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FS Food Stall Maximum Profit

XAT 2024 · Quantitative Ability & Data Interpretation · Medium · Arithmetic

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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?

  1. A.

    Rs. 25,300

  2. B.

    Rs. 28,900

  3. C.

    Rs. 41,400

  4. D.

    Rs. 52,900

  5. E.

    None of the remaining options is correct.

Answer

B

Explanation

Let the increase in price be 10x10x. Selling Price =160+10x= 160 + 10x Number of plates sold =300−10x= 300 - 10x Profit per plate =(160+10x)−120=40+10x= (160 + 10x) - 120 = 40 + 10x Total Profit P(x)=(40+10x)(300−10x)=100(4+x)(30−x)=100(120+26x−x2)P(x) = (40 + 10x)(300 - 10x) = 100(4 + x)(30 - x) = 100(120 + 26x - x^2). To maximize 26x−x226x - x^2, the maximum occurs at x=13x = 13. Max Profit P(13)=100(4+13)(30−13)=100×17×17=28,900P(13) = 100(4 + 13)(30 - 13) = 100 \times 17 \times 17 = 28,900. Capacity check: Number of plates =300−130=170≤400= 300 - 130 = 170 \le 400. Selling price =290≥120= 290 \ge 120. Thus, the maximum profit is Rs. 28,900.

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