To solve this problem, we need to find the dimensions that minimize the surface area (which represents the material used) of a cone with a given volume.

Step-by-Step Solution:

1. Volume of a Cone Formula: The volume $V$ of a cone with radius $r$ and height $h$ is given by: $V = \frac{1}{3} \pi r^2 h$ Since the volume is given as $36\pi$ cubic inches, we can set up the equation: $\frac{1}{3} \pi r^2 h = 36 \pi$ Simplifying, we find: $r^2 h = 108$ or $h = \frac{108}{r^2}$

2. Surface Area of a Cone: The surface area $S$ of a cone (excluding the base, since it's a thin-walled cup) is given by: $S = \pi r \sqrt{r^2 + h^2}$ Substitute $h = \frac{108}{r^2}$ into this equation to express $S$ in terms of $r$ only.

3. Substitute $h$ and Differentiate $S$ with Respect to $r$: Substitute $h$ in terms of $r$: $S = \pi r \sqrt{r^2 + \left(\frac{108}{r^2}\right)^2}$ Now, differentiate $S$ with respect to $r$ and set the derivative to zero to find the critical points, which will give the value of $r$ that minimizes $S$.

   After calculating, we find: $r = 3 \sqrt{2} \text{ inches}$ and substituting this back into $h = \frac{108}{r^2}$, we get: $h = 6 \text{ inches}$

Answer:

• 2a. Radius: $r = 3\sqrt{2}$ inches, which matches option (b).
• 2b. Height: $h = 6$ inches, which matches option (b).

Would you like further details on any step, or do you have any questions?

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Here are five related questions to expand on this problem:

1. How would the minimum surface area change if the volume was different?
2. How do we apply optimization techniques to find minimum or maximum values in general?
3. What would the effect be on the surface area if the cone had a base?
4. How does changing the radius affect the height to maintain the same volume?
5. Can we apply a similar approach to find minimum material for other shapes like cylinders?

Tip: For optimization problems, express all variables in terms of one variable to simplify differentiation.