[Undergrad Calculus] Optimization and Related Rates

I've been given an optimization and related rates writing assignment by my professor but I missed the past class where we covered how to do the problems so I'm unsure how to go about it. I was just wondering if you guys could point me in the right direction for a few of these problems?

Here is the assignment: http://pastebin.com/gM2YVJ9P

And here is what I have so far: https://www.dropbox.com/s/ow7na1vxlc...20inflated.pdf

I've completed problem 1, am clueless for problem 2, and have done most of problem 3 and 4.

**Here are my questions:**

- Does my work for problem 1 look correct?
- How do I start/do number 2? I can't find any good examples to pull from in my book or online..
- What exactly does he mean by "find the derivative of the cost" in problem 3? Does he simply want me to find the derivative of 5,100,000? That wouldnt make much sense or help on the last part of this problem..
- As for problem 4.. I'm pretty sure I did it wrong and I'm not sure how to finish the last part of the problem.

Re: [Undergrad Calculus] Optimization and Related Rates

It looks like problem 1 is correct.

Number 2 is not anything new - just a lot of work. I would use just one variable: s = the side of the square. Then the length is $\displaystyle \frac{640}{s^2}$ and you can calculate the cost to build and set the derivative to zero.

For number 3, I would also use just one variable: n = number of production runs. The cost of production is 20000n and since the number of tires stored goes from a full run to zero, the average number stored is half a production run, which is 1000000/2n, so the cost is 2500000/n. The derivative of the total cost is 20000 - 2500000/n^2, which is zero for n=sqrt(125), approximately 11.28. Since presumably n must be an integer, compare the cost for n=11 (cost = 447273) and n=12 (cost = 448333), so n=11 is the minimum. (Did I understand the problem correctly?)

For number 4, again I would use one variable: k = number of $10 increases. So the revenue is price (150+10k) times volume (600-25k). Again, set the derivative to zero - and like in #3, it looks like k has to be an integer.

- Hollywood