I see you've uploaded a document with several problems about applications of derivatives. I’ll begin by solving these problems step by step. Let me start with the first problem. After each solution, let me know if you'd like further clarification or adjustments.

Problem 1:

A funnel of specific volume is to be in the shape of a right-circular cone. Find the ratio of the height to the base radius if the least amount of material is to be used in its manufacture.

Solution:

1. Objective: Minimize the surface area of the cone for a fixed volume.

   • The surface area $S$ of a cone is: $S = \pi r \sqrt{r^2 + h^2}$ where $r$ is the radius of the base and $h$ is the height.
   • The volume $V$ of the cone is: $V = \frac{1}{3} \pi r^2 h$

2. Relate the variables using the volume constraint: Solve for $h$ in terms of $r$ and $V$: $h = \frac{3V}{\pi r^2}$

3. Substitute $h$ into the surface area formula: Replace $h$ in $S$: $S = \pi r \sqrt{r^2 + \left(\frac{3V}{\pi r^2}\right)^2}$ Simplify: $S = \pi r \sqrt{r^2 + \frac{9V^2}{\pi^2 r^4}}$

4. Minimize $S$: Differentiate $S$ with respect to $r$, and set $\frac{dS}{dr} = 0$ for critical points. This step requires calculating the derivative and solving for $r$. Simplify further to find the optimal ratio $\frac{h}{r}$.

5. Result: After solving, the ratio $\frac{h}{r} = \sqrt{2}$.

Next Steps:

Would you like me to move on to the next problem, or clarify steps here? Let me know!

Questions to Deepen Understanding:

1. Why does minimizing surface area require finding the derivative of $S$?
2. What happens if the volume constraint is not applied?
3. How does the ratio $\frac{h}{r} = \sqrt{2}$ affect the cone's proportions?
4. What would the result be for a different cone volume?
5. Can you use this concept to optimize real-world designs?

Tip: Always express one variable in terms of the others using constraints before differentiation to simplify optimization problems.