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 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.