Question

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?

• A. 240
• B. 300
• C. 400
• D. None of these

Hint:

Use differentiation to find the maximum or minimum of functions representing revenue.

Answer 1

The Correct Option is B

Let the number of samosas in a box be \( x \). The price per samosa for boxes larger than 200 samosas decreases by 10 paise for every additional 20 samosas. - For \( x \leq 200 \), the price is Rs. 2 per samosa. - For \( x>200 \), the price per samosa decreases by 10 paise for every 20 samosas. Revenue for \( x \leq 200 \) is given by: \[ R = 2x. \] For \( x>200 \), the price per samosa is \( 2 - \frac{10}{100} = 1.90 \), and the revenue is: \[ R = 1.90 \times x. \] We differentiate this revenue equation with respect to \( x \) and find the maximum revenue occurs at \( x = 300 \). Therefore, the maximum size of the box is 300.