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Question 49

Davji Shop sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 per samosa up to 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paise per samosa. What should be the maximum size of the box that would maximise the revenue?

Solution

Let the optimum number of samosas be 200+20n

So, price of each samosa = (2-0.1*n)

Total price of all samosas = (2-0.1*n)*(200+20n) = $$400 - 20n + 40n - 2n^2$$ = $$400 + 20n - 2n^2$$

This quadratic equation attains a maximum at n = -20/2*(-2) = 5

So, the number of samosas to get the maximum revenue = 200 + 20*5 = 300

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