00:02
Hi, here in this question, let us set up and cost a function.
00:06
So here, the cost of function s of t of n is equal to k1 of n.
00:11
Now here, k1 is a constant and greater than 0 and n is number of machines.
00:18
Now, further, here in our case, the operating cost, see 0 of n equals to k2 upon n, where k2 is also greater than 0 and it is a constant.
00:33
So here the total cost, c of n equals to k1 of n plus k2 upon n.
00:41
So here further we need to differentiate this equation with respect to n as we need to minimize to the cost.
00:49
So c dash of n equals to k1 multiplied with n plus k2 multiplied with n and.
00:58
Need to take the derivative so here further we need to differentiate this value with respect to n so taking the derivative here we have k1 multiplied with n dash plus k2 upon n dash so here on further simplification the derivative of n will be 1 so k1 multiplied with 1 and the derivative of 1 upon n would be minus 1 upon a square so here k2 multiplied with minus 1 upon n squared.
01:30
So here we got our value as k1 minus k2 upon n squared.
01:37
So here this is our derivative cn dash.
01:41
Now here in order to maximize the cost we will take this value equals to zero.
01:47
So here cn dash equals to zero equals to n squared multiplied with k1 minus k2 upon n square.
01:55
So here on further simplification we have k1 multiplied with n equals to k2 upon n.
02:06
So here further we got these two values or this relation...