Question 2

# Mother Dairy sells milk packets in boxes of different sizes to its vendors. 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. What should be the maximum size of the box that would maximize the revenue per box for Mother Dairy?

Solution

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,
$$(20-x)(2000+200x) = 40000 - 2000x + 4000x -200x^2$$
=> $$-200x^2+2000x+40000$$
We need to maximize this
The minimum/maximum value of a quadratic equation is when $$x = -b/2a$$
Thus, the maximum value = $$-2000/400 = 5$$
Thus, the maximum size of the box that would maximize the revenue per box for Mother Dairy = 2000+200*5 = 3000

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