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Question
A company receives a shipment of 500 scooters every 30 days. From experience, it is known that the inventory on hand is related to the number of days x. Since the shipment, I(x) = 500 – 0.03x2, the daily holding cost per scooter is ₹ 0.3. Determine the total cost for maintaining inventory for 30 days
Solution
Here I(x) = 500 – 0.03x2
C1 = ₹ 0.3
T = 30
Total inventory carrying cost
= `"C"^1 int_0^"T""I"(x) "d"x`
= `0.3 int_0^30 (500 - 0.03x^2) "d"x`
= `0.3[500x - 0.03 (x^3/3)]_0^30`
= `0.3[500x - 0.01x^3]_0^30`
= 0.3[500(30) – 0.01 (30)3] – [0]
= 0.3[15000 – 0.01(27000)]
= 0.3[15000 – 270]
= 0.3[14730]
= ₹ 4,419
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