Jeremy R. answered 04/10/15
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Given the information in the problem statement, we must assume that the rate of consumption is constant at 2400 pairs per year, throughout the year, or 2400 units per unit time which is a year. If there are S pairs of socks in store to begin with, then at time t (in years, or a fraction thereof), there are
( S – 2400*t )
units. As the cost of storage is proportional to this, as well as time ("$0.04 to store one pair of white socks for one year"), in a small incremental time, dt, it is (in dollars)
( S – 2400*t )*0.04*dt
In a best-case scenario, the stocking shelves are replenished just as they become empty, n times a year, with n TBD. Therefore, once replenished, there would immediately be one n-th the amount needed per year on the storage shelves, such that
S = 2400 / n
The cost, in dollars, of storing per restocking period (1/n year) is then found by integration:
int_(t=0)^(t=1/n)(( 2400/n – 2400*t )*0.04)*dt = 0.04*2400*(1/n2 - 0.5/n2) = 48/n2
The cost of reordering per restocking period is, in dollars,
(2400/n)*0.02+0.75 = 48/n+0.75
Therefore, the total cost per year is
n*(48/n2+48/n+0.75) = 48/n + 48 + 0.75n
We minimize this by differentiating w.r.t. n and equating the result to zero, then solving for n:
-48/n2 + 0.75 = 0, n = sqrt(48/0.75) = 8,
exactly.
Answer: To minimize cost, the store should order (2400/8=) 300 socks 8 times per year.