Previous Year Paper
CAT-2002-Previous Years Paper
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 maximize the revenue?
|D||None of these|
Number of samosas = 200 + 20n,
n is a natural number.
Price per samosa = Rs (2 − 0.ln)
Revenue = (200 + 20n) (2 – 0.ln) = 400 + 20n − 2n2
For maxima 20 − 4n = 0; by differentiation n = 5
⇒ Maximum revenue will be at (200 + 20 × 5) = 300 samosas