Question

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?

• A. Rs. 25,300
• B. Rs. 28,900
• C. Rs. 41,400
• D. Rs. 52,900
• E. None of the remaining options is correct.

Answer 1

The Correct Option is B

Step 1: Variable Declaration. Assign 'x' to the quantity of plates sold and 'P' to the price per plate. Initial conditions: P = 160, x = 300.

Step 2: Price-Quantity Relationship. For every Rs. 10 price increase, plate sales decrease by 10. Let 'y' represent the number of Rs. 10 price increments above Rs. 160.

P = 160 + 10y

x = 300 - 10y

Step 3: Profit Calculation. The cost per plate is Rs. 120, yielding a profit per plate of:

Profit per plate = P - 120 = (160 + 10y) - 120 = 40 + 10y

Total profit is calculated as:

Total profit = (40 + 10y)(300 - 10y)

Step 4: Profit Maximization. Expand the profit function:

Profit = (40 + 10y)(300 - 10y) = 12000 + 400y - 120y - 100y² = 12000 + 280y - 100y²

To find the maximum profit, differentiate with respect to 'y' and set the result to zero:

d/dy(12000 + 280y - 100y²) = 280 - 200y

Equating the derivative to zero:

280 - 200y = 0 > y = 1.4

As 'y' must be an integer, round to y = 1.

Step 5: Maximum Profit Determination. With y = 1, the price per plate is:

P = 160 + 10(1) = 170

The number of plates sold is:

x = 300 - 10(1) = 290

The total profit is:

Profit = (170 - 120)(290) = 50 * 290 = 14,500

Answer: Rs. 41,400