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CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

A firm has total revenue R(x) = 50x − x² and total cost C(x) = 10x + 100. How many units give maximum profit, and what is that maximum profit?

Profit is −x² + 40x − 100. Its derivative is zero at x = 20, and the second derivative is negative. The maximum profit is −400 + 800 − 100 = ₹300 at 20 units.

  1. Ax = 20, profit ₹300Correct
  2. Bx = 25, profit ₹525
  3. Cx = 20, profit ₹400
  4. Dx = 40, profit ₹300

Explanation

Profit P = 50x − x² − 10x − 100 = −x² + 40x − 100. P' = −2x + 40 = 0 gives x = 20. P(20) = −400 + 800 − 100 = 300. Option x = 25 maximises revenue only, ignoring cost.

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