FS food stall sells only chicken biryani. If FS fixes a selling price of Rs. 160 per plate, 300 plates of biriyani are sold. For each increase in the selling price by Rs. 10 per plate, 10 fewer plates are sold. Similarly, for each decrease in the selling price by Rs. 10 per plate, 10 more plates are sold. FS incurs a cost of Rs. 120 per plate of biriyani, and has decided that the selling price will never be less than the cost price. Moreover, due to capacity constraints, more than 400 plates cannot be produced in a day. If the selling price on any given day is the same for all the plates and can only be a multiple of Rs. 10, then what is the maximum profit that FS can achieve in a day?
Explanation:
The selling price of each plate will always be a multiple of 10. So, let the number of selling price of a plate be 160 + 10x.
∴ Number of plates sold will be 300 - 10x.
Total revenue = (300 - 10x) × (160 + 10x) = -100x2 + 1400x + 48000
Total cost = 120 × (300 - 10x)
⇒ Net profit = Total revenue - Total cost
= -100x2 + 1400x + 48000 - 120 × (300 - 10x)
= -100x2 + 1400x + 48000 - 36000 + 1200x
= -100x2 + 2600x + 12000
Now, net profit is a quadratic expression -100x2 + 2600x + 12000.
This will be maximum when x = - (2600/2 × -100) = 13
∴ Maximum profit = -100 × (13)2 + 2600 × (13) + 12000
= - 16900 + 33800 + 12000
= 28,900
Hence, option (b).
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