costs $9, and there is a fixed charge of$200 per order. If it costs $3 to store a bottle for a year, how many bottles should be ordered at a time and how many orders should the warehouse place in a year to minimize inventory costs. 2. please read http://www.mathhelpforum.com/math-he...y-problem.html and your course notes, and then post as much of the answer as you can do yourself 3. I found something similar to what your post was about in my book and course notes, these are the answers I came up with, please let me know if they are correct: C=200y + 1.5x xy= 30,000, thus y=30000/x So, C(x)= 200(30000/x)+1.5x Then, C(x)= 6,000,000/x + 1.5x C '(x)= -6,000,000/x^2 + 1.5 -6,000,000/x^2 + 1.5 = 0 x^2= 6,000,000/1.5 Divide, then square root to cancel out x^2= 4,000,000, the square root of 4,000,000= 2,000 Thus, x=2,000, so y= 30,000/2,000 = 15 The final answer: The company will minimize its inventory cost by ordering 2,000 bottles, 15 times during the year at a time. 4. its the right answer, some minor comments in red you have not defined any of your variables so i have to guess what you mean... i assume: C = total annual cost y = number of orders per year x = number of bottles per order In that case: C=200y+ 1.5x You have not included the purchase price of the bottles in your cost function. This is actually ok, because the annual cost is a constant (30000*9) that is not affected by your choice of x and y. Since it is a constant, it will disappear when you differenciate anyway. Different professors will have their own preferences about whether this fixed cost should be included or not you have to do quite a bit of simplification to get the cost of storage, 1.5x • if there are y orders per year, then the time between orders is 1/y. • So the average time each bottle spends in storage is 0.5(1/y) = 1/2y • So the cost of storage between orders is x * 3 * (1/2y) = 3x/2y • Multiply by the number of orders: y * (3x/2y) = 3x/2 = 1.5x xy = 30000, so y=30000/x$\displaystyle C(x) = 200(30000/x) + 1.5x = 6000000x^{-1} + 1.5x\displaystyle C'(x) = -6000000x^{-2} + 1.5$Solve C'(x) =0$\displaystyle 6000000 = 1.5x^2\displaystyle = +/-2000\$