a cross-section of the sphere and cylinder. We have a rectangle (cylinder) inscribed in a circle (sphere). The vertices of the rectangle lie on the circle. The diagonal of the rectangle is the diameter of the circle.
• Relate $r$, $h$, and $R$ using the Pythagorean theorem. We have a right triangle with hypotenuse $R$, one leg $r$, and the other leg $h/2$. Thus, $R^2 = r^2 + (h/2)^2$.
• Substitute $R=3$: $3^2 = r^2 + \left(\frac{h}{2}\right)^2$ $9 = r^2 + \frac{h^2}{4}$
• Solve for $r^2$ in terms of $h$ (or vice versa): $r^2 = 9 - \frac{h^2}{4}$
  • Explanation: We express $r^2$ in terms of $h$, so we can later express the volume as a function of $h$ only.

Step 2: Express Volume in Terms of a Single Variable

• The volume of the cylinder is $V = \pi r^2 h$.
• Substitute the expression for $r^2$ from Step 1: $V = \pi \left(9 - \frac{h^2}{4}\right) h$ $V = \pi \left(9h - \frac{h^3}{4}\right)$
  • Explanation: We now have the volume $V$ as a function of a single variable $h$.

Step 3: Apply Calculus for Maxima/Minima

• To find the maximum volume, differentiate $V$ with respect to $h$ and set the derivative to zero ($dV/dh = 0$).
• Calculate the derivative $dV/dh$: $\frac{dV}{dh} = \frac{d}{dh} \left[\pi \left(9h - \frac{h^3}{4}\right)\right]$ $\frac{dV}{dh} = \pi \left(9 - \frac{3h^2}{4}\right)$
• Set $dV/dh = 0$ to find critical points: $\pi \left(9 - \frac{3h^2}{4}\right) = 0$ $9 - \frac{3h^2}{4} = 0$ $\frac{3h^2}{4} = 9$ $h^2 = \frac{4 \cdot 9}{3}$ $h^2 = 12$ $h = \pm \sqrt{12} = \pm 2\sqrt{3}$
• Since height must be positive, we take the positive root: $h = 2\sqrt{3}$
• Explanation: We find the critical points by setting the first derivative to zero. We discard the negative root because height cannot be negative.

Step 4: Verify Maximum

• To confirm that $h = 2\sqrt{3}$ corresponds to a maximum, we can use the second derivative test.
• Calculate the second derivative $d^2V/dh^2$: $\frac{d^2V}{dh^2} = \frac{d}{dh} \left[\pi \left(9 - \frac{3h^2}{4}\right)\right]$ $\frac{d^2V}{dh^2} = \pi \left(0 - \frac{6h}{4}\right) = -\frac{3\pi h}{2}$
• Evaluate the second derivative at $h = 2\sqrt{3}$: $\frac{d^2V}{dh^2}\Big|_{h=2\sqrt{3}} = -\frac{3\pi (2\sqrt{3})}{2} = -3\pi\sqrt{3}$
• Since $\frac{d^2V}{dh^2} < 0$ at $h = 2\sqrt{3}$, this corresponds to a maximum.

Step 5: Find the height of the cylinder

• The height of the cylinder that maximizes the volume is $h = 2\sqrt{3}$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when differentiating and solving equations.
• Units: Although not explicitly stated, remember that the radius is in some unit of length, so the height will be in the same unit.
• Second Derivative Test: Always verify that you have found a maximum (or minimum) using the second derivative test.

Summary

To find the height of the right circular cylinder of maximum volume inscribed in a sphere of radius 3, we expressed the volume of the cylinder as a function of its height using the relationship between the cylinder's radius, height, and the sphere's radius. We then found the critical points by setting the derivative of the volume function to zero. Finally, we used the second derivative test to confirm that the critical point corresponds to a maximum volume, giving us a height of $2\sqrt{3}$.

Final Answer The final answer is \boxed{2\sqrt{3}}, which corresponds to option (B).