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CMA Foundation · Fundamentals of Business Mathematics and Statistics · Calculus - Application in Business

A manufacturer sells x units at price p = 80 - x/5 rupees per unit with average cost AC = 20 + 2000/x rupees. What is the profit-maximising quantity and the maximum profit?

Profit is maximised at 150 units with a maximum profit of 2,500 rupees. Total cost is 20x + 2,000, so profit is 60x - x^2/5 - 2,000; its derivative is zero at x = 150, giving 9,000 - 4,500 - 2,000.

  1. Ax = 150, profit 4,000
  2. Bx = 150, profit 2,500Correct
  3. Cx = 300, profit 7,000
  4. Dx = 120, profit 1,600

Explanation

Revenue = 80x - x^2/5. Total cost = 20x + 2000. Profit = 60x - x^2/5 - 2000. P' = 60 - 2x/5 = 0 gives x = 150. P = 9000 - 4500 - 2000 = 2,500. Check: p = 50, margin over variable cost = 30, 30 x 150 = 4500, less 2000 = 2,500. The 4,000 option ignores fixed cost incorrectly.

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