I am helping my brother with his math homework and this problem has stumped me.
I tried (150+4n)(490000/n)+0.75n as the cost function but that doesn't get me anywhere when I take second derivative.
A company needs 490,000 items per year. It costs the company \$150 to prepare a production run of these items and \$4 to produce each item. If it also costs the company $0.75 per year for each item stored, find the number of items that should be produced in each run so that total costs of production and storage are minimized.
The correct answer is 14,000 items/run. How?
Best Answer
[Edit, sorry, mis-read the numbers. Fixing them below.]
So if x is the number of production runs, then 490000/x is the number of items produced in a production run. Since the cost of producing each item is the same no matter how many production runs we use, it doesn't really matter for the minimization problem. (If you wanted to include this in the cost function, you would just add the constant value 4*490000.) Since every production run costs 450 then that part of the cost is 150x. And presumably we will at some point store every one of these units per production run, at a cost of (0.75)490000/x. So the total cost function is
$150x + \frac{(0.75)\cdot 490000}{x}$
And you want to minimize this, so you take the derivative and set it to 0, then solve for x:
$150 + 0.75\cdot 490000 \cdot (-1\cdot x^{-2})=0$
so
$x^{2} = \frac{0.75\cdot 490000}{150}$
so
$x \approx 49.5$ (rejecting the negative solution as meaningless in the problem)
We then need to know the number for the lowest cost which is close to this value, so we test 49 and 50. If you do 49 production runs then the cost is 14850 = 150*49 + 0.75*490000/49 but if you do 50 runs the cost is the same, 14850 = 150*50 + 0.75*490000/50. So it doesn't matter which.
If you pick 49, then the number to produce is 490000/49 = 10,000.
(I see this disagrees with your teacher's result so I'll try to see if I missed anything.)