Iift 2008 Question 2
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IIFT 2008
Explanation: We are given that; the vendors are charged Rs. 20 per packet up to 2000 packets in a box. Additions can be made only in a lot size of 200 packets. Each addition of one lot to the box results in a discount of one rupee an all the packets in the box. Let x be number of additional lots. Thus,
(20 − x)(2000 + 200x) = 40000 − 2000x + 4000x − 200x2
−200x2 + 2000x + 40000
We need to maximize this
The minimum/maximum value of a quadratic equation is when
Thus, the maximum value = Thus, the maximum size of the box that would maximize the revenue per box for Mother Dairy = 2000 + 200 x 5 = 3000